NCERT Solutions for Class 11 Maths Chapter 9 - Straight Lines

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.

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.

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).

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

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

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.