# NCERT Solutions for Class 11 Maths Chapter 9 - Straight Lines

## NCERT Solutions for Class 11 Maths Chapter 9 Free PDF Download

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Exercise 10.1

1. Draw a quadrilateral in the cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, – 2). Also, find its area.

Sol. Given vertices A(– 4, 5) and B(0, 7), C(5, – 5) and D(– 4, – 2) are shown below in the X-Y plane.  Joining all the vertices a quadrilateral ABCD is formed.

Area of quadrilateral ABCD

= Area of ΔADC+ Area of ΔABC …(i)

Now, area of ΔADC

$$=\frac{1}{2}|\lbrack-4(-2+5)-4(-5-5)+5(5+2)\rbrack|\\\begin{bmatrix}\because\space\text{Area}=\frac{1}{2}x_1(y_2-y_3)+\\x_2(y_3-y_1)+x_3(y_1-y_2)\end{bmatrix}$$

Here, (x1, y1) = (– 4, 5), (x2, y2)

= (– 4, – 2), (x3, y3)

= (5, – 5)

$$=\frac{1}{2}|\lbrack-4×3-4(-10)+5×7\rbrack|\\=\frac{1}{2}|\lbrack-12+40+35\rbrack|\\=|[75-12]|=\frac{63}{2}\\\therefore\space\text{Area of ΔABC =}\\\frac{1}{2}|[-4(7+5)+0(-5+5)+5(5-7)]|$$

[Here, (x1, y1) = (– 4, 5), (x2, y2) = (0, 7) and (x3, y3) = (5, – 5)]

$$=\frac{1}{2}|\lbrack-4×12+0×(-10)+5(-2)\rbrack|\\=\frac{1}{2}|-48-10|\\=\frac{|-58|}{2}=29$$

∴ From equation (i) area of quadrilateral ABCD

$$=\frac{63}{2}+29\\=\frac{63+58}{2}=\frac{121}{2}\space\text{sq unit.}\\\text{Thus, the area of}\space\square \text{ABCD is}\\\frac{121}{2}\text{sq. unit.}$$

2. The base of an equilatearl triangle with side 2a lies along the Y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Sol. Let ABC be the equilateral triangle withside 2a which lies on Y-axis and third vertex may be A(h,0) or A′.

Since, ΔABC is an equilateral, then

AB = BC

AB2 = BC2

⇒ (h – 0)2 + (0 – a)2 = (2a)2

[∵ Distance between two points (x1, y1) and

$$(x_2,y_2)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$$

For distance AB, (x1, y1) = (a, 0), (x2, y2) = (0, a)]

⇒ h2 + a2 = 4a2

⇒ h2 = 3a2

$$\Rarr\space h=\pm\sqrt{3a}\space\text{(taking square root)}\\\text{Hence, the vertices of triangles are}\\(\sqrt{3a},0),(0,a),(0,-a)\\\text{or\space}(-\sqrt{3a},0),(0,a),(0,-a)$$

3. Find the distance between P(x1, y1) and Q(x2, y2) when:

(i) PQ is parallel to the Y-axis.

(ii) PQ is parallel to the X-axis.

Sol. (i) When PQ is parallel to the Y-axis it means

x1 = x2.

∴ Distance between two points PQ

$$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\=\sqrt{(x_1-x_1)^2+(y_2-y_1)^2}\\(\because x_1=x_2)\\=\sqrt{(y_2-y_1)^2}=|y_2-y_1|$$

(ii) When PQ is parallel to X-axis, it means y2 = y1.

∴ Distance between two points PQ

$$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\=\sqrt{(x_2-x_1)^2+(y_1-y_1)^2}\\\text{(}\because {y_1 = y_2)}\\=\sqrt{(x_2-x_1)^2}=|x_2-x_1|$$

4. Find a point on the X-axis which is equidistant from the points (7, 6) and (3, 4).

Sol. Let any point P on the X-axis is (x, 0) and the given points are A(7, 6) and B(3, 4).

Given, PA = PB

⇒ PA2 = PB2

⇒ (x1 – x2)2 + (y1 – y2)2 = (x1 – x3)2 + (y1 – y3)2

where, x1 = x, x2 = 7, y1 = 0, y2 = 6, x3 = 3, y3 = 4

⇒ (x –7)2 + (0 – 6)2 = (x – 3)2 + (0 – 4)2

[∵ Distance between two points

$$=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}]$$

⇒ x2 + 49 – 14x + 36 = x2 + 9 – 6x + 16

⇒ – 14x + 6x = 25 – 36 – 49

⇒ – 8x = 25 – 85

⇒ – 8x = – 60

$$\Rarr\space x=\frac{60}{8}\\\Rarr\space x=\frac{15}{2}\\\therefore\space\text{The Point P on the X-axis =}\\\bigg(\frac{15}{2},0\bigg).$$

5. Find the slope of a line, which passes through the origin and mid-point of the line segment joining the points P(0, – 4) and B(8, 0).

Sol. Let R be the mid point of PB.

Given points are P(0, – 4) and B(8, 0).

∴ x1 = 0, y1 = – 4, x2 = 8, y2 = 0

These points are plotted in XY plane given below.

Mid-point of PB is

$$\text{R}=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\=\bigg(\frac{0+8}{2},\frac{-4+0}{2}\bigg)=(4,-2)\\\therefore\space\text{Slope of OR =}\\\frac{y_2-y_1}{x_2-x_1}=\frac{-2-0}{4-0}\\\frac{-2}{4}=-\frac{1}{2}\\\bigg(\because\space x_1=0, y_1=0\\x_2=4,y_2=-2\bigg)\\\text{Hence, the slope of OR is }\space-\frac{1}{2}.$$

6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (– 1, – 1) are the vertices of a right angled triangle.

Sol. Let the vertices of given triangle are A(4, 4), B(3, 5) and C(– 1, – 1).

Here (x1, y1) = A(4, 4), (x2, y2) = B(3, 5), (x3, y3) = C(– 1, – 1)

Hence A, B and C are the vertices of a triangle

$$\text{Slope of AB, m}_1 =\frac{y_2-y_1}{x_2-x_1}\\=\frac{5-4}{3-4}=\frac{1}{-1}=-1\\\bigg(\because\space \text{Slope}=\frac{y_2-y_1}{x_2-x_1}\bigg)\\\text{Slope of BC, m}_2 =\frac{y_3-y_2}{x_3-x_2}\\=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$$

$$\text{Slope of CA},\text{m}_\text{3}=\frac{y_3-y_1}{x_3-x_1}\\=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1$$

∵ m1 × m3 = (– 1) × 1 = (– 1)

∴ ΔABC is a right angled triangle at A.

7. Find the slope of a line, which makes an angle of 30° with the positive direction of Y-axis measured anticlockwise.

Sol. Given, ∠YPQ = 30°

To find ∠PAX = ?

Here, ∠YPQ = ∠OPA

(vertically opposite angle)

In ΔPOA

∵ ∠OPA + ∠POA + ∠PAO = 180°

(∵ Sum of all angles of triangle is 180°)

⇒ 30° + 90° + ∠PAO = 180°

⇒ ∠PAO = 180° – 120° = 60°

⇒ ∠PAX = 180° – 60° = 120°

∴ Slope of line AQ = m = tan 120° (∵ m = tan θ)

= tan (180° – 60°)

[∵ tan (180° – θ) = – tan θ]

$$\text{= – tan 60° = −}\sqrt{3}$$

8. Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.

Sol. Given, points are A(x, – 1), B(2, 1) and C(4, 5) are collinear.

Here, x1 = x, y1 = – 1, x2 = 2, y2 = 1, x3 = 4 and y3 = 5

Slope of AB = Slope of BC

$$\Rarr\space\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}\\\Rarr\space\frac{1+1}{2-x}=\frac{5-1}{4-2}\\\bigg(\because\space\text{Slope of line}=\frac{y_2-y_1}{x_2-x_1}\bigg)\\\Rarr\space\frac{2}{2-x}=\frac{4}{2}$$

⇒ 2 – x = 1

⇒ x = 2 – 1 = 1

∴ The value of x is 1.

9. Without using distance formula, show that the points (– 2, – 1), (4, 0), (3, 3) and (– 3, 2) are the vertices of a parallelogram.

Sol. Let ABCD be a parallelogram, where vertices are A(– 2, – 1), B(3, 0), C(3, 3) and D(– 3, 2)

$$\text{Mid-point of AC =}\bigg(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\bigg)\\=\bigg(\frac{-2+3}{2},\frac{-1+3}{2}\bigg)\\=\bigg(\frac{1}{2},\frac{2}{2}\bigg)=\bigg(\frac{1}{2},1\bigg)\\\bigg[\because\space\text{Mid-point of two points}=\\\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\text{Mid-point of BD =}\\\bigg(\frac{x_2+x_4}{2},\frac{y_2+y_4}{2}\bigg)$$

$$=\bigg(\frac{4-3}{2},\frac{0+2}{2}\bigg)=\bigg(\frac{1}{2},1\bigg)$$

∵ Mid-point of AC = Mid-point of BD

∴ ABCD be a parallelogram. Hence Proved.

10. Find the angle between the X-axis and the line joining the pionts (3, – 1) and (4, – 2).

Sol. Given, (x1, y1) = (3, – 1)

and (x2, y2) = (4, – 2)

$$\text{Slope of line m}=\\\frac{y_2-y_1}{x_2-x_1}=\frac{-2+1}{4-3}$$

$$\text{tan}\space\theta=\frac{-1}{1}$$

⇒ tan θ = – 1

⇒ tan θ = – tan 45° (∵ 1 = tan 45°)

⇒ tan θ = tan (180° – 45°)

[∵ tan (180° – θ) = – tan θ]

⇒ tan θ = tan 135°

⇒ θ = 135°

Hence, the angle between the x-axis and line joining the points is 135°.

11. The slope of a line is double the slope of another line. If tangent of the angle between them is $$\frac{\textbf{1}}{\textbf{3}},$$

find the slope of the lines.

Sol. Let slope of one line is m, then slope of another line is 2m. Given, the tangent of the angle between

$$\text{them is tan}\space\theta\\ =\frac{1}{3}\text{and m}_1 \text{= m, m}_\text{2}\text{ = 2m.}\\\therefore\space\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|=\frac{1}{3}\\\Rarr\space\frac{1}{3}=\bigg|\frac{m-2m}{1+m×2m}\bigg|\\\Rarr\space\frac{1}{3}=\bigg|\frac{-m}{1+2m^2}\bigg|$$

⇒ (1 + 2m2) = 3m

⇒ 2m2 – 3m + 1 = 0

Factorize it by splitting the middle term

⇒ 2m2 – 2m – m + 1 = 0

⇒ 2m(m – 1) –1(m – 1) = 0

⇒ (2m – 1) (m – 1) = 0

⇒ 2m – 1 = 0 or m – 1 = 0

$$\Rarr\space m=\frac{1}{2}, m=1.$$

12. A line passes through (x1, y1) and (h, k). If slope of the line is m, show that

k – y1 = m(h – x1).

Sol. Let the given points are P(x1, y1) and Q(h, k).

$$\text{Then, slope m =}\frac{k-y_1}{h-x_1}\\\bigg(\because\space m=\frac{y_2-y_1}{x_2-x_1}\bigg)$$

The equation of line is

k – y1 = m(h – x1).

13. If three points (h, 0), (a, b) and (0, k) lie on a line.

$$\textbf{Show that}\space\frac{\textbf{a}}{\textbf{h}}+\frac{\textbf{b}}{\textbf{K}}\space\textbf{=\space1}.$$

Sol. Given points are A(h, 0), B(a, b) and C(0, k). If A, B and C are collinear, then

slope of AB = slope of BC = slope of CA

Now, slope of AB = slope of BC = slope of BC = slope of CA

$$\therefore\space\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}=\frac{y_1-y_3}{x_1-x_3}\\\frac{b-0}{a-h}=\frac{k-b}{0-a}=\frac{0-k}{h-0}\\\text{Taking first two terms,}\\\Rarr\space\frac{b}{a-h}=\frac{k-b}{-a}$$

⇒ – ab = (a – h) (k – b)

⇒ – ab = ak – ab – hk + bh

⇒ ak + bh = hk

Dividing each term by (hk),

$$\Rarr\space\frac{ak}{hk}+\frac{bh}{hk}=\frac{hk}{hk}\\\Rarr\space\frac{a}{h}+\frac{b}{k}=1.\\\space\textbf{Hence Proved.}$$

14. Consider the following population and year graph (in figure), find the slope of the line AB using it, find what will be the population in the year 2010?

Sol. Let the population in the year 2010 will be k crore. Since, A, B, C are collinear.

∴ Slope of AB = Slope of BC

$$\Rarr\space\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}\\\Rarr\space\frac{97-92}{1995-1985}=\frac{k-97}{2010-1995}$$

(∵ x1 = 1985, x2 = 1995, x3 = 2010, y1 = 92, y2 = 97, y3 = k)

$$\Rarr\space\frac{5}{10}=\frac{k-97}{15}\\\Rarr\space\frac{1}{2}=\frac{k-97}{15}\\\Rarr\space k-97=\frac{15}{2}\\\Rarr\space k=97+\frac{15}{2}$$

= 97 + 7.5 = 104.5

Hence, slope of

$$\text{AB}=\frac{1}{2}\space\text{and the poulation in the year}$$

2010 = 104.5 crore.

Exercise 10.2

In Exercises 1 to 8, find the equation of the line which satisfy the given condition.

1. Write the equations for the X-and Y-axes.

Sol. Any point on X-axis is (X, 0) as y-coordinate on X-axis is zero. Also slope of X-axis is m = 0

∵ Equation of line is

y – y1 = m(x – x1)

∴ y – 0 = 0(x – x) (∵ x1 = x, y1 = 0)

⇒ y = 0

Hence, equation of X-axis, y = 0

Similarly, the x-cordinate on the y-axis is 0.

Thus, the equation of y-axis, x = 0.

2. Passing through the point (– 4, 3) with slope

$$\frac{\textbf{1}}{\textbf{2}}\textbf{.}$$

Sol. Equation of line in one point form is

y – y1 = m(x – x1)

$$\text{Here, (x}_1, y_1) = (– 4, 3)\space\text{and}\\ \text{slope m} =\frac{\text{1}}{\text{2}}\\\therefore\space y-3=\frac{1}{2}(x+4)$$

⇒ 2y – 6 = x + 4

⇒ x – 2y + 10 = 0.

3. Passing through (0, 0) with slope m.

Sol. Here, (x1, y1) = (0, 0), slope = m

Equation of line in one point form is

y – y1 = m(x – x1)

⇒ y – 0 = m(x – 0)

⇒ y = mx.

$$\textbf{4. Passing through}\textbf{(2,2}\sqrt{\textbf{3}}\textbf{)}\\\textbf{and inclined with X-axis}\\\textbf{at an angle of 75°.}\\\textbf{Sol.}\space\text{Here},\space(x_1,y_1) = (2,2\sqrt{3})$$

⇒ m = tan 75°

= tan (45° + 30°)

$$=\frac{\text{tan}\space45\degree+\text{tan}\space 30\degree}{1-\text{tan 45}\degree \text{tan 30\degree}}\\\bigg(\because\space\text{tan}(\text{A+B})=\frac{\text{tan A+tan B}}{\text{1 - tan A tan B}}\bigg)\\\Rarr\space m=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\\Rarr\space m=\frac{\sqrt{3}+1}{\sqrt{3}-1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\\=\frac{(\sqrt{3}+1)^2}{3-1}\\\Rarr\space m=\frac{3+1+2\sqrt{3}}{2}\\=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$$

Equation of line in one point form is

y – y1 = m(x – x1)

$$\Rarr\space y-2\sqrt{3}=(2+\sqrt{3})(x-2)\\\Rarr\space y-2\sqrt{3}=(2+\sqrt{3})x-2(2+\sqrt{3})\\\Rarr\space y-2\sqrt{3}=(2+\sqrt{3})x-4-2\sqrt{3}\\\Rarr\space(2+\sqrt{3})x-y-4=0$$

5. Intersecting the X-axis at a distance of 3 units to the left of origin with slope – 2.

Sol. Since, the line intersect the X-axis to the left of origin. It means it intersect the negative X-axis. Clearly, line AB passes through the point (– 3, 0) and m = – 2.

∴ The equation of line is

y – y1 = m(x – x1)

⇒ y – 0 = – 2(x + 3)

⇒ y = – 2x – 6

⇒ 2x + y + 6 = 0.

6. Intersecting the Y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of X-axis.

Sol. Since, the line intersect the Y-axis, above the origin. It means it intersect the positive Y-axis.

Here, c = 2

m = tan 30°

$$m=\frac{1}{\sqrt{3}}$$

∴ Equation of line is

y = mx + c

$$\Rarr\space y=\frac{1}{\sqrt{3}}x+2\\\Rarr\space \sqrt{3}y= x+2\sqrt{3}\\\Rarr\space x-\sqrt{3}y+2\sqrt{3}=0.$$

7. Passing through the points (– 1, 1) and (2, – 4).

Sol. Given points are A(– 1, 1) and B(2, – 4), then equation of the AB is

$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\Rarr\space y-1=\frac{-4-1}{2+1}(x+1)\\(\because\space x_1=-1,y_1=1,x_2=2,y_2=-4)\\\Rarr\space y-1=\frac{-5}{3}(x+1)$$

⇒ 3y – 3 = – 5x – 5

⇒ 5x + 3y + 2 = 0.

8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive X-axis is 30°.

Sol. Here, p = 5 and α = 30°

∴ x cos 30° + y sin 30° = 5

$$\Rarr\space\frac{\sqrt{3}}{2}x+\frac{1}{2}y=5\\\Rarr\space \sqrt{3}x+y=10.$$

9. The vertices of ΔPQR are P(2, 1), Q(– 2, 3) and R(4, 5). Find equation of the medim thriugh the vertex R.

Sol. Since, median bisects the opposite side i.e., M is the mid-point of PQ.

$$\text{M\space=}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\=\bigg(\frac{2-2}{2},\frac{1+3}{2}\bigg)\\=\bigg(0,\frac{4}{2}\bigg)=(0,2)$$

(∵ x1 = 2, y1 = 1, x2 = – 2, y1 = 3)

∴ The equation of median RM is

$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$

$$\Rarr\space y-5=\frac{2-5}{0-4}(x-4)\\\text{(∵ x}_1 = 4, y_1 = 5, x_2 = 0, y_2 = 2)\\\Rarr\space y-5=\frac{-3}{-4}(x-4)$$

⇒ 4y – 20 = 3x – 12

⇒ 3x – 4y + 8 = 0.

10. Find the equation of the line passing through (– 3, 5) and perpendicular to the line through the points (2, 5) and (– 3, 6).

Sol. Let the points P and Q are (2, 5) and (– 3, 6). We have to find equation of AB.

Let slope of AB be m.

Since, AB ⊥ PQ

(∵ m1 × m2 = – 1)

$$\Rarr\space m×\frac{y_2-y_1}{x_2-x_1}=-1$$

(Here x1 = 2, y1 = 5, x2 = – 3, y2 = 6)

$$\Rarr\space m×\frac{6-5}{-3-2}=-1\\\Rarr\space m×\frac{1}{-5}=-1$$

⇒ m = 5

∴ The equation of line AB passing through the point (– 3, 5) is

y – y1 = m(x – x1)

y – 5 = 5(x + 3) (∵ x1 = – 3, y1 = 5)

⇒ y – 5 = 5x + 15

⇒ 5x – y + 20 = 0.

11. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.

Sol. Le the given points are A(1, 0) and B(2, 3). Let the line PQ divide AB in the ratio 1 : n at R.

Using internally ratio

$$\text{R}=\bigg(\frac{1×x_2+n×x_1}{1+n},\frac{1×y_2+n×y_1}{1+n}\bigg)\\=\bigg(\frac{1×2+n×1}{1+n},\frac{1×3+n×0}{1+n}\bigg)$$

(∵ x1 = 1, y1 = 0, x2 = 2, y2 = 3)

$$=\bigg(\frac{n+2}{n+1},\frac{3}{1+n}\bigg)$$

Also,, PQ ⊥ AB

Let slope of line PQ is m.

∴ Slope of line PQ × Slope of line AB = – 1

(∵ m1m2 = – 1)

$$\Rarr\space m×\frac{y_2-y_1}{x_2-x_1}=-1\\\Rarr\space m×\frac{3-0}{2-1}=-1\\(\because\space x_1 = 1, y_1 = 0, x_2 = 2, y_2 = 3))\\\Rarr\space m×3=-1\\\Rarr\space m=-\frac{1}{3}$$

Now, equation of line PQ by using

y – y1 = m(x – x1)

$$\Rarr\space y-\frac{3}{1+n}\\=\frac{-1}{3}\bigg(x-\frac{n+2}{n+1}\bigg)\\\bigg[\because\space\text{R}\bigg(\frac{n+2}{n+1},\frac{3}{1+n}\bigg)=(x_1,y_1)\bigg]\\\Rarr\space\frac{3(n+1)y-9}{1+n}\\=\frac{-x(n+1)+(n+2)}{n+1}$$

⇒ 3(n + 1)y – 9 = – x(n + 1) + (n + 2)

⇒ x(n + 1) + 3(n + 1)y = n + 2 + 9

⇒ x(n + 1) + 3(n + 1)y = n + 11.

12. Find the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Sol. Equation of line in intercept form is

$$\frac{x}{a}+\frac{y}{b}=1\space\text{...(i)}$$

Since, the line makes equal intercepts on the axes i.e.,

a = b

∴ From Eq. (i),

$$\frac{x}{a}+\frac{y}{b}=1$$

⇒ x + y = a …(ii)

Line (ii) passes through the point (2, 3) i.e., it will satisfy the line (ii).

⇒ 2 + 3 = a

⇒ a = 5

Put the value of a in equation (ii), we get

x + y = 5

Hence, the required equation of line is x + y = 5.

13. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Sol. The equation of line in intercept form is

$$\frac{x}{a}+\frac{y}{b}=1\space\text{...(i)}$$

Given, a + b = 9 …(ii)

And line (i) passes through the point (2, 2), i.e., it will satisfy the line (i) i.e., put x = 2, y = 2 in equation (i)

$$\Rarr\space\frac{2}{a}+\frac{2}{b}=1\space\text{...(iii)}$$

Solve equations (ii) and (iii), to find the values of a and b.

From equation (ii), b = 9 – a put in equation (iii), we get

$$\frac{2}{a}+\frac{2}{9-a}=1$$

⇒ 2(9 – a) + 2a = a(9 – a)

⇒ 18 – 2a + 2a = 9a – a2

⇒ a2 – 9a + 18 = 0

Factorize it by splitting the middle term,

⇒ a2 – 6a – 3a + 18 = 0

⇒ a(a – 6) – 3(a – 6) = 0

⇒ (a – 6) (a – 3) = 0

⇒ a = 6 or 3

When a = 3, b = 6

and a = 6, b = 3

Hence, equation (i) becomes

$$\frac{x}{6}+\frac{y}{3}=1\space\text{or}\\\frac{x}{3}+\frac{y}{6}=1$$

⇒ 3x + 6y = 18

⇒ 6x + 3y = 18

⇒ x + 2y = 6

⇒ 2x + y = 6 (divide by 3)

Hence, the equation of line are 2x + y – 6 = 0 or x + 2y – 6 = 0.

14. Find equation of the line through the point (0, 2) making an angle

$$\frac{\textbf{2}\pi}{\textbf{3}}\space\textbf{with the positive X-axis.}$$

Also, find the equation of the line parallel to it and crossing the Y-axis at a distance of 2 units below the origin.

$$\textbf{Sol.}\space \text{m= tan}\space\theta=\text{tan}\frac{2\pi}{3}\\=\text{tan}\bigg(\pi-\frac{\pi}{3}\bigg)=-\text{tan}\frac{\pi}{3}$$

[∵ tan (π – θ) = – tan θ]

$$m=-\sqrt{3}$$

Hence, equation of line AC through the point A(0, 2) by using y – y1 = m(x – x1)

$$\Rarr\space y-2=-\sqrt{3}(x-0)\\(\because x_1=0, y_1=2)\\\Rarr\space y-2=-\sqrt{3x}\\\Rarr\space \sqrt{3}x+y-2=0$$

Also, the line BD intersect below Y-axis. It means the coordinate of y is negative and x-coordinate is 0 i.e., point is (0, – 2).

Hence, equation of line through the point B(0, – 2) is

$$\text{y+2}=-\sqrt{3}(x-0)$$

(∵  y – y1 = m(x – x1)]

$$\Rarr\space \sqrt{3}x+y+2=0.$$

15. The perpendicular from the origin to a line meets it at the point (– 2, 9). Find the equation of the line.

Sol. Since, PQ ⊥ OR

∴ Slope of PQ × Slope of OR = – 1

(∵ m1 × m2 = – 1)

$$\Rarr\space m×\frac{y_2-y_1}{x_2-x_1}=1\\\Rarr\space m×\frac{9-0}{-2-0}=1$$

(∵ x1 = 0, y1 = 0, x2 = – 2, y2 = 9)

$$\Rarr\space m×\frac{9}{-2}=-1\\\Rarr\space m=\frac{2}{9}$$

Now, equation of PQ by using

y – y1 = m(x – x1)

$$y-9=\frac{2}{9}(x+2)$$

(∵ x1 = – 2, y = 9)

⇒ 9y – 81 = 2x + 4

⇒ 2x – 9y + 85 = 0.

16. The length L (in centimeters) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 where C = 20 and L = 125.134 when C = 110, express L in terms of C.

Sol. Given, L = 124.942 where C = 20 and L = 125.134 where C = 110

∴ In coordinate form (20, 124.942) and (110, 125.134) are two pints.

Equation of linear function i.e., line can be taken as

$$\text{L-L}_1=\frac{\text{L}_2-\text{L}_1}{\text{C}_2-\text{C}_1}(\text{C-C}_1)\\\bigg[\because\space y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\bigg]\\\Rarr\space\text{L - 124.942}=\\\frac{1.25.134-124.942}{110-20}(\text{C-20})\\\Rarr\space\text{L - 124.942}=\frac{0.192}{90}(\text{C-20})\\\Rarr\space \text{L}=\frac{0.192}{90}\text{(C - 20)+124.942.}$$

17. The onwer of a milk store finds that he can sell 980 L of milk each week at ₹14/L and 1220 L of milk each week at ₹16/L. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17 L?

Sol. Let price and ltire be denoted in ordered pair (x, y) where x denotes the ₹ per litre and y denotes the quantity of milk in litre.

Given, (14, 980) and (16, 1220) are two points.

Let linear relations i.e., linear equation is

$$\text{y-y}_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\Rarr\space y-980=\frac{1220-980}{x_2-x_1}(x-14)\\\Rarr\space y-980=\frac{240}{2}(x-14)$$

(∵ x1 = 14, y1 = 980, x2 = 16, y2 = 1220)

⇒ y – 980 = 120(x – 14)

⇒ y – 980 = 120x – 120 × 14

⇒ 120x – y = 1680 – 980

⇒ 120x – y = 700

⇒ 120 × 17 – y = 700

⇒ y = 2040 – 700

y = 1340

He will sell weekly 1340 L milk at the rate of ₹17 L.

18. P(a, b) is the mid-point of a line segment between axes. Show that the equation of the line is

$$\frac{\textbf{x}}{\textbf{a}}\textbf{+}\frac{\textbf{y}}{\textbf{b}}\textbf{=2.}$$

Sol. Let equation of line AB in intercept form is

$$\frac{x}{h}+\frac{y}{k}=1\space\text{...(i)}$$

where, h and k are the intecepts made by the line on X-axis and Y-axis, respectively.

$$\text{Mid-point of AB =}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\=\bigg(\frac{h+0}{2},\frac{0+k}{2}\bigg)$$

(∵ x1 = h, y1 = 0, x2 = 0, y2 = k)

∵ P(a, b) is the mid-point of AB i.e.,

$$\frac{h+0}{2}=a\space\text{and}\space\frac{0+k}{2}=b$$

⇒ h = 2a and k = 2b

On putting the values of h and k in equation (i), we get

$$\frac{x}{2a}+\frac{y}{2b}=1\\\Rarr\space\frac{x}{a}+\frac{y}{b}=2.\\\textbf{Hence Proved.}$$

19. Point R(h, k) divides a line segment betwen the axes in the ratio 1 : 2. Find equation of the line.

Sol. Let equation of line AB is

$$\frac{x}{a}+\frac{y}{b}=1\space\text{...(i)}\\\text{By using internally ratio}\\\text{R(h, k) =}\bigg(\frac{1x_2+2x_1}{1+2},\frac{1y_2+2y_1}{1+2}\bigg)\\\Rarr\space h=\frac{1×0+2×a}{1+2},\\k=\frac{1×b+2×0}{1+2}$$

[∵ P(x, y) divide the line A(x1, y1) and B(x2, y2) in the ratio m : n internally

$$\therefore\space\text{P(x,y)}=\bigg(\frac{nx_2+mx_1}{n+m},\frac{ny_2+my_1}{n+m}\bigg)]\\\Rarr\space h=\frac{2a}{3}, k=\frac{b}{3}\\\Rarr\space a=\frac{3h}{2},b=3k$$

On putting the value of a and b in equation (i), we get

$$\frac{x}{\frac{3h}{2}}+\frac{y}{3k}=1\\\Rarr\space\frac{2x}{3h}+\frac{y}{3k}=1\\\Rarr\space\frac{2kx+hy}{3hk}=1$$

⇒ 2kx + hy = 3hk.

20. By using the concept of equation of a line, prove that three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

Sol. The given points are A(3, 0), B(– 2, – 2) and C(8, 2).

Equation of line AB through A(3, 0) and B(– 2, – 2) is

$$\text{y-y}_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\text{y-0}=\frac{-2-0}{-2-3}(x-3)$$

(∵ x1 = 3, y1 = 0, x2 = – 2, y2 = – 2)

$$\Rarr\space y=\frac{-2}{-5}(x-3)$$

⇒ 5y = 2x – 6

⇒ 2x – 5y – 6 = 0 …(i)

On putting the point C(8, 2) in equation (i), we get

2 × 8 – 5 × 2 – 6 = 0

⇒ 16 – 10 – 6 = 0

⇒ 16 – 16 = 0

⇒ 0 = 0

Hence, points A, B and C are collinear.

Exercise 10.3

1. Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.

(a) x + 7y = 0

(b) 6x + 3y – 5 = 0

(c) y = 0

Sol. (a) The Given equation x + 7y = 0 can be written as

⇒ 7y = – x

$$\Rarr\space y=-\frac{1}{7}x+0$$

On comparing with y = mx + c, we get

$$m=-\frac{1}{7}, c=0$$

(b) The given equation 6x + 3y – 5 = 0 can be written as

⇒ 3y = – 6x + 5

$$\Rarr\space y=\frac{-6}{3}x+\frac{5}{3}\\\Rarr\space y=-2x+\frac{5}{3}\\\text{On comparing with y = mx + c, we get}\\\text{m = – 2, c =}\frac{5}{3}$$

(c) The given equation y = 0 can be written as

⇒ y = 0x + 0

On comparing with y = mx + c, we get

⇒ m = 0, c = 0.

2. Reduce the following equations into intercept form and find their intercepts on the axes.

(a) 3x + 2y – 12 = 0

(b) 4x – 3y = 6

(c) 3y + 2 = 0

Sol. (a) The given 3x + 2y = 12 can be written as

$$\Rarr\space\frac{3x}{12}+\frac{2y}{12}=1\\\Rarr\space \frac{x}{4}+\frac{x}{6}=1\\\text{On comparing with}\\\frac{x}{a}+\frac{y}{b}=1$$

⇒ a = 4, b = 6

∴ X-intercept = 4, Y-intercept = 6

(b) The given equation 4x – 3y = 6 can be written as equation

$$\Rarr\space\frac{4x}{6}-\frac{3y}{6}=1\\\Rarr\space\frac{x}{6}-\frac{y}{2}=1\\\Rarr\frac{x}{\frac{3}{2}}+\frac{y}{(-2)}=1\\\text{On comparing with}\frac{x}{a}+\frac{y}{b}=1\\\text{we get x-intercept}=\frac{3}{2}\\\text{and Y-intercept = -2}$$

(c) The given equation 3y + 2 = 0 can be written as

⇒ 3y = – 2

$$\Rarr\space y=-\frac{2}{3}\text{or 0x}+\frac{y}{(\frac{-2}{3})}=1\\\text{On comparing with}\space\frac{x}{a}+\frac{y}{b}=1,\\\text{we get Intercept with Y-axis}=-\frac{2}{3}\\\text{and no intercept on X-axis.}$$

3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive X-axis.
$$\textbf{(a) x-}\sqrt{\textbf{3}}\textbf{y}\textbf{+8=0}$$

(b) y – 2 = 0

(c) x – y = 0

Sol. (a) The given, equation of line is

$$x-\sqrt{3}y+8=0\\\Rarr\space x-\sqrt{3}y=-8\\\Rarr\space -x+\sqrt{3}y=8\space\text{...(i)}\\\text{Now, divide by}\\\sqrt{(\text{coefficient of x})^2+(\text{coefficient of y})^2}\\\sqrt{(-1)^2+(\sqrt{3})^2}\\=\sqrt{1+3}=2,\\\text{we get}\\-\frac{1}{2}x+\frac{\sqrt{3}}{2}y=\frac{8}{2}$$

⇒ – cos 60° x + sin 60° y = 4

(∵ convert in the form of x cos α + y sin α = p)

[∵ cos x is negative and sin x is positive, it is possible in second quadrant.]

⇒ x cos (180° – 60°) + y sin (180° – 60)°) = 4

⇒ x cos 120° + y sin 120° = 4 is the normal form

[∵ cos (180° – θ) = – sin θ and sin (180° – θ) = cos θ]

On comparing with x cos α + y sin α = p, we get

α = 120°, p = 4

(b) The given, equation of line is

y – 2 = 0 …(i)

Now, divide by

$$\sqrt{(\text{coefficient of x})^2+(\text{coefficient of y})^2}\\\text{i.e.,}\space\sqrt{0+1^{2}}=\text{we get}\\\frac{0x}{1}+\frac{y}{1}=\frac{2}{1}$$

or x cos 90° + y sin 90° = 2 is the normal form
On comparing with x cos a = y sin a = p, we get

α = 90° and p = 2.

(c) The given, equation of line is

x – y = 4

On adding above equation by

$$\sqrt{(\text{coefficient of x})^2+(\text{coeeficient of y})^2}\\\text{i.e.}\sqrt{(1)^2+(-1)^2}\space\text{we get}\\\frac{1}{\sqrt{2}}x-\frac{1}{\sqrt{2}}y=\frac{4}{\sqrt{2}}\\\text{⇒ cos 45° x – sin 45° y =}2\sqrt{2}$$

[ ∵ cos x is positive and sin x is negative, it is possible only in fourth quadrant.]

⇒ x cos (360° – 45°) + y sin (360° – 45°)

$$=2\sqrt{2}$$

[∵ cos (360° – θ) = cos θ and sin (360° – θ) = – sin θ]

$$\Rarr\space \text{x cos 315}\degree+y\space\text{sin 315}\degree=2\sqrt{2}\\\text{is the normal form.}$$

On comparing with x cos α + y sin α = p, we get

$$\alpha=315\degree\space\text{and}\space p=2\sqrt{2}$$

4. Find the distance of the point (– 1, 1) from the line 12(x + 6) = 5(y – 2).

Sol. The given, equation of line is

ax+by+c=0

12(x+6)=5(y-2)

⇒ 12x + 72 = 5y – 10

⇒ 12x – 5y + 82 = 0

Hence, distance of the point (– 1, 1) from the line is

$$d=\bigg|\frac{12×(-1)-5×1+82}{\sqrt{(12)^2+(-5)^2}}\bigg|\\\bigg[\because\space d=\bigg|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\bigg|;\\x_1=-1,y_1=1; a=12, b=-5, c=82\bigg]\\=\bigg|\frac{-12-5+82}{144+25}\bigg|\\=\bigg|\frac{82-17}{\sqrt{169}}\bigg|=\frac{65}{13}=5.$$

5. Find the points on the x-axis, whose distances from the line

$$\frac{\textbf{x}}{\textbf{3}}+\frac{\textbf{y}}{\textbf{4}}\textbf{=1}\space\textbf{are 4 units.}$$

Sol. Let the coordinates on the X-axis is (h, 0) on as X-axis is y-coordinate will be zero.

Given, equation of line is

$$\frac{x}{3}+\frac{y}{4}=1$$

⇒ 4x + 3y = 12 …(i)

Distance of (h, 0) from the line (i) is

$$d=\bigg|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\bigg|\\\Rarr\space 4=\bigg|\frac{4×h+3×0-12}{\sqrt{4^2+3^2}}\bigg|$$

(∵ a = 4, b = 3, c = 12, x1 = h, y1 = 0)

$$\Rarr\space\bigg|\frac{4h-12}{5}\bigg|=4$$

⇒ |4h – 12| = 0

⇒ 4h – 12 = ± 20

On taking positive sign,

4h – 12 = 20

⇒ 4h = 32

⇒ h = 8

On taking negative sign,

4h – 12 = – 20

⇒ 4h = – 8

⇒ h = – 2

Hence, points on the X-axis are (8, 0) and (– 2, 0).

6. Find the distance between parallel lines

(a) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(b) l(x + y) + p = 0 and l(x + y) – r = 0.

Sol. (i) Given, lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 are parallel.

∴ Distance between two parallel lines

$$=\bigg|\frac{c-d}{\sqrt{a^2+b^2}}\bigg|\\=\bigg|\frac{-34-31}{\sqrt{15^2+8^2}}\bigg|\\\text{(}\because\space\text{c = -34, d=31, a=15, b=8}\text{)}\\=\frac{|-65|}{\sqrt{225+64}}\\=\frac{65}{\sqrt{289}}\\=\frac{65}{17}\space\text{units}$$

(ii) Given, lines lx + ly + p = 0 and lx + ly – r = 0 are parallel.

∴ Requried distance

$$=\bigg|\frac{\text{c-d}}{\sqrt{a^2+b^2}}\bigg|\\=\bigg|\frac{p+r}{\sqrt{l^2+l^2}}\bigg|\\=\bigg|\frac{\text{p+r}}{\sqrt{2l}}\bigg|$$

(∵ a = l, b = l, c = p, d = – rI)

7. Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (– 2, 3).

Sol. Given equation of a line is

3x – 4y + 2 = 0 …(i)

Equation of line parallel to equation (i) is

3x – 4y + 2 = 0 …(ii)

∵ Line (ii) passes through the points (– 2, 3) i.e., put x = – 2, y = 3 in equation (ii).

∴ 3(– 2) – 4 × 3 + k = 0

⇒ – 6 – 12 + k = 0

⇒ k = 18

Put k = 18 in equation (ii), we get

3x – 4y + 18 = 0.

8. Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having X-intercept 3.

Sol. Given, equation of line is

x – 7y + 5 = 0 …(i)

Equation of line perpendicular to equation (i) is

7x + y + k = 0 …(ii)

∵ Line (ii) passes through the point (3, 0).

i.e., put x = 3, y = 0 in equation (ii),

∴ 7 × 3 + 0 + k = 0

⇒ k = – 21

Put k = – 21 in equation (ii), we get

7x + y – 21 = 0.

9. Find angles between the lines

$$\sqrt{\textbf{3}}\textbf{x+y=1}\space\textbf{and}\space \textbf{x+}\sqrt{\textbf{3}}\textbf{y=1.}$$

Sol. Given, equation of lines are

$$\sqrt{3}x+y=1\\\Rarr\space y=-\sqrt{3}x+1\\\text{and}\space x+\sqrt{3}y=1\\\Rarr\space\sqrt{3}y=-x+1\\\Rarr\space y=\frac{-1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}$$

On comparing with y = m1x + c1 and y = m2x + c2, we get

$$m_1=-\sqrt{3}\space\text{and}\space m_2=-\frac{1}{\sqrt{3}}\\\therefore\space\text{tan}\space\theta=\bigg|\frac{-\sqrt{3}-\bigg(-\frac{1}{\sqrt{3}}\bigg)}{1+(-\sqrt{3})\bigg(\frac{1}{\sqrt{3}}\bigg)}\bigg|\\\bigg[\because\space \text{tan}\space\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\bigg]\\\Rarr\space \text{tan}\space\theta=\bigg|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+1}\bigg|\\=\bigg|\frac{\frac{-3+1}{\sqrt{3}}}{2}\bigg|\\\Rarr\space \text{tan}\space\theta=\frac{1}{\sqrt{3}}$$

⇒ tan θ = tan 30°

⇒ θ = 30°

Hence, the angle between two lines is 30°.

10. The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.

Sol. Equation of line through (h, 3) and (4, 1) by using

$$\text{y-y}_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\Rarr\space y-3=\frac{1-3}{4-h}(x-h)\\(\because\space x_1=h,y_1=3,x_2=4, y_2=1)\\\Rarr\space y-3=\frac{-2}{4-h}(x-h)\\\Rarr\space y-3=-\frac{2x}{4-h}+\frac{2h}{4-h}\\\Rarr\space y=-\frac{2x}{4-h}+\bigg(3+\frac{2h}{4-h}\bigg)=0\\\text{...(i)}$$

Given equation of line is

7x – 9y – 19 = 0

$$\Rarr\space y=\frac{7}{9}x-\frac{19}{9}\\\text{...(ii)}$$

Compare lines, equation (i) and (ii) are with y = mx + c

$$\therefore\space m_1=-\frac{2}{4-h}\\\text{and}\space m_2=\frac{7}{9}$$

∵ Lines (i) and (ii) are perpendicular to each other, i.e.,

Slope of line (i) × Slope of line (ii) = – 1

∴ m1m2 = – 1

$$\Rarr\space-\frac{2}{4-h}×\frac{7}{9}=1$$

⇒ 14 = 9(4 – h)

⇒ 14 = 36 – 9h

⇒ 9h = 22

$$\therefore\space h=\frac{22}{9}.$$

11. Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is

A(x – x1) + B(y – y1) = 0.

Sol. Given equation of line is

Ax + By + C = 0 …(i)

Equation of line parallel to equation (i) is

Ax + By + K = 0 …(ii)

Line (ii) passes through the point (x1, y1) i.e., it will satisfy the line (ii).

∴ Ax1 + By1 + K = 0

⇒ K = – (Ax1 + By1)

Put the value of K in equation (ii),

⇒ Ax + By – (Ax1 + By1) = 0

⇒ Ax + By – Ax1 – By1 = 0

⇒ A(x – x1) + B(y – y1) = 0

12. Two lines passing through the point (2, 3) intersect, each other at an angle of 60°. If slope of one line is 2, find the equation of the other line.

Sol. Equation of line l1 by using y – y1 = m(x – x1)

where (x1, y1) = (2, 3) and m = 2 is

y – 3 = 2(x – 2)

⇒ y – 3 = 2x – 4

⇒ 2x – y – 1 = 0

⇒ y = 2x – 1 …(i)

Here, slope of line l2 is

m1 = 2

Let slope of line l2 is m.

Given, θ = 60°

Angle between two lines is

$$\text{tan}\space\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\\\Rarr\space \text{tan 60}\degree=\bigg|\frac{2-m}{1+2m}\bigg|\\\Rarr\space\sqrt{3}=\bigg|\frac{2-m}{1+2m}\bigg|\\\Rarr\space\frac{2-m}{1+2m}=\pm\sqrt{3}$$

Taking positive sign, we get

$$\frac{2-m}{1+2m}=\frac{\sqrt{3}}{1}$$

$$\Rarr\space 2-m=\sqrt{3}(1+2m)\\\Rarr\space 2-m=\sqrt{3}+2\sqrt{3}m\\\Rarr\space 2-\sqrt{3}=m(1+2\sqrt{3})\\\Rarr\space m=\frac{2-\sqrt{3}}{1+2\sqrt{3}}$$

Again taking negative sign, we get

$$\frac{2-m}{1+2m}=-\sqrt{3}$$

$$\Rarr\space 2-m=-\sqrt{3}(1+2m)\\\Rarr\space m-2\sqrt{3m}=(\sqrt{3}+2)\\\Rarr\space\text{m}=-\bigg(\frac{\sqrt{3}+2}{2\sqrt{3}-1}\bigg)$$

Now, equation of line l2 by using

y – y1 = m(x – x1) is

$$\text{y-3}=\frac{2-\sqrt{3}}{1+2\sqrt{3}}(x-2)\\\bigg(\text{when m=}\frac{2-\sqrt{3}}{1+2\sqrt{3}}\bigg)\\\Rarr\space y(1+2\sqrt{3})-3-6\sqrt{3}\\=x(2-\sqrt{3})-4+2\sqrt{3}\\\Rarr\space x(2-\sqrt{3})-y(1+2\sqrt{3})+8\sqrt{3}-1=0\\\text{Again,}\space \text{y-3}=-\bigg(\frac{\sqrt{3}+2}{2\sqrt{3}-1}\bigg)(x-2)\\\begin{bmatrix}\text{when m = -}\bigg(\frac{\sqrt{3}+2}{2\sqrt{3}-1}\bigg)\\\text{and\space }x_1=2, y_1=3\end{bmatrix}$$

$$\Rarr\space y(2\sqrt{3}-1)-6\sqrt{3}+3\\=-(\sqrt{3}+2)+2\sqrt{3}+4\\\Rarr\space (2+\sqrt{3})x+(2\sqrt{3}-1)-8\sqrt{3}-1=0.$$

13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (– 1, 2).

Sol. Let the given points are P(3, 4) and Q(– 1, 2) then B is the mid-point of PQ, i.e.,

$$\text{B\space=}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\=\bigg(\frac{3-1}{2},\frac{4+2}{2}\bigg)=(1,3)$$

(∵ x1 = 3, y1 = 4, x2 = – 1, y2 = 2)

Now,

AB ⊥ PQ,

i.e., Slope of AB × Slope of PQ = – 1

$$\Rarr\space m×\frac{y_2-y_1}{2}=1\\\Rarr\space m×\frac{2-4}{-1-3}=-1\\\Rarr\space m×\frac{-2}{-4}=-1\\\Rarr\space m×\frac{1}{2}=-1$$

⇒ m = – 2

Now, equation of right bisector AB by using

y – y1 = m(x – x1) where (x1 – y1) = (1, 3) and m = – 2 is

y – 3 = – 2(x – 1)

⇒ y – 3 = – 2x + 2

⇒ 2x + y – 5 = 0.

14. Find the coordinates of the foot of perpendicular from the point (– 1, 3) to the line 3x – 4y – 16 = 0.

Sol. Given equation of line AB is

3x – 4y – 16 = 0

$$\Rarr\space y=\frac{3}{4}x-4\space\text{...(i)}\\\text{Slope of line (i) is}\\m_1=\frac{3}{4}$$

Let slope of line PQ is m.

Since  PQ ⊥ AB

∴ Slope of PQ × Slope of AB = – 1

$$\Rarr\space m×\bigg(\frac{3}{4}\bigg)=-1\\(\therefore\space \text{m}_1\text{m}_2=-1)\\\Rarr\space m=-\frac{4}{3}$$

Hence, equation of line PQ by using

y – y1 = m(x – x1)

where (x1, y1) = – 1, 3) and

$$m=-\frac{4}{3}\space\text{is}$$

$$\text{y-3}=-\frac{4}{3}(x+1)$$

⇒ 3y –9 = – 4x – 4

⇒ 4x + 3y – 5 = 0 …(ii)

Now, solving equation (i) and (ii), to find thecorodinates of Q.

From equation (i),

$$y=\frac{3}{4}x-4\\\text{On putting the value of y}\\\text{ in equation (ii), we get}\\\text{4x+3}\bigg(\frac{3}{4}x-4\bigg)-5=0\\\Rarr\space 4x+\frac{9x}{4}-12-5=0\\\Rarr\space \frac{25x}{4}=17\\\Rarr\space 25x=17×4\\\Rarr\space x=\frac{68}{25}$$

∴ From equation (i),

$$y=\frac{3}{4}×\frac{68}{25}-4\\=\frac{3×17}{25}=\frac{14}{1}=\frac{51}{25}-\frac{4}{1}\\=\frac{51-100}{25}=-\frac{49}{25}$$

Hence, co-ordinates of foot of perpendicular is

$$\bigg(\frac{68}{25},-\frac{49}{25}\bigg)$$

15. The perpendicular from the origin to the line y = mx + c meets at the point (– 1, 2). Find the values of m and c.

Sol.

$$\because\space\text{line OP}\perp\text{line AB}\\\therefore\text{Slope of OP × Slope of AB = – 1}\\\Rarr\space \frac{2-0}{-1-0}×=-1\\\Rarr\space\text{-2m=-1}\\\Rarr\space m=\frac{1}{2}\\\text{Also, point (– 1, 2) satisfies the line}\\ \text{y = mx + c i.e., 2 = m(– 1) + c} \\\text{(Putting x = – 1, y = 2)}$$

$$\Rarr\space 2=-\frac{1}{2}+c\\\Rarr\space c=2+\frac{1}{2}=\frac{5}{2}\\\text{Hence,}\space m=\frac{1}{2},c=\frac{5}{2}$$

16. If p and q are lengths of perpendicular from origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k respectively, prvoe that p2 + 4q2 = k2.

Sol.

$$\text{For Fig. I, x cos}\space\theta\space–\space\text{y sin}\space\theta = \text{k cos }\theta \text{…(i)}\\\text{For Fig. II, x sec}\space\theta\text{ + y cosec}\space\theta \text{= k …(ii)}\\\therefore\space\text{Perpendicular distance from (0, 0) to}\\\text{ the equation (i) is}\\\text{P =}\bigg|\frac{0-0-kcos 2\theta}{\sqrt{\text{cos}^{2}\theta+ sin^{2}\theta}}\bigg|=\text{k cos 2}\theta\space\\\text{…(iii)}\\\bigg[\because \text{d}=\bigg|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\bigg|\bigg]\\\text{Again, equation (ii) can be written as}\\\frac{x}{\text{cos}\space\theta}+\frac{y}{sin \theta}=k\\\text{⇒ x sin}\space\theta + y\space cos \space\theta = \text{k cos }\theta\space\text{sin}\space\theta\\\Rarr\space\text{x sin}\space\theta+y\space cos\theta$$

$$=\frac{k}{2}(2\space\text{cos}\theta\space sin\theta)\\\Rarr\space\text{ x sin}\theta + y cos\theta=\frac{k}{2}\text{sin}\theta\\\Rarr\space x sin\theta + tcos\theta-\frac{k}{2}\text{sin}2\theta=0\\\text{...(iv)}$$

Now, perpendicular distance from (0, 0) to equation (iv) is

$$q=\bigg|\frac{0+0-\frac{k}{2}\text{sin 2}\theta}{\sqrt{\text{sin}^{2}\theta+cos^{2}\theta}}\bigg|$$

$$\begin{pmatrix}\because\space a=sin\theta, b=cos\theta,\\ c=-\frac{k}{2}\text{sin2}\theta, x_1=0, y_1=0\end{pmatrix}$$

$$\Rarr\space q=\frac{k}{2}\text{sin 2}\theta$$

(∴ sin2 θ + cos2 θ = 1) …(v)

⇒ 2q = k sin 2θ

On squaring and adding equations (ii) (v),

p2 + 4q2 = k2 cos2 2θ + k2 sin2

= k2 (cos2 2θ + sin2 θ)

∴ p2 + 4q2 = k2 (∵ sin2 θ + cos2 θ = 1)

Hence Proved.

17. In the DABC with vertices A(2, 3), B(4, – 1) and C(1, 2) find the equation and length of altitude from the vertex A.

Sol. Given, vertices of a DABC are A(2, 3), B(4, – 1) and C(1, 2).

∴ Line AD ⊥ line BC

∴ Slope of AD × Slope of BC = – 1

$$\Rarr\space m×\frac{y_2-y_1}{x_2-x_1}=-1\\\Rarr\space m×\frac{2+1}{1-4}=-1\\(\because\space x_1=4, y_1=-1,x_2=1,y_2=2)\\\Rarr\space m×\frac{3}{-3}=-1$$

⇒ m = 1

Hence, equation of  AD, by using y – y1 = m(x – x1) where (x1, y1) = (2, 3) and m = 1 is

y – 3 = (1x – 2)

⇒ x – y – 2 + 3 = 0

⇒ x – y + 1 = 0

Now, equation of BC by using

$$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\space\text{is}$$

$$\text{y+1}=\frac{2+1}{1-4}(x-4)\\(\because\space x_1=4, y=-1, x_2=1, y_2=2)\\\Rarr\space y+1=\frac{3}{-3}(x-4)$$

⇒ y + 1 = – x + 4

⇒ x + y – 3 = 0

∴ Length of altitude AD = Perpendicular distance from (2, 3) to the line BC.

i.e., x + y – 3 = 0

$$=\bigg|\frac{2+3-3}{\sqrt{1^2+1^2}}\bigg|=\frac{2}{\sqrt{2}}\\=\sqrt{2}\\\bigg[\because d=\bigg|\frac{ax_1+by_1+c}{\sqrt{a^{2}+b^{2}}}\bigg|\bigg]$$

18. If p is the length of perpendicular from origin to the line whose intercepts on the axes are a and b,

$$\textbf{then show that}\space\frac{\textbf{1}}{\textbf{p}^{2}}=\frac{\textbf{1}}{\textbf{a}^{\textbf{2}}}\textbf{+}\frac{\textbf{1}}{\textbf{b}^{\textbf{2}}}$$

Sol. Equation of line in intercept form is

$$\frac{x}{a}+\frac{y}{b}=1\\\Rarr\space \frac{x}{a}+\frac{y}{b}-1=0\\\text{Its distance from origin is}\\\bigg|\frac{0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\bigg|=p\\\bigg(\because\space d=\bigg|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\bigg|\bigg)$$

$$\Rarr\space p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\\\Rarr\space\frac{1}{p}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}\\\text{On squaring both sides, we get}\\\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\\\textbf{Hence Proved.}$$

Miscellaneous Exercise

1. Find the values of k for which the line (k – 3)x – (4 – k2)y + k2 – 7k + 6 = 0 is:

(a) parallel to the X-axis

(b) parallel to the Y-axis

(c) passing through the origin

Sol. Equation of given line is

(k – 3)x – (4 – k2)y + k2 – 7k + 6 = 0 …(i)

(a) When line (i) is parallel to X-axis, then slope = 0

$$\text{i.e.}\space \frac{(k-3)}{(4-k^2)}=0\\\bigg(\because\space\text{Slope of line (i) is}\space\frac{k-3}{4-k^2}\bigg)$$

⇒ k – 3 = 0

⇒ k = 3

(b) When line (i) is parallel to Y-axis,

$$\text{then slope}=\infty=\frac{1}{0}\\\text{i.e.}\space \frac{k-3}{4-k^2}=\frac{1}{0}\\\bigg(\because \text{Slope of line (i) is}\frac{k-3}{4-k^2}\bigg)$$

⇒ 4 – k2 = 0

⇒ k = ± 2

(c) When line (i) passing through origin (0, 0), then it will satisfy it

0 – 0 + k2 – 7k + 6 = 0

(Put x = 0, v = 0 in equation (i))

⇒ k2 – 6k – k + 6 = 0

⇒ k(k – 6) – 1(k – 6) = 0

⇒ (k – 6) (k – 1) = 0

∴ k = 1, 6

2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line

$$\sqrt{\textbf{3}}\textbf{x+y+2=0.}$$

Sol. Given, equation of line is

$$\sqrt{3}x+y+2=0\\\Rarr\space\sqrt{3}x+y=-2\\\Rarr\space \sqrt{3}x-y=2$$

On dividing above equation by

$$\sqrt{(\text{coefficient of x})^{2}+(\text{coefficient of y})^2}\\\text{i.e.,}\space\sqrt{(-\sqrt{3})^2+(-1)^2}\\=\sqrt{3+1}\\=\sqrt{4}=2\\\Rarr\space\frac{-\sqrt{3}}{2}x-\frac{1}{2}y=\frac{2}{2}$$

⇒ – cos 30° x – sin 30° y = 1

(∵ for convert in the form of x cos θ + y sin θ = p)

⇒ cos (180° + 30°)x + sin (180° + 30°)y = 1

[since cos θ sin θ both are negative in third quadrant]

⇒ (cos 210°)x + (sin 210°)y = 1 is the normal form.

On comparing with x cos θ + y sin θ = p, we get

θ = 210° and p = 1.

3. Find the equation of the lines, which cut off intercepts on the axes whose sum and product are 1 and – 6, respectively.

Sol. Let the intercept form of a line is

$$\frac{x}{a}+\frac{y}{b}=1\space\text{...(i)}$$

Given a + b = 1 …(ii)

and

ab = – 6 …(iii)

For solving equation (ii) and equation (iii) to find the value of a and b.

$$\text{Put} b=\frac{-6}{a}\text{from equation (iii) in equation (ii),}\\a-\frac{6}{a}=1\\\Rarr\space \frac{a}{1}-\frac{6}{a}=1$$

⇒ a2 – 6 = a

⇒ a2 – a – 6 = 0

⇒ a2 – (3 – 2)a – 6 = 0

⇒ a(a – 3) + 2((a – 3) = 0

⇒ (a + 2) (a – 3) = 0

⇒ If a = 3, – 2

then b = – 2, 3

Put a = 3, – 2 respectively in equation (i), we get the required equations of lines

$$\frac{x}{3}+\frac{y}{-2}=1\space\text{or}\space\\\frac{x}{-2}+\frac{y}{3}=1$$

⇒ 2x – 3y = 6 or – 3x + 2y = 6

⇒ 2x – 3y – 6 = 0 or 3x – 2y + 6 = 0.

4. What are the points on the Y-axis whose distances from the line

$$\frac{\textbf{x}}{\textbf{3}}\textbf{+}\frac{\textbf{y}}{\textbf{4}}\textbf{=1}\textbf{ is 4 units}$$

Sol. Let any point on the Y-axis is (0, k) on as Y-axis the x coordinate will be zero.

Equation of given line is

$$\frac{x}{3}+\frac{y}{4}=1\\\text{or 4x + 3y = 12}$$

⇒ 4x + 3y – 12 = 0 its distance from (0, k), is 4 units.

$$\text{i.e.,}\space\bigg|\frac{0+3k-12}{\sqrt{4^2+3^2}}\bigg|=4\\\begin{pmatrix}\because\space\text{Perpendicular distance}\\ d=\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\\\text{and a = 4, b = 3, c = – 12, x}_1 = 0, y_1 = k\end{pmatrix}\\\Rarr\space \bigg|\frac{3k-12}{5}\bigg|=4$$

⇒ |3k – 12| = 20

⇒ 3k – 12 = ± 20

Taking positive sign,

3k = 20 + 12

⇒ 3k = 32

$$\Rarr\space k=\frac{32}{3}$$

Taking negative sign,

3k = – 20 + 12

⇒ 3k = – 8

$$\Rarr\space k=-\frac{8}{3}$$

Hence, the required points on the Y-axis are

$$\bigg(0,-\frac{8}{3}\bigg),\bigg(0,\frac{32}{3}\bigg).$$

5. Find the perpendicular distance from the origin of the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ).

Sol. Equation of the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ) is

$$\text{y-sin}\theta=\frac{sin\phi- sin\space\theta}{cos\phi-cos\space\theta}(x-cos\theta)\\\begin{pmatrix}\because\space y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\text{Here x}_1= cos\theta, y_1= sin\theta\\x^2= cos\phi, y_2= sin\phi\end{pmatrix}\\\Rarr y-sin\theta=\frac{2 cos\frac{\phi+\theta}{2}\text{sin}\frac{\phi-2}{2}}{-2\space\text{sin}\frac{\phi+\theta}{2}\text{sin}\frac{\phi-\theta}{2}}(c-cos \theta)$$

$$\begin{bmatrix}\because\space\text{cos C - cos D}=-2\text{sin}\frac{\text{C+D}}{2}\text{sin}\frac{\text{C-D}}{2}\\\text{and}\space\text{sin C - sin D}=2 \text{cos}\frac{\text{C+D}}{2}\text{sin}\frac{\text{C-D}}{2}\end{bmatrix}$$

$$\Rarr\space\text{y- sin}\space\theta=\frac{\text{cos}\frac{\phi+\theta}{2}}{-\text{sin}\frac{\phi+\theta}{2}}(x-cos\space\theta)\\\Rarr\space -y\space\text{sin}\frac{\phi+\theta}{2}+\text{sin}\space\theta sin\frac{\phi+\theta}{2}\\=x \text{cos}\frac{\phi + \theta}{2}-\text{cos}\space\theta\text{cos}\frac{\phi+\theta}{2}\\\Rarr\space \text{x cos}\frac{\phi+\theta}{2}+y\space\text{sin}\frac{\phi+\theta}{2}\\=\text{sin}\space\theta \text{sin}\frac{\phi+\theta}{2}+\text{cos}\theta\text{cos}\frac{\phi+\theta}{2}\\\Rarr\space \text{x cos}\frac{\phi+\theta}{2}+\text{y sin}\frac{\phi+\theta}{2}\\=\text{cos}\bigg(\theta-\frac{\theta+\phi}{2}\bigg)$$

[∵ cos (A – B) = cos A cos B + sin A sin B]

$$\Rarr\space\text{x cos}\frac{\phi+\theta}{2}+y\space \text{sin}\frac{\phi+\theta}{2}\\-\text{cos}\bigg(\frac{\theta-\phi}{2}\bigg)=0$$

Its distance from origin (0, 0) is

$$d=\begin{vmatrix}\frac{0-0-\text{cos}\bigg(\frac{\theta-\phi}{2}\bigg)}{\sqrt{\text{cos}^2\bigg(\frac{\phi+\theta}{2}\bigg)+\text{sin}^2}\bigg(\frac{\phi+\theta}{2}\bigg)}\end{vmatrix}\\\bigg[\because\space d=\bigg|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\bigg|\bigg]\\\Rarr\space\text{d= cos}\bigg(\frac{\theta-\phi}{2}\bigg)\\(\because\space cos^2\theta+ sin^2\theta=1)$$

6. Find the equation of the parallel to Y-axis and drawn through the point of intersection of the lines x –7y + 5 = 0 and 3x + y = 0.

Sol. Given, equation of lines are

x – 7y + 5 = 0 …(i)

and 3x + y = 0 …(ii)

On solving equations (i) and (ii), we find the point of intersection, from equation (i)

x = 7y – 5

On putting the value of x in equation (ii), we get

3(7y – 5) + y = 0

⇒ 21y – 15 + y = 0

⇒ 22y = 15

$$\Rarr\space y=\frac{15}{22}\\\therefore \text{From equation (i),}\\ x = \frac{7×15}{22}-5=\frac{105-110}{22}\\\Rarr\space x=-\frac{5}{22}\\\text{Hence, the point is}\\\bigg(-\frac{5}{22},\frac{15}{22}\bigg)$$

Hence, equation of required line by using

y – y1 = m(x – x1)

$$\text{where (x}_1,y_1)=\bigg(-\frac{5}{22},\frac{15}{22}\bigg)\\\text{and m}=\frac{1}{0}\\\text{as line parallel to Y-axis, is}\\y-\frac{15}{22}=\frac{1}{0}\bigg(x+\frac{5}{22}\bigg)\\\Rarr\space 0=x+\frac{5}{22}$$

⇒ 22x + 5 = 0

Hence, the required equation of the line is 22x + 5 = 0.

7. Find the equation of a line drawn perpendicular to the line

$$\frac{\textbf{x}}{\textbf{4}}+\frac{\textbf{y}}{\textbf{6}}\textbf{=1}$$ through the point where it meets the Y-axis.

Sol. Given, equation of line is

$$\frac{x}{4}+\frac{x}{6}=1\\\Rarr\space \frac{3x+2y}{12}=1$$

⇒ 3x + 2y = 12 …(i)

$$\text{Its slope}=\frac{-3}{2}$$

If line (i) meet the Y-axis put x = 0 in equation (i), we get

0 + 2y = 12

⇒ y = 6

∴ Point is (0, 6).

Now, equation of the line perpendicular to equation (i) is

y – y1 = m(x – x1)

$$y-6=\frac{2}{3}(x-0)$$

⇒ 3y – 18 = 2x

=  2x – 3y + 18 = 0

[∵ m1m2 = – 1]

$$m_2=-1×\frac{\normalsize-2}{3}=\frac{2}{3}$$

Hence, equation (ii) becomes

2x – 3y + 18 = 0.

8. Find the area of the triangle formed by the line y – x = 0, x + y = 0 and x – k = 0.

Sol. Let the given lines represent the equation of sides of the triangle are

AB : y – x = 0 …(i)

BC : x + y = 0 …(ii)

and CA : x – k = 0 …(iii)

For point A, on solving equations (i) and (iii), we get A(k, k).

For point B, on solving equations (i) and (ii), we get B(0, 0).

For point C, on solving equations (ii) and (iii), we get C(k, – k).

So, the required points are

A(k, k), B(0, 0), C(k, – k)

Hence area of ΔABC

$$=\frac{1}{2}[k(0+k)+0(-k-k)+k(k-0)]\\\begin{bmatrix}\because \text{Area of triangle}\\=\frac{1}{2}[x_1(y_2-y_3)+\\x_2(y_3-y_1)+x_3(y_1-y_2)]\end{bmatrix}\\=\frac{1}{2}[k^2+k^2]\\=\frac{2k^2}{2}=k^2\text{sq unit}$$

Hence, the area of the triangle is k2 sq. units.

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Sol. Given, equation of three lines are

3x – y – 2 = 0 …(i)

px + 2y – 3 = 0 …(ii)

and

2x – y – 3 = 0 …(iii)

Adding equations (i) and (iii), we get

5x – 5 = 0

⇒ x = 1

∴ From equation (i),

3 × 1 + y – 2 = 0 (∵ put x = 1)

⇒ y = – 1

So, the points is (1, – 1).

Above point will satisfy equation (ii) i.e.,

p × 1 + 2(– 1) – 3 = 0

⇒ p – 2 – 3 = 0 (put x = 1, y = – 1)

⇒ p = 5

Hence, the value of P is 5.

10. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) = 0.

Sol. Given, equation of three lines are'

y = m1x + c1 …(i)

y = m2x + c2 …(ii)

y = m3x + c3 …(iii)

On solving equations (i) and (ii),

m1x + c1 = m2x + c2

⇒ m1x – m2x = c2 – c1

$$\therefore\space x=\frac{c_2-c_1}{m_1-m_2}\\\text{From equation (i),}\\\text{y = m}_1\bigg(\frac{c_2-c_1}{m_1-m_2}\bigg)+\frac{c_1}{1}\\\Rarr\space y=\frac{m_1c_2-m_1c_1+m_1c_1-m_2c_1}{m_1-m_2}\\\Rarr\space y=\frac{m_1c_2-m_2c_1}{m_1-m_2}$$

∴ Point of intersection of line is (i) and (ii)

$$p\bigg(\frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2}\bigg)$$

On putting the values of x and y in equation (iii), we get

$$\frac{m_1c_2-m_2c_1}{m_1-m_2}\\=m_3\bigg(\frac{c_2-c_1}{m_1-m_2}\bigg)+\frac{c_3}{1}\\\frac{m_1c_2-m_2c_1}{m_1-m_2}\\=\frac{m_3c_2-m_3c_1+m_1c_3-m_2c_3}{m_1-m_2}$$

⇒ m1c2 – m2c1 = m3c2 – m3c1 + m1c3 – m2c3

⇒ m1c2 – m2c1 – m3c2 + m3c1 – m1c3 + m2c3 = 0

⇒ m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) = 0.

Hence Proved.

11. Find the equation of the lines through the point (3, 2) which makes an angle of 45° with the line x – 2y = 3.

Sol. Equation of line l1 is x – 2y = 3 …(i)

Slope of line l1

$$m_1=-\bigg(\frac{1}{-2}\bigg)\\m_1=\frac{1}{2}\\\text{Angle between two lines is}\\\text{tan}\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\\\text{Given,}\space\theta=45\degree,\\m_1=\frac{1}{2},m_2=m\space (\text{let})\\\therefore\space \text{tan}45\degree=\begin{vmatrix}\frac{\frac{1}{2}-m}{1+\frac{1}{2}m}\end{vmatrix}\\\Rarr\space 1=\bigg|\frac{1-2m}{2+m}\bigg|$$

$$\Rarr\space \frac{1-2m}{2+m}\pm1\\\text{Taking positive sign}\\\frac{1-2m}{2+m}=1$$

⇒ 1 – 2m = 2 + m

⇒ 1 – 2 = 3m

$$\Rarr\space m=-\frac{1}{3}\\\text{Taking negative sign,}\\\frac{1-2m}{2+m}=1$$

⇒ 1 – 2m = – 2 – m

⇒ 1 + 2 = 2m – m

⇒ m = 3

∴  Slope of line l2,

$$m=3,\frac{-1}{3}$$

Now, equation of line l2 by using

y – y1 = m(x – x1)

$$\Rarr\space\text{y-2}=-\frac{1}{3}(x-3)\\\begin{pmatrix}\text{when m}=-\frac{1}{3}\\x_1=3, y_1=2\end{pmatrix}$$

⇒ 3y – 6 = – x + 3

⇒ x + 3y –9 = 0

Again,

y – 2 = 3(x – 3)

$$\begin{pmatrix}\text{when m=3,}\\x_1=3,y_1=2\end{pmatrix}$$

⇒ y – 2 = 3x –9

⇒ 3x – y –9 + 2 = 0

⇒ 3x – y – 7 = 0

Hence, the equation of the lines are

x + 3y = 9, 3x – y = 7.

12. Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axis.

Sol. Given equation of lines are

4x + 7y –3 = 0 …(i)

and

2x – 3y + 1 = 0 …(ii)

On solving equations (i) and (ii),

From equation (ii),

3y = 2x + 1

$$\Rarr\space y=\frac{2}{3}x+\frac{1}{3}\\\text{On putting in equation (i), we get}\\\text{4x+7}\bigg(\frac{2}{3}x+\frac{1}{3}\bigg)-3=0\\\Rarr\space 4x+\frac{14}{3}x+\frac{7}{3}-3=0\\\Rarr\space\frac{12x+14x}{3}+\frac{7-9}{3}=0\\\Rarr\space\frac{26x-2}{3}=0\\\Rarr\space x=\frac{2}{26}\\\Rarr\space x=\frac{1}{13}$$

On putting the value of x in equation (i), we get

$$\frac{4}{13}+7y-3=0\\\Rarr\space 7y=3-\frac{4}{13}\\\Rarr\space 7y=\frac{39-4}{13}\\\Rarr\space 7y=\frac{35}{13}\\\Rarr\space y=\frac{5}{13}\\\text{Point is}\bigg(\frac{1}{13},\frac{5}{13}\bigg)\\\text{Now, equation of line in intercept form is}\\\frac{x}{a}+\frac{y}{b}=1\space\text{...(iii)}$$

Since, line (iii) has equal intercepts on the axes i.e., a = b.

$$\Rarr\space \frac{x}{a}+\frac{y}{b}=1\\\Rarr\text{x + y = a}\space\text{...(iv)}$$

Above line passes through the point

$$\bigg(\frac{1}{13},\frac{5}{13}\bigg)\\\text{i.e., point will satisfy it.}\\\frac{1}{13}+\frac{5}{13}=a\\\Rarr\space a=\frac{6}{13}\\\text{Hence, required equation of line (iv) becomes}\\\text{x + y =}\frac{6}{13}$$

⇒ 13x + 13y = 6.

13. Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is

$$\frac{\textbf{y}}{\textbf{x}}=\frac{\textbf{m}\space\pm\space \textbf{tan}\space\theta}{\textbf{1}\mp \textbf{m tan}\space\theta}$$

Sol. Given, equation of line y = mx + c and slope of given line is m and slope of another line is m2.

Angle between two lines is

$$\text{tan}\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\\\Rarr\space\text{tan}\theta=\bigg|\frac{m-m_2}{1+mm_2}\bigg|\\(\because m_1=m)\\\Rarr\space \frac{m-m_2}{1+mm_2}=\pm\text{tan}\space\theta\\\text{T aking postivie sign,}\\\frac{m-m_2}{1+mm_2}=\frac{\text{tan}\space\theta}{1}$$

⇒ m – m2 = tan  θ + mm2 tan  θ

⇒ m – tan  θ = m2 + mm2 tan  θ

⇒ m2 (1 + m tan  θ) = m – tan  θ

$$\Rarr\space m_2=\frac{\text{m-tan}\space\theta}{1+ m\text{tan}\theta}\space\text{...(i)}\\\text{Taking negative sign,}\\\frac{m-m_2}{1+mm_2}\\=\frac{-\text{tan}\theta}{1}$$

⇒ m – m2 = – tan θ – mm2 tan θ

⇒ m + tan θ = m2 – mm2 tan θ

⇒ m + tan θ = m2 (1 – m tan θ)

$$\Rarr\space m_2=\frac{m+\text{tan}\space\theta}{1- m\text{tan}\space\theta}\space\text{...(ii)}\\\therefore\space\text{From equations (i) and (ii)}\\m_2=\frac{m+\text{tan}\space\theta}{1\mp \text{m tan}\space\theta}$$

Equation of line through (0, 0) by using

y – y1 = m(x – x1)

where m = m2 is

$$\text{y-0}=\frac{m\pm\text{tan}\theta}{1\mp\text{m tan}\space\theta}(x-0)\\\Rarr\space\frac{y}{x}=\frac{m\pm\text{tan}\theta}{1\mp \text{m tan}\theta}.$$

14. In what ratio, the line joining (– 1, 1) and (5, 7) is divided by the line x + y = 4?

Sol. Let the point P divides the line joining the points A(– 1, 1) and B(5, 7) in the ratio is K : 1.

$$\text{Then},\space\text{P}=\bigg(\frac{k×x_2+1×x_1}{\text{K+1}},\frac{K×y_1+1×y_1}{\text{K+1}}\bigg)$$

(∵ x1 = – 1, y1 = 1, x2 = 5, y2 = 7)

$$=\bigg(\frac{K×5+1×(-1)}{K+1},\frac{K×7+1×1}{K+1}\bigg)\\=\bigg(\frac{5K-1}{K+1},\frac{7K+1}{K+1}\bigg)$$

Point P will satisfy the line x + y = 4

$$\text{i.e,}\space\frac{5K-1}{K+1}+\frac{7K+1}{K+1}=\frac{4}{1}\\\Rarr\space\frac{5K-1+7K+1}{K+1}=4$$

⇒ 12K = 4K + 4

⇒ 8K = 4

$$\Rarr\space\text{K}=\frac{4}{8}=\frac{1}{2}$$

Hence, the required ratio is K : 1 = 1 : 2 (internally).

15. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Sol. The equation of line AB is

4x + 7y + 5 = 0 …(i)

and the equation of line PQ is

2x – y = 0 …(ii)

On solving equations (i) and (ii), to obtain the coordinate of Q.

From equation (ii) put y = 2x in equation (i)

4x + 7(2x) + 5 = 0

⇒ 18x + 5 = 0

$$\Rarr\space x=-\frac{5}{18}\\\text{Put x = }-\frac{5}{18}\space\text{in equation (ii),}\\2\bigg(-\frac{5}{18}\bigg)-y=0\\\Rarr\space y=-\frac{5}{9}\\\therefore\space\text{Coordinate of Q}=\bigg(-\frac{5}{18},-\frac{5}{9}\bigg)\\\text{Length of QP}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\=\sqrt{\bigg(1+\frac{5}{18}\bigg)^2+\bigg(2+\frac{5}{9}\bigg)^2}$$

$$\begin{pmatrix}\because\space x_1=-\frac{5}{18},y_1=-\frac{5}{9},\\ x_1=2,\space y_2=2\end{pmatrix}\\=\sqrt{\bigg(\frac{23}{18}\bigg)^2+\bigg(\frac{23}{9}\bigg)^2}\\=\frac{23}{9}\sqrt{\frac{1}{4}+1}\\=\frac{23}{9}×\sqrt{\frac{1}{4}+1}\\=\frac{23}{9}×\frac{\sqrt{5}}{2}\\=\frac{23}{18}\sqrt{5}\space\text{sq unit}\\\text{Thus, the required distance is}\\\frac{23\sqrt{5}}{18}\space\text{units}.$$

16. Find the direction in which a straight line must be drawn through the piont (– 1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from the point.

Sol. A line passing through point P(– 1, 2) and having slope m is

y – 2 = m(x + 1)

$$\begin{pmatrix}\because\space y=y_1=m(x+1)\\\text{where,}\space x_1=-1, y_1=2\end{pmatrix}$$

y = mx + m + 2 …(i)

Now, determine the point of intersection of line AB and line PQ.

On putting the value of y from equation (i) in x + y = 4, we get

x + mx + m + 2 = 4

⇒ x(1 + m) = 2 – m

$$\Rarr\space x=\frac{2-m}{1+m}\\\therefore\text{From equation (i),}\\\text{y=m}\bigg(\frac{2-m}{1+m}\bigg)+m+2\\=\frac{2m-m^2+m+m^2+2+2m}{1+m}\\=\frac{2+5m}{1+m}\\\therefore\space\text{Coordinate of Q}\\=\bigg(\frac{2-m}{1+m},\frac{2+5m}{1+m}\bigg)\\\because\space \text{Distance between PQ}$$

PQ = 3 (given)

⇒ (PQ)2 = 9

⇒ (x2 – x1)2 + (y2 – y1)2 = 9

$$\Rarr\space\bigg(\frac{2-m}{1+m}+1\bigg)^2+\bigg(\frac{2+5m}{1+m}-2\bigg)^2=9\\\begin{bmatrix}\because\space x_1=-1, y_1=2\\ x_2=\frac{2-m}{1+m}, y_2=\frac{2-5m}{1+m}\end{bmatrix}\\\Rarr\space\bigg(\frac{3}{1+m}\bigg)^2+\bigg(\frac{3m}{1+m}\bigg)^2=9$$

⇒ 9 + 9m2 = 9(1 + m)2

⇒ 1 + m2 = 1 + 2m + m2

⇒ 2m = 0

⇒ m = 0

∴ Slope of PQ is zero, i.e., it is parallel to X-axis.

17. The hypotenuse of a right-angled traingle has its ends at the points (1, 3) and (– 4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Sol. First we plot the points A(1, 3) and B(– 4, 1) in the XY-plane. From the point A(1, 3), we draw a line parallel to Y-axis. And the point B(– 4, 1), we draw a line parallel to X-axis. The point of intersectin of two lines is on C, which is right angled at C.

∴ The coordinate of C will be (1, 1).

∴ Equation of line AC passing through A(1, 3) and C(1, 1) is

$$\text{y-y}_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\therefore\space y-3=\frac{1-3}{1-1}(x-1)\\\Rarr\space y-3=\frac{-2}{0}(x-1)$$

⇒ x = 1

Equation of line BC is

$$\text{y-1}=\frac{1-1}{1+4}(x-1)\\\Rarr\space y-1=\frac{0}{1+4}(x-1)$$

⇒ y – 1 = 0

⇒ y = 1

Hence , the legs of a triangle are x = 1 and y = 1.

18. Find the image of the point (3, 8) w.r.t. the line x + 3y = 7 assuming the line to be plane mirror.

Sol. Let equation of line AB is

x + 3y = 7 …(i)

Here slope of a line is

$$m_1=\frac{1}{3}$$

Let P(3, 8) be the given point for which image to be found.

Here, PQ ⊥ AB

Slope of PQ × Slope of AB = – 1

Let slope of PQ is m.

$$\therefore\space m×\bigg(\frac{-1}{3}\bigg)=-1\space\\(\because m_1m_2=-1)$$

⇒ m = 3

Now, equation of line PQ by using

y – y1 = m(x – x1)

where, (x1, y1) = (3, 8) and m = 3 is

y – 8 = 3(x – 3)

⇒ y – 8 = 3x – 9

⇒ 3x – y – 1 = 0 …(ii)

To determine point Q, solving equations (i) and (ii)

From equation (i) x = 7 – 3y put in equation (ii), we get

⇒ 3(7 – 3y) – y – 1 = 0

⇒ 21 – 9y – y – 1 = 0

⇒ 20 – 10y = 0

⇒ 10y = 20

⇒ y = 2

∴ From equation (i), x = 7 – 3 × 2 = 7 – 6 = 1

∴ Coordinate of Q = (1, 2).

Let the image be R(h, k), since the line is a plane mirror, so point Q will be the mid-point i.e., Q(1, 2).

Now, mid-point of P(3, 8) and R(h, k) is

$$=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\=\bigg(\frac{3+h}{2},\frac{8+k}{2}\bigg)$$

On comparing with coordinate Q(1, 2),

$$\therefore\space \frac{3+h}{2}=1\text{and}\\\frac{8+k}{2}=2$$

⇒ 3 + h = 2 and 8 + k = 4

⇒ h = 2 – 3 and k = 4 – 8

⇒ h = – 1 and k = – 4

Hence, image of a point P is R(– 1, – 4).

19. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

Sol. Let the angle between the lines y = 3x + 1 and y = mx + 4 is θ.

Here, slopes of the lines are

m1 = 3, m2 = m

$$\text{tan}\space\theta=\bigg|\frac{3-m}{1+3m}\bigg|\space\text{...(i)}\\\bigg(\because \text{tan}\theta=\bigg|\frac{m_1-m_2}{1+m_1m_2}\bigg|\bigg)$$

Again, it is given that angle between the lines 2y = x + 3 and y = mx + 4 is also θ.

$$\text{Here slopes are m}_1=\frac{1}{2}\space\text{and}\space m_2=m.\\\therefore\space\text{tan}\space\theta=\begin{vmatrix}\frac{\frac{1}{2}-m}{1+\frac{1}{2}×m}\end{vmatrix}\space\text{...(iii)}$$

From equations (i) and (ii), we get

$$\begin{vmatrix}\frac{3-m}{1+3m}\end{vmatrix}=\begin{vmatrix}\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\end{vmatrix}\\\Rarr\space\frac{3-m}{1+3m}=\pm\bigg(\frac{1-2m}{2+m}\bigg)$$

Taking positive signs,

⇒ (3 – m) (2 + m) = (1 – 2m) (1 + 3m)

⇒ 6 + 3m – 2m – m2 = 1 + 3m – 2m – 6m2

⇒ 6 = 1 – 5m2

⇒ 5m2 = 1 – 6

$$\text{m}^2=\frac{-5}{5}$$

which is not possible as m2 ≥ 0.

Again, taking negative sign, we get

$$\frac{3-m}{1+3m}=-\frac{1-2m}{2+m}$$

⇒ (3 – m) (2 + m) = (2m – 1) (1 + 3m)

⇒ 6 + 3m – 2m – m2 = 2m + 6m2 – 1 – 3m

⇒ 6 + m – m2 = 6m2 – m – 1

⇒ 6m2 – m – 1 – 6 – m + m2 = 0

⇒ 7m2 – 2m – 7 = 0

$$\Rarr m=\frac{2\pm\sqrt{4+4×7×7}}{2×7}\\\bigg[\because x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\bigg]\\\Rarr\space =\frac{2\pm2\sqrt{1+49}}{2×7}\\m=\frac{1\pm\sqrt{50}}{7}\\m=\frac{1\pm5\sqrt{2}}{7}$$

20. If sum of perpendicular distances of a variable point P(x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Sol. Given equations of lines are

x + y – 5 = 0 …(i)

and 3x – 2y + 7 = 0 …(ii)

Given, the sum of perpendicular distances of a variable point P(x, y) from lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is 10.

$$\therefore\space\begin{vmatrix}\frac{x+y-5}{\sqrt{(1)^2 +(1)^2}}\end{vmatrix}+\begin{vmatrix}\frac{3x-2y+7}{\sqrt{3^2+(-2)^2}}\end{vmatrix}\\=10\\\begin{pmatrix}\because \text{Perpendicular distance d}=\\\bigg|\frac{ax_1+by_1+x}{\sqrt{a^2+b^2}}\bigg|\end{pmatrix}\\\Rarr\space\frac{x+y-5}{\sqrt{2}}+\frac{3x-2y+7}{\sqrt{13}}\\=10$$

$$\Rarr\space{\sqrt{13}}(x+y-5)+\sqrt{2}(3x-2y+7)\\=10\sqrt{2}\sqrt{13}\\\Rarr\space x (\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})-\\5\sqrt{13}+7\sqrt{2}=10\sqrt{26}\\\Rarr\space x(\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})+\\7\sqrt{2}-5\sqrt{13}-10\sqrt{26}=0$$

which is the equation of line.

Hence, point P must move on a line.

21. Find the equation of a line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Sol. Equation of given lines are

9x + 6y –7 = 0

$$\text{or}\space 3x+2y-\frac{7}{3}=0\space\text{...(i)}$$

and 3x + 2y + 6 = 0 …(ii)

Equation of line parallel to either equation (i) or equation (ii) is

3x + 2y + k = 0 …(iii)

Now, given line (iii) is equidistant from lines (i) and (ii).

i.e., Distance between lines (i) and (iii) = Distance between lines (ii) and (iii), d1 = d2.

$$\therefore\space\begin{vmatrix}\frac{k+\frac{7}{3}}{\sqrt{3^2+2^2}}\end{vmatrix}=\begin{vmatrix}\frac{k-6}{\sqrt{3^2}+\sqrt{2^2}}\end{vmatrix}\\\begin{pmatrix}\because\space\text{Distance between parallel lines is}\space\\\bigg|\frac{c-d}{\sqrt{a^2+b^2}}\bigg|\end{pmatrix}\\\Rarr\space\bigg|k+\frac{7}{3}\bigg|=|k-6|\\\Rarr\space k+\frac{7}{3}=\pm(k-6)$$

Taking negative signs,

$$\Rarr\space k+\frac{7}{3}=-k+6\\\Rarr\space 2k=6-\frac{7}{3}\\\Rarr\space 2k=\frac{11}{3}\\\Rarr\space k=\frac{11}{6}\\\text{Hence, equation (iii) becomes}\\\text{3x+2y}+\frac{11}{6}=0\\\bigg(\text{Put}\space k=\frac{11}{6}\bigg)$$

∴ 18x + 12y + 11 = 0 is the required line.

22. A ray of light passing through the point (1, 2) reflects on the X-axis at point A and the reflected ray passes through the point (5, 3). Find coordinates of A.

Sol. In a figure PA is the incident ray and AR is the reflected ray, which makes an angle q from the X-axis.

It is clear from the figure that

AS ⊥ OX

It means AS bisect the angle PAR.

Then,  ∠PAS = ∠RAS

⇒ ∠RAX = ∠PAQ = θ (Let)

⇒ ∠XAP = 180 – θ

Slope of AR = tan (180° – θ)

$$=\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{5-k}\space\text{...(i)}$$

[where point A is (k, 0)]

∴ From equations (i) and (ii),

$$\frac{3}{5-k}=-\frac{2}{1-k}$$

⇒ 3 – 3k = – 10 + 2k

⇒ 5k = 13

$$\Rarr\space k=\frac{13}{5}$$

$$\text{Hence, the coordinate of A is}\\\bigg(\frac{13}{5},0\bigg).$$

23. Prove that the product of the lengths of the perpendicular drawn the points

$$\textbf{(}\sqrt{\textbf{a}^\textbf{2}-\textbf{b}^\textbf{2}}\textbf{,0}\textbf{)}\space\textbf{and}\space\textbf{(}-\sqrt{\textbf{a}^\textbf{2}\textbf{-b}^\textbf{2}\textbf{,0}}\textbf{)}\\\textbf{to the line}\\\frac{\textbf{x}}{\textbf{a}}\textbf{cos}\space\theta+\frac{\textbf{y}}{\textbf{b}}\textbf{sin}\space\theta=1\textbf{ is b}^{\textbf{2}}.$$

Sol. The equation of given line is

$$\frac{x}{a}\text{cos}\theta+\frac{y}{b}\text{sin}\space\theta =1\\\Rarr\space\frac{bx\space cos\theta + ay\space sin\theta}{ab}=1$$

⇒ bx cos θ + ay sin θ = ab

⇒ bx cos θ + ay sin θ – ab = 0 …(i)

$$\text{At point}(\sqrt{a^2-b^2},0)\\\text{the perpendicular distance from line (i)}\\\text{p}=\bigg|\frac{Ax_2+By_1+C}{\sqrt{A^2+B^2}}\bigg|\\p_1=\bigg|\frac{b\sqrt{a^2-b^2}\space cos \theta-0-ab}{\sqrt{b^2\space cos^{2}\theta+ a^2\space sin^2\theta}}\bigg|$$

$$(\because\space \text{A=b cos}\theta, B=a sin\space\theta, C=-ab;\\x_1=\sqrt{a^2-b^2}, y_1=0)$$

$$\text{and point}\space (-\sqrt{a^2-b^2},0)\\\text{the perpendicular distance from line (i),}\\\text{p}_2=\bigg|\frac{-b\sqrt{a^2-b^2} \text{cos}\theta -ab}{\sqrt{b^2 \text{cos}^2\theta+a^2\text{sin}^2\theta}}\bigg|\\\begin{pmatrix}\because\space \text{A = b cos}\theta, \text{B=a sin}\theta\\\text{C}=-ab,x_1=-\sqrt{a^2-b^2},\\y_1=0\end{pmatrix}$$

On multiplying p1 and p2, we get

$$\therefore\space p_1p_2=\bigg|\frac{b\sqrt{a^2-b^2}\text{cos}\space\theta-ab}{\sqrt{b^2}\text{cos}^2\theta + a^2\space sin^2\theta}\bigg|×\\\bigg|\frac{-b\sqrt{a^2-b^2}\text{cos}\theta-ab}{\sqrt{b^2\space \text{cos}^2\theta + a^2\space\text{sin}^2\theta}}\bigg|$$

$$=\frac{b|\sqrt{a^2-b^2}\text{cos}\space\theta-a|×b|\sqrt{a^2-b^2}\space\text{cos}\space\theta+a|}{(b^2 \text{cos}^2\theta+a^2\text{sin}^2\theta)}$$

$$=\frac{b^2|(a^2-b^2)\space\text{cos}^2\theta-a^2|}{(b^2\space \text{cos}^2\theta+a^2\space\text{sin}^2\theta)}\\(\because\space\text{(A+B)(A-B)}=\text{A}^2-\text{B}^2)\\=\frac{b^2|a^2\space\text{cos}^2\theta-b^2\space\text{cos}^2\theta-a^2|}{(b^2\space \text{cos}^2\theta+a^2\text{sin}^2\theta)}\\=\frac{b^2|a^2(\text{cos}^2\theta-1)-b^2\space\text{cos}^2\theta|}{(b^2\text{cos}^2\theta+a^2\space \text{sin}^2\theta)}$$

$$\Rarr\space\text{P}_1\text{P}_2=\frac{b^2|-a^2\space\text{sin}^2\theta-b^2\space\text{cos}^2\theta|}{(b^2\text{cos}^2\theta+a^2\space\text{sin}^2\theta)}\\(\because\space\text{sin}^2\theta + \text{cos}^2\theta=1)\\=\frac{b^2|a^2\space \text{sin}^2\theta+b^2\space \text{cos}^2\theta|}{(b^2\space\text{cos}^2\theta+a^2\space\text{sin}^2\theta)}$$

(∵ |– A| = |A|)

∴ p1p2 = b2.   Hence Proved.

24. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x –7y + 8 = 0 in the least time. Find equation of the path that he should follow.

Sol. Given equation of straight path is

2x – 3y + 4 = 0 …(i)

and

3x + 4y – 5 = 0 …(ii)

On solving equations (i) and (ii) for crossing point.

From equation (i),

2x = – 4 + 3y

$$\Rarr\space x=\frac{-4+3y}{2}\\\text{Put}\space x=\frac{-4+3y}{2}\space\text{in equation (ii),}\\3\bigg(\frac{-4+3y}{2}\bigg)+4y-5=0$$

⇒ – 12 + 9y + 8y – 10 = 0

⇒ 17y – 22 = 0

$$\Rarr\space y=\frac{22}{17}$$

$$\text{Put}\space y=\frac{22}{7}\space\text{in equation (i), we get}\\2x-3\bigg(\frac{22}{17}\bigg)+4=0\\\Rarr\space 2x=-4+\frac{66}{17}\\\Rarr\space 2x=\frac{-68+66}{17}\\\Rarr\space x=-\frac{1}{17}\\\therefore\space x=-\frac{1}{17}, y=\frac{22}{17}$$

∵ Line PQ ⊥ line AB

then find the equation of PQ.

∴ Slope of PQ × Slope of AB = – 1

$$\Rarr\space m×\bigg(\frac{6}{-7}\bigg)=-1\\\begin{pmatrix}\because\space\text{Slope of line is 6x-7y+8=0}\\\text{is}\bigg(\frac{6}{-7}\bigg)\end{pmatrix}\\\Rarr\space m×\frac{6}{7}=-1\\\Rarr\space m=-\frac{7}{6}$$

The equation of line PQ by using

y – y1 = m(x – x1)

$$\text{and passing through}\\\bigg(-\frac{1}{17},\frac{22}{17}\bigg)\space\text{and m}=-\frac{7}{6}\space\text{is}\\y-\frac{22}{7}=-\frac{7}{6}\bigg(x+\frac{1}{17}\bigg)\\\Rarr\space \frac{17y-22}{17}=-\frac{7}{6}\bigg(\frac{17x+1}{17}\bigg)$$

⇒ 6 × 17y – 22 × 6 = –7 × 17x – 7

⇒ 102y – 132 = – 119x – 7

⇒ 119x + 102y – 132 + 7 = 0

⇒ 119x + 102y – 125 = 0.