# NCERT Solutions for Class 10 Maths Chapter 7 - Coordinate Geometry

Important Points

1. Coordinate Geometry : It is one of the branch of geometry where the position of a point is defined using coordinates.

2. Coordinate Axes : Let be two lines XOX′ and YOY′ are intersect each other at a point O.

XOX′ line is called X-axis or abscissa.

YOY′ line is called Y-axis or ordinate. Coordinate plane divided into four quadrant :

(i) Quadrant-I : (x, y) (XOY plane)

(ii) Quadrant-II : (– x, y) (X′OY plane)

(iii) Quadrant-III : (–x, –y) (X′OY′ plane)

(iv) Quadrant-IV : (x, –y) (XOY′ plane)

Note : (i) y = 0, on x-axis

(ii) x = 0, on y-axis

3. Distance between Two Points : The distance between two points A(x1, y1) and B(x2, y2) is given by

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

4. Some Useful Results

(i) A triangle is equilateral iff its all sides are equal.

(ii) A triangle is isosceles iff its two sides are equal.

(iii) A quadrilateral is parallelogram iff opposite sides are equal and diagonals are not equal.

(iv) A quadrilateral is a rectangle iff opposite sides are equal and diagonals are equal.

(v) A quadrilateral is a rhombus iff all the four sides are equal and diagonals are not equal.

(vi) A quadrilateral is a square iff all the four sides are equal and the diagonals are equal.

(vii) In parallelogram, rectangle, square and rhombus all have diagonals bisecting each other.

5. Section Formulae

(i) The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) internally, in the ratio m : n are :

$$x=\frac{mx_2+nx_1}{m+n},y=\frac{my_2+ny_1}{m+n}$$

(ii) Divides externally in ratio m : n

$$x=\frac{mx_2-nx_1}{m-n},y=\frac{my_2-ny_1}{m-n}$$

(iii) If G be the centroid of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) then

$$\text{G}=\bigg(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\bigg)$$

(iv) If P is the mid-point of A and B, then

$$\text{P}=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)$$

6. Area of a Triangle : If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a ΔABC, then the area of ΔABC is given by the following formula $$\text{Area of}\space\Delta\text{ABC}=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$$

Note : Area of triangle is taken always positive.

7. Condition of Collinearity of Three Points : Three points A(x1, y1), B(x2, y2)and C(x3, y3) will be collinear if and only if area of ΔABC = 0.

$$\text{i.e.,}\space\frac{1}{2}[x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)]=0\\\text{i.e}[x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)]=0$$

Exercise 7.1

1. Find the distance between the following pairs of points

(i) (2, 3), (4, 1)

(ii) (– 5, 7), (– 1, 3)

(iii) (a, b), (– a, – b)

Sol. (i) Let A(2, 3) and B(4, 1) be the given points.

Here, x1 = 2, y1 = 3 and x2 = 4, y2 = 1

$$\therefore\space\text{AB}=\sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}\\=\sqrt{(4-2)^{2} + (1-3)^{2}}\\=\sqrt{(2)^{2}+(-2)^{2}}\\=\sqrt{4+4}\\=\sqrt{8}\space=2\sqrt{2}$$

(ii) Let A(– 5, 7) and B(– 1, 3) be the given points.

Here, x1 = – 5, y1 = 7 and x2 = – 1, y2 = 3

$$\therefore\space\text{AB}=\sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}\\=\sqrt{(-1+5)^{2} + (3-7)^{2}}\\=\sqrt{(4)^{2} + (\normalsize-4)^{2}}\\=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$$

(iii) Let A(a, b) and B(– a, – b) be the given points.

Here, x1 = a, y1 = b and x2 = – a, y2 = – b

$$\therefore\space\text{AB}=\sqrt{(x_2-x_1)^{2}+ (y_2-y_1)^{2}}\\=\sqrt{(-a-a)^{2} + (-b-b)^{2}}\\=\sqrt{(-2a)^{2} + (-2b)^{2}}\\=\sqrt{4a^{2}+4b^{2}}\\=2\sqrt{a^{2} + b^{2} }$$

2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B in given figure. Sol. Let M(0, 0) and N(36, 15) be the given points.

Here, x1 = 0, y1 = 0 and x2 = 36, y2 = 15

$$\therefore\space\text{MV}=\sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}\\=\sqrt{(36-0)^{2} + (15-0)^{2}}\\\sqrt{1296 + 225}\\=\sqrt{1521}=39$$

Since, the position of towns A and B are given (0, 0) and (36, 15), respectively and so, the distance between them is 39 km.

3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Sol. Let A(1, 5), B(2, 3) and C(–2, –1) be the given points.

For AB

x1 = 1, y1 = 5, x2 = 2, y2 = 3

Then,

$$\text{AB}=\sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}\\=\sqrt{(2-1)^{2}+(3-5)^{2}}\\=\sqrt{1+4}\\=\sqrt{5}\\\text{For BC}\\\text{x}_1 = 2, y_1 = 3, x_2 = – 2, y_2 = – 11\\\text{BC}=\sqrt{(-2-2)^{2} + (-11-3)^{2}}\\=\sqrt{16+196}\\=\sqrt{212}\\=2\sqrt{53}$$

For AC

x1 = 1, y1 = 5, x2 = –2, y2 = –11

$$\text{AC}=\sqrt{(-2-1)^{2}+(-11-5)^{2}}\\=\sqrt{9+256}\\=\sqrt{265}\\\text{Now AB + AC =}\sqrt{5}+\sqrt{265}\\\text{and \space BC}=\sqrt{212}$$

AB + AC ≠ BC

Similarly,

BC + AC ≠ AB

AB + BC ≠ AC

Hence, points are not collinear.

4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Sol. Let P(5, – 2), Q(6, 4) and R(7, – 2) are the given points.

$$\text{Then, PQ}=\sqrt{(6-5)^{2} + (4+2)^{2}}\\=\sqrt{36+0}=6\\\text{and}\space \text{BD}=\sqrt{(6-6)^{2} + (1-7)^{2}}\\=\sqrt{0+36}=\sqrt{36}=6$$

$$\text{Then, PQ}=\sqrt{(6-5)^{2} + (4+2)^{2}}\\=\sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}\\=\sqrt{1+36}=\sqrt{37}\\\text{QR}=\sqrt{(7-6)^{2} + (-2-4)^{4}}\\=\sqrt{1+36}=\sqrt{37}$$

Since, PQ = QR

∴ ΔPQR is an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A,B, C and D is shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, ‘‘Don’t you think ABCD is a square?’’ Chameli disagrees. Using distance formula, find which of them is correct. Sol. From the given figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4), and (6, 1).

$$\therefore\space\text{AB}=\sqrt{(6-3)^{2} + (7-4)^{2}}\\\lbrack \because\space\text{Distance} = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}\rbrack\\=\sqrt{3^{2} + 3^{2}}\\\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\\\text{BC}=\sqrt{(9-6)^{2} + (4-7)^{2}}\\=\sqrt{3^{2} + 3^{2}}\\\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\\\text{CD}=\sqrt{(6-9)^{2} + (1-4)^{2}}\\=\sqrt{(-3)^{2} + (-3)^{2}}\\\sqrt{9+9}=\sqrt{18}=3\sqrt{2}\\\text{and DA =}\sqrt{(3-6)^{2} + (4-1)^{2}}\\\sqrt{(-3)^{2}+3^{2}}\\\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$$

∴ AB = BC = CD = DA

$$\text{Now ,}\space \text{AC}=\sqrt{(9-3)^{2} + (4-4)^{2}}\\=\sqrt{36+0}=6\\\text{and BD =}\sqrt{(6-6)^{2} + (1-7)^{2}}\\=\sqrt{0+36}=\sqrt{36}=6$$

∴ AC = BD = 6

Since, the four sides and diagonals are equal. Hence, ABCD is a square. So, Champa is correct. Play Video about Chapter 7 Coordinate Geometry in One Shot

6. Name the type of quadrilateral formed, if any, by the following points and give reasons for your answer.

(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)

(ii) (– 3, 5), (3, 1), (0, 3), (– 1, – 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Sol. (i) Let A(– 1, – 2), B(1, 0), C(– 1, 2) and D(– 3, 0) be the given points.

$$\text{Then,\space AB}=\sqrt{(1+1)^{2} + (0+2)^{2}}\\=\sqrt{2^{2}+2^{2}}\\=[\because\text{Disance} = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}]\\=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\text{BC}=\sqrt{(-1-1)^{2} + (2-0)^{2}}\\=\sqrt{(-2)^{2} + 2^{2}}\\=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\text{BC}=\sqrt{(-1-1)^{2} + (2-0)^{2}}\\=\sqrt{(-2)^{2} +(-2)^{2}}\\=\sqrt{4+4}\\=\sqrt{8}=2\sqrt{2}\\\text{CD}=\sqrt{(-3+1)^{2} +(0-2)^{2}}\\=\sqrt{(2)^{2} + (\normalsize-2)^{2}}\\=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$$

$$\text{DA}=\sqrt{(-1+3)^{2} + (-2-0)^{2}}\\=\sqrt{(2)^{2} + (\normalsize-2)^{2}}\\=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$$

∴ AB = BC = CD = DA

$$\therefore\space\text{AB = BC = CD = DA}\\\text{Now, AC =}\sqrt{(-1+1)^{2} + (2+2)^{2}}\\=\sqrt{0+4^{2}}=4\\\text{and \space BD}=\sqrt{(-3-1)^{2} + (0-0)^{2}}\\=\sqrt{(-4)^{2}+0}=4\\\therefore\space\text{AC = BD}$$

Since, the four sides AB, BC, CD and DA are equal and also diagonals AC and BD are equal.

∴ The quadrilateral ABCD is a square.

(ii) Let A(– 3, 5), B(3, 1), C(0, 3) and D(– 1, – 4) be the given ponts.

From the figure, the points A, B and C are collinear. So, no quadrilateral is formed by given points. (iii) Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2) be the given points

$$\text{Then, AB =}\space\sqrt{(7-4)^{2} + (6-5)^{2}}\\=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10}\\\lbrack\because\space\text{Distance =}\sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}} \rbrack\\\text{BC}=\sqrt{(4-7)^{2} + (3-6)^{2}}\\=\sqrt{(-3)^{2} + (-3)^{2}}\\=\sqrt{9+9}\\=\sqrt{18}\\\text{CD}=\sqrt{(1-4)^{2}+(2-3)^{2}}\\=\sqrt{(-3)^{2} + (-1)^{2}}\\=\sqrt{9+1}\\=\sqrt{10}\\\text{DA}=\sqrt{(4-1)^{2}+(5-2)^{2}}\\=\sqrt{3^{2} + (-3)^{2}}\\=\sqrt{9+9}=\sqrt{18}\\\text{AC}=\sqrt{(4-4)^{2} + (3-5)^{2}}\\\sqrt{0+(-2)^{2}}=-2$$

$$\text{and BD =}\sqrt{(1-7)^{2} + (2-6)^{2}}\\\sqrt{(-6)^{4} + (-4)^{2}}\\=\sqrt{36+16}=\sqrt{52}$$

Since, AB = CD, BC = DA

and AC ≠ BD

∴ The quadrilateral ABCD is a parallelogram.

7. Find the point on the X-axis which is equidistant from (2, – 5) and (– 2, 9).

Sol. Let be point Q on X-axis is P(x, 0)

and given points are A(2, – 5) and B(– 2, 9)

According to question

PA = PB

Squaring both sides,

PA2 = PB2 ...(i)

For PA

x1 = x, y1 = 0, x2 = 2, y2 = 5

$$\text{PA}=\sqrt{(2-x)^{2} + (-5-0)^{2}}\\\text{PA}=\sqrt{(2-x)^{2}+25}$$

Squaring both sides,

PA2 = (2 – x)2 + 25

For PB

x1 = x, y1 = 0, x2 = – 2, y2 = 9

$$\text{PB}=\sqrt{(-2-x)^{2} + (9-0)^{2}}$$

Squaring both sides,

PB2 = (– 2 – x)2 + 81

Put the value PA2 and PB2 in eq. (i)

(2 – x)2 + 25 = (– 2 – x)2 + 81

⇒ 4 + x2 – 4x – 4 – x2 – 4x = 81 – 25

⇒ – 8x = 56

⇒ x = – 7

So, the point equidistant from given points on the X-axis is (– 7, 0).

8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Sol. Given,

P(2, –3) and Q(10, y)

According to question

Then, PQ = 10 ...(i)

x1 = 2, y1 = –3, x2 = 10, y2 = y

$$\text{PQ}=\sqrt{(10-2)^{2} + (y+3)^{2}}$$

Put the value PQ in equation (i)

$$\sqrt{8^{2} + (y+3)^{2}}=10$$

Squaring both sides,

⇒ 64 + (y + 3)2 = 100

⇒ (y + 3)2 = 100 – 64

⇒ (y + 3)2 = 36

⇒ y + 3 = ± 6

Taking (+)

⇒ y = 6 – 3 = 3

Taking (–)

⇒ y = – 6 – 3

⇒ y = – 9

Hence, y = 3, – 9.

9. If Q(0, 1) is equidistant from P(5, – 3) and R(x, 6), find the values of x. Also, find the distance QR and PR.

Sol. Since, the point Q(0, 1) is equidistant from
P(5, – 3) and R(x, 6).

∴ QP = QR

⇒ QP2 = QR2

⇒ (5 – 0)2 + (– 3 – 1)2 = (x – 0)2 + (6 – 1)2

$$[\because\space\text{Distance}=\sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}]$$

⇒ 52 + 42 = x2 + 52

⇒ 25 + 16 = x2 + 25

⇒ x2 = 16

⇒ x = ± 4

Thus, R is (4, 6) or (– 4, 6)

Now, QR = Distance between Q(0, 1) and R(4, 6)

$$=\sqrt{(4-0)^{2} + (6-1)^{2}}\\=\sqrt{4^{2} + 5^{2}}\\=\sqrt{16+25}=\sqrt{41}$$

Also, QR = Distance between Q(0, 1) andR(– 4, 6)

$$=\sqrt{(4-5)^{2} + (6+3)^{2}}$$

$$=\sqrt{(-1)^{2} + 9^{2}}\\=\sqrt{1+81}=\sqrt{82}$$

Also, PR = Distance between P(5, – 3) and R(– 4, 6)

$$=\sqrt{(-4-5)^{2} + (6+3)^{2}}\\=\sqrt{(-9)^{2} + 9^{2}}=9\sqrt{1+1}\\9\sqrt{2}$$

10. Find a relation between x and y such that the ponit (x, y) is equidistant from the points
(3, 6) and (– 3, 4).

Sol. Let the point A(x, y) be equidistant from the points B(3, 6) and C(– 3, 4).

∴ AB = AC

⇒ AB2 = AC2

⇒ (x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2

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

⇒ x2 – 6x + 9 + y2 – 12y + 36

= x2 + 6x + 9 + y2 – 8y +16

⇒ – 6x – 6x – 12y + 8y +36 – 16 = 0

⇒ – 12x – 4y + 20 = 0

⇒ – 4(3x + y – 5) = 0

⇒ 3x + y – 5 = 0 (Q – 4 ≠ 0)

Exercise 7.2

1. Find the coordinates of the point which divides the join of (– 1, 7) and (4, – 3) in the ratio 2 : 3.

Sol. Let P(x, y) be the required point.

Then given,

x1 = – 1, y1 = 7,

x2 = 4, y2 = –3

and m = 2, n = 3

Using the section formula

$$\text{P}\bigg(\frac{mx_2+nx_1}{m+n},\frac{my_2+my_1}{m+n}\bigg)\\\text{P}\bigg(\frac{2×4+3×(\normalsize-1)}{2+3},\frac{2×(\normalsize-3)+3×7}{2+3}\bigg)\\\text{P}\bigg(\frac{8-3}{5},\frac{-6+21}{5}\bigg)\\\text{P}\bigg(\frac{5}{5},\frac{15}{5}\bigg)$$

P(1, 3)

Hence, coordinates of the points are (1, 3).

2. Find the coordinates of the points of trisection of the line segment joining (4, – 1) and (– 2, – 3).

Sol. Let A(4, – 1) and B(– 2, – 3) be the line segments and points of trisection of the line segment be P and Q. Then,

AP = PQ = BQ = k (Say)

∴ PB = PQ + QB = 2k

and AQ = AP + PQ = 2k ⇒ AP : PB = k : 2k = 1 : 2

and AQ : QB = 2k : k = 2 : 1

Since, P divides AB internally in he ratio 1: 2. So, the coordinates of P are

$$\bigg(\text{By using}\space\frac{m_1x_2+m_2x_1}{m_1+m_2}\space\text{and}\space\frac{m_1y_2+ m_2y_1}{m_1+m_2}\bigg)\\=\bigg(\frac{1×(\normalsize-2)+2×4}{1+2},\frac{1×(\normalsize-3)+2×(\normalsize-1)}{1+2}\bigg)\\=\bigg(\frac{-2+8}{3},\frac{-3-2}{3}\bigg)\\=\bigg(\frac{6}{3},\frac{\normalsize-5}{3}\bigg)=\bigg(2,-\frac{5}{3}\bigg)$$

and Q divides AB internally in the ratio 2 : 1.

So, the coordinates of Q are

$$=\bigg(\frac{2×(-2)+1×4}{2+1},\frac{2×(-3)+1×(\normalsize-1)}{2+1}\bigg)\\\bigg(\text{By using}\frac{m_1x_2+m_2x_1}{m_1+m_2}\space\text{and}\space\frac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)\\=\bigg(\frac{-4+4}{3},\frac{-6-1}{3}\bigg)\\=\bigg(0,-\frac{7}{3}\bigg)\\\text{So, the two points of trisection are}\bigg(2,-\frac{5}{3}\bigg)\text{and}\bigg(0,-\frac{7}{3}\bigg)$$

3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in figure. Niharika runs$$\frac{1}{4}\text{th}$$ the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag? Sol. From the above figure, the position of green flag posted by Niharika is

$$\text{M}\bigg(2,\frac{1}{4}×100\bigg)\space\text{i.e.,}\\\text{M (2, 25) and red flag posted by Preet is}\space\text{N}\bigg(8,\frac{1}{5}×100\bigg)\space\text{i.e.}\text{N(8,20).}\\\text{Now,}\space\text{MN}=\sqrt{(6)^{2}+(\normalsize-5)^{2}}\\=\sqrt{(6)^{2}+(-5)^{2}}\\=\sqrt{36+25}=\sqrt{61}$$

Let P be the position of the blue flag posted by Rashmi in the halfwayof line segment MN. $$\therefore\space\text{P is given by =}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\=\bigg(\frac{2+8}{2},\frac{25+80}{2}\bigg)\\=\bigg(\frac{10}{2},\frac{45}{2}\bigg)=(5,22.5)$$

Hence, the blue flag is on the fifth line at a distance = 22.5 m above it.

4. Find the ratio in which the line segment joining the ponts (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Sol. Let (– 1, 6) divide the line joining A(– 3, 10) and B(6, – 8) internally in the ratio m : n.

Then,

$$(\normalsize-1,6)=\bigg(\frac{6m-3n}{m+n},\frac{-8m+10n}{m+n}\bigg)\\-1=\frac{6m-3n}{m+n}$$

– m – n = 6m – 3n

– m – 6m = – 3n + n

– 7m = – 2n

7m = 2n

$$\frac{m}{n}=\frac{2}{7}$$

m : n = 2 : 7

So, the point (– 1, 6), divide the line joining AB in the ratio 2 : 7.

5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the X-axis. Also, find the coordinates of the point of division.

Sol. Let the ratio λ : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio λ : 1 are

$$\bigg(\frac{-4\lambda+1}{\lambda+1},\frac{5\lambda+5}{\lambda+1}\bigg)\space\text{...(i)}$$

This points lies on the X-axis.

So, y-coordinate must be zero.

$$\frac{5\lambda-5}{\lambda+1}=0$$

⇒ 5λ – 5 = 0

⇒ 5λ = 5

⇒ λ = 1

⇒ λ : 1 = 1 : 1

Put the value λ in equation (i),

Then,

$$\text{Coordinates are}\bigg(\frac{-4+1}{1+1},\frac{5-5}{1+1}\bigg)\\\text{Coordinates are}\bigg(\frac{-3}{2},0\bigg)\\\text{The ratio is 1 : 1 and the point of intersection as}\bigg(-\frac{3}{2},0\bigg)$$

6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Sol. Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram. Since, ABCD is a parallelogram.

∴ AC and BD will bisect each other. Hence, mid-point of AC and mid-point of BD are same point.

$$∴ \text{Mid-point of AC is}\bigg(\frac{1+x}{2},\frac{2+6}{2}\bigg)=\text{Mid-point of BD is}\space\bigg(\frac{4+3}{2},\frac{y+5}{2}\bigg)\\\bigg[\because\text{Mid-point}=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\therefore\frac{1+x}{2}=\frac{4+3}{2}\text{and}\frac{2+6}{2}=\frac{5+y}{2}$$

⇒ 1 + x = 7 and 8 = 5 + y

∴ x = 6 and y = 3

7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Sol. Let be coordinates of A(x, y).

AB be a diameter of the circle and C is centre of circle.

We know that centre of circle is a mid-point of diameter. ∴ C is the mid-point of AB.

Then,

x1 = x, y1 = y, x2 = 1, y2 = 4

$$\text{Coordinates of C =}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\(2,-3)=\bigg(\frac{x+1}{2},\frac{y+4}{2}\bigg)\\\text{Then,}\\2=\frac{x+1}{2},x=3\\-3=\frac{y+4}{2},y=-10$$

So, the required coordinates of A are (3, – 10).

8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that $$\textbf{AP}=\frac{\textbf{3}}{\textbf{7}}\textbf{AB}$$and P lies on the line segment AB.

Sol. According to the question, $$\text{AP}=\frac{\text{3}}{\text{7}}\text{AB}\\\Rarr\space\frac{\text{AB}}{\text{AP}}=\frac{7}{3}\\\Rarr\frac{\text{AP+PB}}{\text{AP}}=\frac{3+4}{3}\\\Rarr1+\frac{\text{PB}}{\text{AP}}=1+\frac{4}{3}\\\Rarr\space\frac{\text{PB}}{\text{AP}}=\frac{4}{3}\\\Rarr\space\frac{\text{AP}}{\text{PB}}=\frac{3}{4}\\\Rarr\space\text{AP : PB = 3 : 4}$$

Suppose, P(x, y) be the point which divides the line segment joining the points A(– 2, – 2) and B(2, – 4) in the ratio 3 : 4

$$\therefore\space x=\frac{3×2+4×(\normalsize-2)}{3+4}\\=\frac{6-8}{7}=-\frac{2}{7}\space\bigg(\because x=\frac{mx_2+nx_1}{m+n}\bigg)\\\text{and}\space y=\frac{3×(\normalsize-4)+4×(\normalsize-2)}{3+4}\\\bigg(\because y=\frac{my_2+ny_1}{m+n}\bigg)\\=\frac{-12-8}{7}\\=-\frac{20}{7}\\\text{Hence, the required coordinates of the point P are}\bigg(-\frac{2}{7},-\frac{20}{7}\bigg).$$

9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.

Sol. Let P, Q and R be the points of line segment AB such that

AP = PQ = QR = RB

Let AP = PQ = QR = RB = k $$\text{Now,}\space\frac{\text{AP}}{\text{PB}}=\frac{k}{3k}=\frac{1}{3}\\\text{Therefore, P divides AB internally in the ratio 1 : 3.}\\\because\space\text{Internally ratio}\\=\bigg(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}\bigg)\\\text{P}=\bigg(\frac{1×2+3(\normalsize-2)}{1+3},\frac{1×8+3×2}{1+3}\bigg)\\=\bigg(\frac{2-6}{4},\frac{8+6}{4}\bigg)\\=\bigg(\frac{\normalsize-4}{4},\frac{14}{4}\bigg)=\bigg(\normalsize-1,\frac{7}{2}\bigg)\\\text{Again,}\frac{\text{AR}}{\text{RB}}=\frac{3k}{k}=\frac{3}{1}$$

Therefore, R divides AB internally in the ratio 3 : 1.

$$\text{R}=\bigg(\frac{3×2+1×(\normalsize-2)}{3+1},\frac{3×8+1×2}{3+1}\bigg)\\=\bigg(\frac{6-2}{4},\frac{24+2}{4}\bigg)\\=\bigg(\frac{4}{4},\frac{26}{4}\bigg)=\bigg(1,\frac{13}{2}\bigg)\\\text{Also},\frac{\text{AQ}}{\text{QB}}=\frac{2k}{2k}=\frac{1}{1}\\{\therefore\text{Q is the mid-point of AB.}}\\\therefore\text{Q}=\bigg(\frac{-2+2}{2},\frac{2+8}{2}\bigg)\text{i.e., \text{Q}=}\bigg(\frac{0}{2},\frac{10}{2}\bigg)=(0,5)\\\bigg[\because\space\text{Mid-point}=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\text{So, required points are}\bigg(\normalsize-1,\frac{7}{2}\bigg),\bigg(1,\frac{13}{2}\bigg)\text{and}(0,5).$$

10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, 1) taken in order.

$$\text{[Hint : Area of rhombus =}\frac{1}{2}\space(\text{Product of its diagonals})]$$

Sol. Let A(3, 0), B(4, 5), C(– 1, 4) and D(– 2, – 1) be the vertices of the rhombus ABCD.

$$\therefore\space\text{Diagonal, AC =}\sqrt{(-1-3)^{2}+(4-0)^{2}}\\\lbrack\because\space\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} \rbrack\\=\sqrt{(-4)^{2}+4^{2}}\\=\sqrt{16+15}=\sqrt{32}=4\sqrt{2}\\\text{Diagonal, BD =}\sqrt{(-2-4)^{2}+(-1-5)^{2}}\\=\sqrt{(-6)^{2} + (-6)^{2}}\\=\sqrt{36+6}\\=\sqrt{72}=6\sqrt{2}\\\therefore\text{Area of the rhombus ABCD =}\frac{1}{2}×\text{AC}×\text{BD}\\=\frac{1}{2}×4\sqrt{2}×6\sqrt{2}\\=2×6×\sqrt{2}×\sqrt{2}$$

= 12 × 2

= 24 sq. units

Exercise 7.3

1. Find the area of the triangle whose vertices are

(i) (2, 3), (– 1, 0), (2, – 4)

(ii) (– 5, – 1), (3, – 5), (5, 2)

Sol. (i) Let be ΔABC.

Its vertices A(x1, y1), B(x2, y2) and C(x3, y3).

Then,

x1 = 2, y1 = 3, x2 = – 1, y2 = 0, x3 = 2, y3 = – 4

Area of ΔABC

$$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2}[2(0+4)+(\normalsize-1)(-4-3)+2(3-0)]\\=\frac{1}{2}[8+7+6]\\=\frac{21}{2}\text{sq.units}$$

(ii) Let be a ΔABC

Its vertices are A(x1, y1), B(x2, y2) and C(x3, y3)

Then given,

x1 = – 5, y1 = – 1, x2 = 3, y2 = – 5, x3 = 5, y3 = 2

Area of ΔABC

$$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2[-5(-5-2)+3(2+1)+5(-1+5)]}\\=\frac{1}{2}[35+9+20]\\=\frac{1}{2}[35+9+20]\\=\frac{64}{2}\\=32 \text{square units}$$

2. In each of the following, find the value of k, for which the points are collinear

(i) (7, – 2), (5, 1), (3, k)

(ii) (8, 1, (k, – 4), (2, – 5)

Sol. (i) Let A = (x1, y1) = (7, – 2), B = (x2, y2) = (5, 1) and C = (x3, y3) = (3, k)

Since, the points are collinear.

∴ Area of ΔABC = 0

$$\Rarr\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$$

⇒ 7(1 – k) + 5(k + 2) + 3(– 2 – 1) = 0

(Multiply of 2)

⇒ 7 – 7k + 5k + 10 – 9 = 0

⇒ – 2k + 8 = 0

⇒ 2k = 8

⇒ k = 4

(ii) Let A = (x1, y1) = (8, 1), B = (x2, y2) = (k, – 4) and C = (x3, y3) = (2, – 5)

Since, the points are collinear

∴ Area of ΔABC = 0

$$\Rarr\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0\\\Rarr\space8 (– 4 + 5) + k (– 5 – 1) + 2(1 + 4) = 0\\\text{(Multiply of 2)}\\⇒ 8(1) + k(– 6) + 2(5) = 0\\⇒ 8 – 6k + 10 = 0\\⇒ – 6k = – 18\\\therefore\space k=\frac{18}{6}=3$$

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, – 1), (2, 1) and (0, 3). Find the ratio of this area of the area of the given triangle.

$$\textbf{sol.}\space\therefore\text{Mid-point of AC =}\bigg(\frac{0+0}{2},\frac{3-1}{2}\bigg)\\\bigg[\because\space\text{Mid point =}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\text{= (0, 1) = M}\\\text{Mid-point of AB =}\bigg(\frac{0+2}{2},\frac{-1+1}{2}\bigg)=(1,0)=\text{N}\\\text{and Mid-point of BC =}\bigg(\frac{0+2}{2},\frac{3+1}{2}\bigg)=(1,2)=\text{P}$$

Let N(1, 0) = (x1, y1) and P(1, 2) = (x2, y2) and M(0, 1) = (x3, y3)

$$\therefore\space\text{Area of ΔNPM =}\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0\\=\frac{1}{2}[1(2-1)+1(1-0)+0(0-2)]\\=\frac{1}{2}[1(1)+1+0]\\=\frac{2}{2}=1=1$$ Let A = (x1, y1) = (0, – 1), B = (x2, y2) = (2, 1) and C = (x3, y3) = (0, 3)

∴ Area of ΔABC

$$\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0\\\frac{1}{2}[0(1-3)+2(3-1)+0(-1-1)]\\\frac{1}{2}(0+8+0)\\\text{= 4 sq. units}\\{\therefore\text{Required ratio =}\frac{\text{Area of} \Delta \text{NPM}}{\text{Area of}\Delta \text{ABC}}}=\frac{1}{4}$$

4. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

Sol. Let A(– 4, – 2), B(– 3, – 5), C(3, – 2) and D(2, 3) be the vertices of the quadrilateral ABCD.

∴ Area of quadrilateral ABCD = Area of ΔACD + Area of ΔABC

$$=\frac{1}{2}[-4(-2-3)+3(3+2)+2(-2+2)]+\frac{1}{2}[\normalsize-4(\normalsize-5+2)-3(-2+2)+3(-3+5)]\\\bigg(\because\space\text{Area of triangle =}\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\bigg)\\=\frac{1}{2}[-4(-5)+3(5)+2(0)]+\frac{1}{2}[-4(\normalsize-3)-3(0)+3(3)]\\=\frac{1}{2}(20+15+0)+\frac{1}{2}(12-0+9)\\=\frac{1}{2}(35+21)\\=\frac{1}{2}×56=28\space\text{sq.units}$$

5. A median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).

Sol. According to the question, AD is the median of ΔABC, therefore D is the mid-point of BC. $$\therefore\space\text{Coordinates of D are}\bigg(\frac{3+5}{2},\frac{-2+2}{2}\bigg)\text{i.e.,}(4,0)\\\bigg[\because\space\text{Mid-point=}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\therefore\text{Area of ΔADC =}\frac{1}{2}[4(0-2)+4(2+6)+5(-6-0)]\\\bigg(\because\text{Area of triangle =}\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\bigg)\\\text{Here, A(4, 0) =} (x_1, y_1)\\\text{D(4, 0) =} (x_2, y_2) \text{and} \space C(5, 2) = (x_3, y_3)\\=\frac{1}{2}(-8+32-30)\\\frac{1}{2}×(\normalsize-6)=-3$$

= 3 sq. units

$$(\because\space\text{Area of triangle is positive})\\\text{and Area of ΔABD =}\frac{1}{2}[4(-2-0)+3(0+6)+4(-6+2)]$$

[Let A(4, – 6) = (x1, y1), B(3, – 2) = (x2, y2) and D(4, 0) = (x3, y3)]

$$=\frac{1}{2}(\normalsize-8+18-16)\\=\frac{1}{2}(\normalsize-6)=\normalsize-3$$

= 3 sq. units

(∵ Area of triangle is positive)

∵ Area of ΔADC = Area of ΔABD

Hence, the median of the triangle divides it into two triangle of equal areas.

Exercise 7.3 (Optional)

1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

Sol. Let the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7) in the ratio k : 1 at the point P

$$\therefore\text{The coordinates of P are}\bigg(\frac{3k+2}{k+1},\frac{7k-2}{k+1}\bigg)\\\bigg[\because\text{Internally ratio =}\bigg(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)\bigg]\\\text{But, P lies on 2x + y – 4 = 0}\\\therefore\space 2\bigg(\frac{3k+2}{k+1}\bigg)+\frac{7k-2}{k+1}-4=0$$

⇒ 6k + 4 + 7k – 2 – 4k – 4 = 0

⇒ 9k – 2 = 0

⇒ 9k = 2

$$\Rarr\space k=\frac{2}{9}$$

∴ Point P divides the line in the ratio 2 : 9.

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Sol. Given points A(x, y), B(1, 2) and C(7, 0)

If points are collinear.

Then, ΔABC = 0

x1 = x, y1 = y, x2 = 1, y2 = 2, x3 = 7, y3 = 0

12[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0

⇒ x(2 – 0) + 1(0 – y) + 7(y – 2) = 0

⇒ 2x – y + 7y – 14 = 0

⇒ x + 3y – 7 = 0

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

Sol. Let C(x, y) be the cetnre of the circle passing through the points P(6, – 6), Q(3, – 7) and R(3, 3). Then, PC = QC = CR (Radius of circle)

Now, PC = QC

⇒ PC2 = QC2

⇒ (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2

$$[\because\space\text{Distance}=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}]$$

⇒ x2 – 12x + 36 + y2 + 12y + 36 = x2 – 6x + 9+ y2 + 14y + 49

⇒ – 12x + 6x + 12y – 14y + 72 – 68= 0

⇒ – 6x – 2y + 14 = 0

3x + y – 7 = 0

(Divide by – 2) ...(i)

and QC = CR

⇒ QC2 = CR2

⇒ (x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2

⇒ x2 – 6x + 9 + y2 + 14y + 49 = x2 – 6x + 9 + y2– 6y + 9

⇒ – 6x + 6x + 14y + 6y + 58 – 18 = 0

⇒ 20y + 40 = 0

$$\Rarr\space y=-\frac{40}{20}=-2\space\text{(ii)}$$

Putting y = – 2 in Eq. (i), we get

3x – 2 – 7 = 0

⇒ 3x = 9

⇒ x = 3

Hence, centre is (3, – 2).

4. The two opposite vertices of a square are (– 1, 2) and (3, 2). Find the coordinates of the other two vertices.

Sol. Let PQRM be a square and let P(– 1, 2) and R(3, 2) be the vertices.

Let the coordinates of Q be (x, y). ∵ PQ = QR

⇒ PQ2 = QR2

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

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

⇒ x2 + 1 + 2x + y2 + 4 – 4y = x2 + 9 – 6x+ y2 + 4 – 4y

⇒ 2x + 1 = – 6x + 9

⇒ 8x = 8

⇒ x = 1 ...(i)

In ΔPQR, we have

PQ2 + QR2 = PR2

(x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2 = (3 + 1)2 + (2 – 2)2

⇒ x2 + 1 + 2x + y2 + 4 – 4y + x2 + 9 – 6x + y2 + 4 – 4y = 42 + 02

⇒ 2x2 + 2y2 – 4x – 8y + 2 = 0

⇒ x2 + y2 – 2x – 4y + 1 = 0 ...(ii)

Putting x = 1 from Eq. (i) in Eq. (ii), we get

1 + y2 – 2 – 4y + 1 = 0

⇒ y2 – 4y = 0

⇒ y(y – 4) = 0 ⇒ y = 0 or 4

Hence, the required vertices of square are (1, 0) and (1, 4).

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot. (i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of ΔPQR, if C is the origin?
Also, calculate the areas of the triangles in these cases. What do you observer?

Sol. (i) When A is taken as origin, AD and AB as coordinate axes, i.e., X-axis and Y-axis, respectively. Coordinates of P, Q and R are respectively (4, 6), (3, 2) and (6, 5).

(ii) When C is taken as origin and CB as X-axis and CD as Y-axis.

∴ Coordinates of P, Q and R are respectively (12, 2), (13, 6) and (10, 3).

Condition I : When A is taken as origin

P(4, 6), Q(3, 2) and R(6, 5)

Then,

x1 = 4, y1 = 6, x2 = 3, y2 = 2, x3 = 6, y3 = 5

Area of ΔPQR

$$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2}[4(2-5)+3(5-6)+6(6-2)]\\=\frac{1}{2}[-12-3+24]\\\frac{9}{2}\space\text{sq. units}$$

Condition II : When C is taken as origin.

P(12, 2), Q(13, 6) and R(10, 3)

Area of ΔPQR

$$=\frac{1}{2}[x_1(y_2-y_3 + x_2(y_3-y_1)+x_3(y_1-y_2))]\\=\frac{1}{2}[12(6-3)+13(3-2)+10(2-6)]\\\frac{1}{2}[36+13-40]\\\frac{9}{2}\text{sq.units}$$

Hence, we observe that the areas of triangles of both condition are same.

6. The vertices of ΔABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that$$\frac{\textbf{AD}}{\textbf{AB}}=\frac{\textbf{AE}}{\textbf{AC}}=\frac{1}{4}.$$

Calculate the area of the ΔADE and compare it with the area of ΔABC.

$$\text{Sol. Given,}\space\frac{\text{AD}}{\text{AB}}=\frac{1}{4}\\\Rarr\frac{\text{AB}}{\text{AD}}=\frac{4}{1}\\\Rarr\frac{\text{AD+DB}}{\text{AD}}=\frac{4}{1}\\\Rarr 1+\frac{\text{DB}}{\text{AD}}=\frac{4}{1}\\\Rarr\frac{\text{AD}}{\text{DB}}=\frac{1}{3}$$ Hence, D divides AB internally in the ratio 1 : 3.

∴ The coordinates of D are

$$\bigg(\frac{1×1+3×4}{1+3},\frac{1×5+3×6}{1+3}\bigg)\text{i.e,}\bigg(\frac{1+12}{4},\frac{5+18}{4}\bigg)\text{or}\bigg(\frac{13}{4},\frac{23}{4}\bigg)\\\bigg[\because\space\text{Internally ratio =}\bigg(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)\bigg]\\\text{Again,}\space\frac{\text{AE}}{\text{AC}}=\frac{1}{4}\\\Rarr\space\frac{\text{AC}}{\text{AE}}=\frac{4}{1}\\\Rarr\space\frac{\text{AE+EC}}{\text{AE}}=\frac{4}{1}\\\Rarr\space 1+\frac{\text{EC}}{\text{AE}}=4\\\Rarr\space\frac{\text{AE}}{\text{EC}}=\frac{1}{3}$$

Hence, E divides AC internally in the ratio 1 : 3.

∴ The coordinates of E are

$$\bigg(\frac{1×7+3×4}{1+3},\frac{1×2+3×6}{1+3}\bigg)\text{i.e.,}\bigg(\frac{7+12}{y},\frac{2+18}{4}\bigg)\space\text{or}\bigg(\frac{19}{4},5\bigg).\\\bigg[\because\space\text{Area of}\space\Delta=\frac{1}{2}[4(5-2)+1(2-6)+7(6-5)]\bigg]\\\frac{1}{2}[12-4+7]\\=\frac{15}{2}\text{sq. units}\\\text{For} \Delta \text{ADE}\\x_1=4,y_1=6,x_2=\frac{13}{4},y_2=\frac{23}{4},x_3=\frac{19}{4},y_3=5\\\text{and area of} Δ\text{ADE =}\bigg[\frac{1}{2}4\bigg(\frac{23}{4}-5\bigg)+\frac{13}{4}(5-6)+\frac{19}{4}\bigg(6-\frac{23}{4}\bigg)\bigg]\\=\frac{1}{2}\bigg[4×\frac{3}{4}-\frac{13}{4}+\frac{19}{4}×\frac{1}{4}\bigg]\\=\frac{1}{2}\bigg[3-\frac{13}{4}+\frac{19}{16}\bigg]$$

$$=\frac{1}{2}\bigg[\frac{48-52+19}{6}\bigg]\\=\frac{15}{32}\space\text{sq units}\\\therefore\frac{\text{Area of}\Delta \text{ADE}}{\text{Area of}\Delta \text{ABC}}=\frac{15/32}{15/2}=1:16$$

7. Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC.

(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

(iii) Find the coordinates of points Q and R on medians BE and CF respectively, such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

(iv) What do you observe?
Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.

(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ΔABC, find the coordinates of the centroid of the triangle.

Sol. Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC. (i) Since, AD is the median of ΔABC.

∴ D is the mid-point of BC.

$$\therefore\space\text{The coordinates of D are}\space\bigg(\frac{6+1}{2},\frac{5+4}{2}\bigg)\text{i.e,}\bigg(\frac{7}{2},\frac{9}{2}\bigg)\\\bigg[\because\text{Mid\space point}=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\text{(ii) Since, P divides AD in the ratio 2 : 1, so its coordinates are}\\\bigg(\frac{2×\frac{7}{2}+1×4}{2+1},\frac{2×\frac{9}{2}+1×2}{2+1}\bigg)\\\text{or}\bigg(\frac{7+4}{2},\frac{9+2}{3}\bigg)\text{i.e.,}\bigg(\frac{11}{3},\frac{11}{3}\bigg)\\\bigg[\because\space\text{Internally ratio =}\bigg(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)\bigg]$$

(iii) Since, BE is the median of ΔABC, so E is the mid-point of AC and its coordinates are

$$\text{E}\bigg(\frac{4+1}{2},\frac{2+4}{2}\bigg)\text{i.e.,}\space\text{E}\bigg(\frac{5}{2},3\bigg)$$

Since, Q divides BE in the ratio 2 : 1 so, its coordinates are

$$\text{Q}\bigg(\frac{2×\frac{5}{2}+1×6}{2+1},\frac{2×3+1×5}{2+1}\bigg)\text{or}\\\text{Q}\bigg(\frac{5+6}{3},\frac{6+5}{3}\bigg)\text{or}\space\text{Q}\bigg(\frac{11}{3},\frac{11}{3}\bigg)$$

Since, CF is the median of DABC, so F is the mid-point of AB. Therefore, its coordinates are

$$\text{F}\bigg(\frac{4+6}{2},\frac{2+5}{2}\bigg)\space\text{i.e.,}\text{F}\bigg(5,\frac{7}{2}\bigg)$$

Since, R divides in the ratio 2 : 1, so, its coordinates are

$$\text{R}\bigg(\frac{2×5+1×1}{2+1},\frac{2×\frac{7}{2}+1×4}{2+1}\bigg)\\\text{or}\space\text{R}\bigg(\frac{10+1}{3},\frac{7+4}{3}\bigg)\text{or R}\bigg(\frac{11}{3},\frac{11}{3}\bigg)$$

$$\text{(iv) We find that the points P, Q and R coincide}\\\text{at the point}\bigg(\frac{11}{3},\frac{11}{3}\bigg).\text{This point is known as the centroid of the triangle.}$$

(v) Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of ΔABC whose medians are AD, BE and CD respectively, then D, E and F are respectively the mid-points of BC, CA and AB.

$$\therefore\space\text{Coordinates of D are}\bigg(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\bigg)$$ Coordinates of a point G dividing AD in the ratio 2 : 1 are

$$\bigg(\frac{1(x_1)+2\frac{(x_2+x_3)}{2}}{1+2},\frac{1(y_1)+2+\frac{(y_2+y_3)}{2}}{1+2}\bigg)\\=\bigg(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\bigg)\\\text{The coordinates of E are}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_3}{2}\bigg).$$

The coordinates of a point dividing BE in the ratio 2 : 1 are

$$\bigg(\frac{1(x_2)+2\frac{(x_1+x_3)}{2}}{1+2},\frac{1(y_2)+2\frac{(y_1+y_3)}{2}}{1+2}\bigg)\\=\bigg(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\bigg)$$

8. ABCD is a rectangle formed by the points A(– 1, – 1), B(– 1, 4), C(5, 4) and D(5, – 1), P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Sol. Given, vertices of a rectangle are A(– 1, – 1), B(– 1, 4), C(5, 4) and D(5, – 1).

$$\therefore\space\text{Mid-point of AB is P}\bigg(\frac{-1-1}{2},\frac{-1+4}{2}\bigg)\text{or P}\bigg(\normalsize-1,\frac{3}{2}\bigg)\\\bigg[\because\text{Mid-point =}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\text{Mid-point of BC is Q}\bigg(\frac{5-1}{2},\frac{4+4}{2}\bigg)\text{or}\space\text{Q(2,4).}\\\text{Mid-point of CD is R}\bigg(\frac{5+5}{2},\frac{4-1}{2}\bigg)\text{or R}\bigg(5,\frac{3}{2}\bigg).\\\text{Mid-point of AD is S}\bigg(\frac{-1+5}{2},\frac{-1-1}{2}\bigg)\text{or S(2,1)}\\\text{Now,}\space\text{PQ}=\sqrt{(2+1)^{2}+\bigg(4-\frac{3}{2}\bigg)^{2}}$$ $$=\sqrt{3^{2}+\bigg(\frac{5}{2}\bigg)^{2}}\\=\sqrt{9+\frac{25}{4}}\\=\sqrt{\frac{36+25}{4}}=\sqrt{\frac{61}{4}}\\\text{QR}=\sqrt{(5-2)^{2}+\bigg(\frac{3}{2}-4\bigg)^{2}}\\=\sqrt{3^{2}+\bigg(\frac{5}{2}\bigg)^{2}}=\sqrt{9+\frac{25}{4}}\\=\sqrt{\frac{36+25}{4}}=\sqrt{\frac{61}{4}}\\\text{RS}=\sqrt{(2-5)^{2}+\bigg(-1-\frac{3}{2}\bigg)^{2}}$$

$$=\sqrt{(-3)^{2}+\bigg(\frac{-5}{2}\bigg)^{2}}\\=\sqrt{9+\frac{25}{4}}=\sqrt{\frac{36+25}{4}}\\=\sqrt{\frac{61}{4}}\\\text{SP}=\sqrt{(-1-2)^{2}+\bigg(\frac{3}{2}+1\bigg)^{2}}\\=\sqrt{(-3)^{3}+\bigg(\frac{5}{2}\bigg)^{2}}\\=\sqrt{9+\frac{25}{4}}\\=\sqrt{\frac{36+25}{4}}\\=\sqrt{\frac{61}{4}}$$

∵ PQ = QR = RS = SP

$$\text{Now,}\space\text{PR}=\sqrt{(5+1)^{2}+\bigg(\frac{3}{2}-\frac{3}{2}\bigg)^{2}}\\=\sqrt{6^{2}+0}=6\\\text{and\space SQ}=\sqrt{(2-2)^{2}+(4+1)^{2}}\\=\sqrt{0+5^{2}}=5$$

⇒ PR ≠ SQ

Since, all the sides are equal but diagonals are not equal.

∴ PQRS is a rhombus.

$$=\sqrt{\frac{61}{4}}$$

Selected NCERT Exemplar Problems

Exercise 7.1

• Choose the correct answer from the given four options.

1. The distance of the point P(2, 3) from the x-axis is

(a) 2

(b) 3

(c) 1

(d) 5

Sol. (b) 3

Explanation:

The distance of the poiint P(2, 3) from the x-axis = Ordinate of a point P(2, 3)
= 3.

2. The distance between the points A(0, 6) and B(0, – 2) is

(a) 6

(b) 8

(c) 4

(d) 2

Sol. (b) 8

Explanation:

Here x1 = 0, y1 = 6 and x2 = 0, y2 = – 2

$$\therefore\space d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\\=\sqrt{(0-0)^{2}+(-2-6)^{2}}\\=-\sqrt{(-8)^{2}}\\=8$$

3. The distance of the point P(– 6, 8) from the origin is

(a) 8

$$\textbf{(b)\space 2 }\sqrt{\textbf{7}}$$

(c) 10

(d) 6

Sol. (c) 10

Explanation:

Here x1 = – 6, y1 = 8 and x2 = 0, y2 = 0

$$d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\\=\sqrt{0-(-6)^{2}+(0-8)^{2}}\\=\sqrt{(6)^{2}+(\normalsize-8)^{2}}\\=\sqrt{36+64}\\=\sqrt{100}=10$$

4. The distance between the points (0, 5) and (– 5, 0) is

(a) 5

$$\textbf{(b)\space 5}\sqrt{\textbf{2}}$$

$$\textbf{(c)\space 2}\sqrt{\textbf{5}}$$

(d) 10

$$\textbf{Sol.}\space\text{(c)\space 5}\sqrt{\text{2}}$$

Explanation:

Here x1 = 0, y1 = 5 and x2 = – 5, y2 = 0

$$\therefore\space d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\\=\sqrt{(-5-0)^{2}+(0-5)^{2}}\\=\sqrt{25+25}\\=\sqrt{50}\\=5\sqrt{2}$$

5. AOBS is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 5). The length of its diagonal is

(a) 5

(b) 3

$$\sqrt{34}$$

(d) 4

Sol. (c) $$\sqrt{34}$$

Explanation:

Let the fourth vertex of a rectangle be C(x1, y1) The length of the diagonal AB = Distance between the points A(0, 3) and B(5, 0)

Here x1 = 0, y1 = 3 and x2 = 5, y2 = 0

$$\therefore\space d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\\=\sqrt{(5-0)^{2}+(0-3)^{2}}\\=\sqrt{25+9}\\=\sqrt{34}$$

6. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

(a) 5

(b) 12

(c) 11

$$\textbf{(d)}\space \textbf{7}+\sqrt{\textbf{5}}$$

Sol. (b) 12

Explanation: The perimeter of ΔAOB

= d(AO) + d(OH) + d(AB)

$$=4+3+\sqrt{(4)^{2}+(3)^{2}}\\=7+\sqrt{16+9}\\=7+\sqrt{25}$$

= 7 + 5

= 12

7. The area of a triangle with vertices A(3, 0), B(7, 0) and C(8, 4).

(a) 14

(b) 28

(c) 8

(d) 6

Sol. (c) 8

Explanation:

The area of the ΔABC whose points A = (x1, y1), B = (x2, y2) and C = (x3, y3) are given by area of

$$\Delta\text{ABC}=\bigg|\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\bigg|\\\text{Here, x}_1 = 3, y_1 = 0, x_2 = 7, y_2 = 0, x_3 = 8, y_3 = 4\\=\bigg|\frac{1}{2}[3(0-4)+7(4-0)+8(0-0)]\bigg|\\=\bigg|\frac{1}{2}[-12+28+0]\bigg|=\bigg|\frac{1}{2}(16)\bigg|=8$$

8. The points (– 4, 0), (4, 0), (0, 3) are the vertices of a

(a) right triangle

(b) isosceles triangle

(c) equilateral triangle

(d) scalene triangle

Sol. (b) isosceles triangle

Explanation:

Let A = (– 4, 0), B = (4, 0), C = (0, 3)

$$\text{Now,}\space\text{AB}=\sqrt{(4-(-4)^{2})+(0-0)^{2}}\\=\sqrt{(4+4)^{2}}=\sqrt{8^{2}}=8\\\text{BC =}\sqrt{(0-4)^{2}+(3-0)^{2}}\\=\sqrt{16+9}=\sqrt{25}=5\\\text{AC}=\sqrt{(0-(-4)^{2}+(3-0)^{2})}\\=\sqrt{16+9}=\sqrt{25}=5$$

∵ BC = AC

∴ ΔABC is an isosceles triangle.

9. The point which divides the line segment joining the points (7, – 6) and (3, 4) in ratio 1 : 2 internally lies in the

(c) III quaerant

Explanation:

If P(x, y) divides the line segment joining A(x1, y1) andB(x2, y2), internally in the ration m : n,

$$\text{then}\space x=\frac{mx_2+nx_1}{m+n}\text{and}\space y=\frac{my_2+ny_1}{m+n}$$

Here, x1 = 7, y1 = – 6 and x2 = 3, y2 = 4, m = 1, n = 2

$$\therefore\space x=\frac{1(3)+2(7)}{1+2},y=\frac{1(4)+2(\normalsize-6)}{1+2}\\\Rarr\space x=\frac{3+14}{3}, y=\frac{4-12}{3}\\\Rarr\space x=\frac{17}{3},y=-\frac{8}{3}\\\Rarr\space(x,y)=\bigg(\frac{17}{3},-\frac{8}{3}\bigg)\space\text{lies in IV quadrant.}\\\lbrack\because\space \text{x-coordinate is positive and y-coordinate is negative}\rbrack$$

10. The fourth vertex D of a parallelogram ABCD whose three vertices are A(– 2, 3), B(6, 7) and C(8, 3) is

(a) (0, 1)

(b) (0, – 1)

(c) (– 1, 0)

(d) (1, 0)

Sol. (b) (0, – 1)

Explanation:

Let D ≡ (x4, y4). Let L and M be the middle points of AC and BD, respectively.

$$\text{Then,}\space \text{L}=\bigg(\frac{-2+8}{2},\frac{3+3}{2}\bigg)=(3,3)\\\bigg[\because\space\text{Mid-point =}\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\bigg]\\\text{and M}=\bigg(\frac{6+x_4}{2},\frac{7+y_4}{2}\bigg)$$ Since, ABCD is a parallelogram, therefore diagonsis AC and BD will bisect each other.

Hence, L and M are the same points.

$$\therefore\space 3=\frac{6+x_4}{2}\space\text{and}\space 3=\frac{7+y_4}{2}$$

⇒ 6 = 6 + x4 and 6 = 7 + y4

⇒ x4 = 0 and y4 = 6 – 7

⇒ x4 = 0 and y4 = – 1

So, D(0, – 1)

11. If the point P(2, 1) lies on the line segment joining points A(4, 2)and B(8, 4), then

$$\textbf{(a) AP=}\frac{1}{3}\textbf{AB}$$

(b) AP = PB

$$\textbf{(c) PB=}\frac{1}{3}\textbf{AB}\\\textbf{(d) AP=}\frac{1}{2}\textbf{AB}$$

$$\textbf{Sol.}\space(d)\space\text{AP}=\frac{1}{2}\text{AB}$$

Explanation: $$\text{Now, AP =}\sqrt{(2-4)^{2}+(1-2)^{2}}\\=\sqrt{(-2)^{2}+(-1)^{2}}\\=\sqrt{4+1}=\sqrt{5}\\\text{AB}=\sqrt{(8-4)^{2}+(4-2)^{2}}\\\sqrt{(4)^{2}+(2)^{2}}\\\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\\\text{BP}=\sqrt{(8-2)^{2}+(4-1)^{2}}\\=\sqrt{6^{2}+3^{2}}=\sqrt{36+9}\\=\sqrt{45}=3\sqrt{5}\\\therefore\space\text{AB}=2\sqrt{5}=3\sqrt{5}\\\Rarr\space\text{AP}=\frac{\text{AB}}{2}$$

$$\textbf{12. If P}\bigg(\frac{\textbf{a}}{\textbf{3}},\textbf{4}\bigg)\textbf{is the mid-point of the line segment}\\\textbf{joining the points Q(– 6, 5) and R(– 2, 3),}\\\textbf{then the value of a is}$$

(a) – 4

(b) – 12

(c) 12

(d) – 6

Sol. (b) – 12

Explanation:

$$\therefore\space\text{Mid-point of QR is P}\bigg(\frac{-6-2}{2},\frac{5+3}{2}\bigg)\text{or P(– 4, 4).}$$ $$\text{But, mid-point P}\bigg(\frac{a}{3},4\bigg)\text{is given}\\\therefore\space\bigg(\frac{a}{3},4\bigg)≡(\normalsize-4,4)\\\text{On comparing the coordinates, we get}\\\frac{a}{3}=-4\space\text{and}\space 4=4$$

a = – 12

13. A line intersects the y-axis and x-axis at the ponits P and Q respectively. If (2, – 5) is the mid-point of PQ, then the coordinates of P and Q are, respectively.

(a) (0, – 5) and (2, 0)

(b) (0, 10) and (– 4, 0)

(c) (0, 4) and (– 10, 0)

(d) (0, – 10) and (4, 0)

Sol. (d) (0, – 10) and (4, 0)

Explanation:

Let the coordinates of x-axis and y-axis be Q(x, 0) and P(0, y)

∴ Mid-point of P(0, y) and Q(x, 0) is

$$\text{M}\bigg(\frac{0+x}{2},\frac{y+0}{2}\bigg)$$

But it is given M(0, – 5) $$\therefore\space 2=\frac{x+0}{2}\space\text{and}\space-5=\frac{y+0}{2}$$

⇒ 4 = x and – 10 = y

⇒ x = 4 and y = – 10

∴ The coordinates of P and Q are (0, – 10) and (4, 0).

14. The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is

(a) (a + b + c)2

(b) 0

(c) a + b + c

(d) abc

Sol. (b) 0

Explanation:

Let A ≡ (x1, y1) ≡ (a, b + c)

B ≡ (x2, y2) ≡ (b, c + a)

C ≡ (x3, y3) ≡ (c, a + b)

$$\therefore\space\text{Area of ΔABC =}\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2}[a(c+a-a-b)+b(a+b-b-c)+ c(b+c-c-a)]\\=\frac{1}{2}[a(c-b)+b(a-c)+c(b-a)]\\=\frac{1}{2}[ac-ab+ab-bc+bc-ac]\\=\frac{1}{2}=0$$

15. If the distance between the points (4, p) and (1, 0) is 5, then the value of p is

(a) 4 only

(b) ± 4

(c) – 4 only

(d) 0

Sol. (b) ± 4

Explanation:

According to the question, the distance between the ponts (4, p) and (1, 0) = 5.

$$\text{i.e.}\space\sqrt{(1-4)^{2}+(0-p)^{2}}= 5\\\Rarr\space\sqrt{(-3)^{2}+p^{2}}=5\\\Rarr\sqrt{9+p^{2}}=5$$

On squaring both the sides, we get

9 + p2 = 25

⇒ p2 = 16 ⇒ p ± 4

16. If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then

(a) a = b

(b) a = 2b

(c) 2a = b

(d) a = – b

Sol. (c) 2a = b

Explanation:

Let A ≡ (x1, y1) ≡ (1, 2)

B ≡ (x2, y2) ≡ (0, 0)

C ≡ (x3, y3) ≡ (a, b)

$$\therefore\space\text{Area of ΔABC =}\space\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2}[1(0-b)+0(b-2)+a(2-0)]\\\frac{1}{2}[-b+0+2a]\\=\frac{1}{2}(2a-b)$$

Since, the points A(1, 2), O(0, 0) and C(a, b) are collinear, then area of ΔABC should be equal to zero.

i.e., Area of ΔABC = 0

$$\Rarr\space\frac{1}{2}(2a-b)=0$$

⇒ 2a – b = 0

⇒ 2a = b

Exercise 7.2

• State whether the following statements are true or false. Justify your answer.

1. ΔABC with vertices A(0, 2), B(0, 2) and C(2, 0)is similar to ΔDEF with vertices D(– 4, 0),
E(4, 0) and F(0, 4).

Sol. True.

$$\text{Now,}\space\text{AB}=\sqrt{(0+2)^{2} + (2-0)^{2}}\\=\sqrt{4+4}=2\sqrt{2}\\\text{BC =}\sqrt{(2-0)^{2}+(0-2)^{2}}\\=\sqrt{4+4}=2\sqrt{2}\\\text{CA}=\sqrt{(-2-2)^{2}+(0-0)^{2}}\\=\sqrt{(-4)^{2}+0}=4\\\text{FD}=\sqrt{(0+4)^{2}+(0-4)^{2}}\\=\sqrt{4^{2}+(-4)^{2}}=4\sqrt{2}\\\text{FE =}\sqrt{(4-0)^{2}+(0-4)^{2}}\\\sqrt{4^{2}+4^{2}}=4\sqrt{2}\\\text{ED}=\sqrt{(4+(\normalsize-4))^{2}+(0)^{2}}\\=\sqrt{8 ^{2}}=8$$

Here, we see that sides of a ΔABC is twice the side of ΔDEF. Hence, both the triangles are similar.

2. Point P(– 4, 2) lie on the line segment joining the points A(– 4, 6) and B(– 4, –6).

Sol. True. From the figure, point P(– 4, 2) lies oin the line segment joining the points A(– 4, 6) and B(– 4, – 6).

3. The points (0, 5), (0, – 9) and (3, 6) are collinear.

Sol. False.

∴ Area of triangle

$$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2}[0(-9-6)+0(6-5)+3(5+9)]\\=\frac{1}{2}[0+0+3×14]=21\neq0$$

Hence, points are non-collinear.

4. Points A(3, 1), B(12, – 2) and C(0, 2) cannot be the vertices of a triangle.

Sol. False.

Let A = (x1, y1) = (3, 1), B = (x2, y2) = (12, – 2), C = (x3, y3) = (0, 2)

∴ Area of ΔABC

$$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2}[3(-2-2)+12(2-1)+0(1-(-2))]\\=\frac{1}{2}[3-(-4)+12(1)+0]\\=\frac{1}{3}[-12+12]=0$$

Since, the points A(3, 1), B(12, – 2)and C(0, 2) are collinear but area of ΔABC = 0.

So, the points A(3, 1), B(12, – 2) and C(0, 2) cannot be the vertices of a triangle.

5. Points A(4, 3), B(6, 4), C(5, – 6) and D(– 3, 5) are the vertices of parallelogram.

Sol. False.

$$\text{Now,}\space\text{AB}=\sqrt{(6-4)^{2}+(4-3)^{2}}\\\sqrt{2^{2}+1^{2}}=\sqrt{5}\\\text{BC}=\sqrt{(5-6)^{2}+(-6-4)^{2}}\\=\sqrt{(-1)^{2}+(-10)^{2}}=\sqrt{101}\\\text{CD}=\sqrt{(-3-5)^{2}+(5+6)^{2}}\\=\sqrt{(-8)^{2}+11^{2}}\\=\sqrt{64+121}=\sqrt{185}\\\text{DA}=\sqrt{(4+3)^{2}+(3-5)^{2}}\\=\sqrt{7^{2}+(\normalsize-2)^{2}}\\=\sqrt{49+4}=\sqrt{53}$$

Here, we see that opposite side AB and CD, BC and DA are not equal.

Hence, given vertices are not the vertices of a parallelogram.

6. A circle has its centre at the origin and a point (5, 0) lies on it. The point Q(6, 8) lies outside the circle.

Sol. True. $$\text{Now, OP}=\sqrt{(5-0)^{2}+(0-0)^{2}}\\=\sqrt{5^{2}+0^{2}}=5\text{radius of circle}\\\text{OQ =}\sqrt{(6-0)^{2}+(8-0)^{2}}\\=\sqrt{6^{2}-8^{2}}=\sqrt{36+64}\\=\sqrt{100}=10$$

∴ OQ >OP

Hence, it is true that point Q(6, 8), lies outside the circle.

7. Points A(– 6, 10), B(– 4, 6) and C(3, – 8) are collinear such that $$\textbf{AB}=\frac{\textbf{2}}{\textbf{9}}\textbf{AC.}$$

Sol. True.

$$\text{Area of ΔABC = }\frac{1}{2}[-6(6-(-8))+(-4)(-8-10)+3(10-6)]\\=\frac{1}{2}[\normalsize-6(14)+(\normalsize-4)(\normalsize-18)+3(4)]\\\frac{1}{2}[-84+72+12]=0\\\text{Now,}\space\text{AB}=\sqrt{(-4+6)^{2}+(6-10)^{2}}\\=\sqrt{2^{2}+4^{2}}=\sqrt{4+16}\\=\sqrt{20}=2\sqrt{5}\\\text{AC}=\sqrt{(3+6)^{2}+(-8-10)^{2}}\\=\sqrt{9^{2}+18^{2}}=\sqrt{81+324}\\=\sqrt{405}\\=\sqrt{81×5}=9\sqrt{5}\\\Rarr\space\text{AB}=\frac{2}{9}\text{AC}$$

8. The point P(– 2, 4) lies on a circle of radius 6 and centre (3, 5).

Sol. False.

Now, distance between P(– 2, 4) and centre (3, 5) is

$$=\sqrt{(3-2)^{2}+(5-4)^{2}}\\=\sqrt{5^{2}+1^{2}}\\=\sqrt{25+1}=\sqrt{26}$$

Which is not equal to the radius of the circle.

9. The points A(– 1, – 2), B(4, 3), C(2, 5) and D(– 3, 0) in that order form a rectangle.

Sol. True. $$\text{Now,\space AB}=\sqrt{(4+1)^{2}+(3+2)^{2}}\\=\sqrt{5^{2}+5^{2}}\\=\sqrt{25+25}\\\sqrt{5^{2}+ 5^{2}}=5\sqrt{2}\\\text{CD =}\sqrt{(-3-2)^{2}+(0-5)^{2}}\\=\sqrt{(\normalsize-5)^{2}+(\normalsize-5)^{2}}\\=\sqrt{25+25}=5\sqrt{2}\\\text{AD}=\sqrt{(-3+1)^{2}+(0+2)^{2}}\\=\sqrt{(-2)^{2}+2^{2}}\\=\sqrt{4+4}=2\sqrt{2}\\\text{BC}=\sqrt{(4-2)^{2}+(3-5)^{2}}\\=\sqrt{2^{2}+(\normalsize-2)^{2}}\\=\sqrt{4+4}=2\sqrt{2}$$

Since, AB = CD and AD = BC

$$\text{Also,}\space\text{AC}=\sqrt{(2+1)^{2}+(5+2)^{2}}\\=\sqrt{3^{2}+7^{2}}\\\sqrt{9+49}=\sqrt{58}\\\text{DB}=\sqrt{(4+3)^{2}+(3-0)^{2}}\\=\sqrt{7^{2}+3^{2}}\\=\sqrt{49+9}=\sqrt{58}$$

∴ Diagonals AC and BD are equal.

∴ The points A(– 1, – 2), B(4, 3), C(2, 5) and D(– 3, 0) form a rectangle.

10. Point P(5, – 3) is one of the two points of trisection of line segment joining the points A(7, – 2) and B(1, – 5).

Sol. True. Given, A ≡ (7, – 2) and B ≡ (1, – 5)

Let P and Q trisect the line segment AB.

$$\text{Then,}\space\frac{\text{AP}}{\text{PB}}=\frac{1}{2}\space\text{and}\space\frac{\text{AQ}}{\text{QB}}=\frac{2}{1}$$

P divides AB intrenally in the ratio 1 : 2.

$$\therefore\space\text{P}=\bigg(\frac{1(1)+2(7)}{1+2},\frac{1(\normalsize-5)+2(\normalsize-2)}{1+2}\bigg)\text{i.e.,}\\\bigg(\frac{15}{3},\frac{\normalsize-9}{3}\bigg)$$

P ≡ (5, – 3).

Exercise 7.3

1. Find the value of a, if the distance between the points A(– 3, – 14) and B(a, – 5) is 9 units.

Sol. According to the question,

AB = 9

$$\Rarr\space\sqrt{(a+3)^{2}+(-5+14)^{2}}=9\\\Rarr\space\sqrt{(a+3)^{2}+(9)^{2}}=9$$

On squaring both the sides, we get

(a + 3)2 + 81 = 81 ⇒ (a + 3)2 = 0 ⇒ a = – 3

2. Find the value of m, if the points (5, 1), (– 2, – 3) and (8, 2m) are collinear.

Sol. Let A ≡ (x1, y1) ≡ (5, 1), B ≡ (x2, y2) ≡ (– 2, – 3), C ≡ (x3, y3) ≡ (8, 2m)

Since, the points A = (5, 1), B = (– 2, – 3) and C = (8, 2m) are collinear.

∴ Area of ΔABC = 0

$$\therefore\space\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0\\\Rarr\frac{1}{2}[5[(\normalsize-3-2m)+(\normalsize-2)(2m-1)+8(1-(\normalsize-3))]]=0\\\Rarr\space\frac{1}{2}[-15=10m-4m+2+32]=0\\\Rarr\frac{1}{2}(-14m+19)=0\\\Rarr\space m=\frac{19}{14}$$

3. If the point A(2, – 4) is equidistant from P(3, 8) and Q(– 10, y), find the value of y. Also, find distance PQ.

Sol. According to the question,

PA = QA

$$\sqrt{(2-3)^{2}+(-4-8)^{2}}=\sqrt{(\normalsize-1)^{2}+(\normalsize-12)^{2}}\\\sqrt{(\normalsize-1)^{2}+(\normalsize-12)^{2}}\\=\sqrt{(12)^{2}+(4+y)^{2}}\\\sqrt{1+144}=\sqrt{144+16+y^{2}+8y}\\\sqrt{145}=\sqrt{160+y^{2}+8y}$$

On squaring both the sides, we get

145 = 160 + y2 + 8y

y2 + 8y + 160 – 145 = 0

y2 + 8y + 15 = 0

y2 + 5y + 3y + 15 = 0

y(y + 5) + 3(y – 5) =0

(y + 5)(y + 3) = 0

If y + 5 = 0 ⇒ y = – 5

If y + 3 = 0 ⇒ y = – 3

∴ y = – 3, – 5

$$\text{Now,}\space\text{PQ}=\sqrt{(-10-3)^{2}+(-3-8)^{2}}\\\sqrt{169+121}\\=\sqrt{290}\\\sqrt{(-13)^{2}+(-5-8)^{2}}\\\text{(Putting y = – 5)}\\=\sqrt{169+169}\\\sqrt{338}$$

Hence, value of y are – 3 and – 5 and

$$\text{PQ}=\sqrt{290}\space\text{and}\sqrt{338}$$

4. Find the ratio in which the point

$$\textbf{P}\bigg(\frac{\textbf{3}}{\textbf{4}},\frac{\textbf{5}}{\textbf{12}}\bigg)\\\textbf{divides the line segment joining the points A}\bigg(\frac{\textbf{1}}{\textbf{2}},\frac{\textbf{3}}{\textbf{2}} \bigg)\\\textbf{and B(2, – 5).}$$

$$\textbf{Sol.}\space\text{Let P}\bigg(\frac{3}{5},\frac{5}{12}\bigg)\space\text{divide AB internally in the ratio m : n.}\\\text{Using the section formula, we get}\\\bigg(\frac{3}{4},\frac{5}{12}\bigg)=\bigg(\frac{2m-\frac{n}{2}}{m+n},\frac{-5m+\frac{3}{2}n}{m+n}\bigg)\\\text{On equating, we get}\\\Rarr\frac{3}{4}=\frac{2m-\frac{n}{2}}{m+n}\space\text{and}\space\frac{5}{12}=\frac{-5m+\frac{3}{2}n}{m+n}\\\Rarr\space\frac{3}{4}=\frac{4m-n}{2(m+n)}\space\text{and}\space\frac{5}{12}=\frac{-10m+3n}{2(m+n)}\\\Rarr\space\frac{3}{2}=\frac{4m-n}{m+n}\space\text{and}\space\frac{5}{6}=\frac{-10m+3n}{m+n}$$

⇒ 3m + 2n = 8m – 2n

and 5m + 5n = – 60m + 18n

⇒ 5n – 5m = 0 and 65m – 13n = 0

⇒ n = m and 13(5m – n) = 0

⇒ n = m and 5m – n = 0

Q m = n does not satisfy

∴ 5m – n = 0

⇒ 5m = n

$$\Rarr\space\frac{m}{n}=\frac{1}{5}$$

5. If (9a – 2,– b) divides line segment joining A(3a + 1, – 3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

Sol. Let P(9a – 2, – b) divides AB internally in the ratio 3 : 1.

∴ By section formula, we get

$$\text{9a-2}=\frac{3(8a)+1(3a+1)}{3+1}\\\text{and}\space-b=\frac{3(5)+1(\normalsize-3)}{3+1}\\9a-2=\frac{24a+3a+1}{4}\\\text{and}-b=\frac{15-3}{4}\\9a-2=\frac{27a+1}{4}\\\text{and}\space-b=\frac{12}{4}$$

36a – 8 = 27a + 1

and b = – 3

36a – 27a – 8 – 1 = 0

9a – 9 = 0

⇒ a = 1

6. If (a, b) is the mid-point of the line segment joining the points A(10, – 6) and B(k, 4) and a – 2b = 18, find the value of k and the distance AB.

Sol. ∵ (a, b) is the mid-point of AB.

$$\therefore\space(a,b)=\bigg(\frac{10+k}{2},\frac{6+4}{2}\bigg)\\(a,b)=\bigg(\frac{10+k}{2},-1\bigg)\\\text{a}=\frac{10+k}{2}\space\text{...(i)}$$

b = – 1 ...(ii)

Given, a – 2b = 18

From equation (ii),

a – 2(– 1) = 18

⇒ a + 2 = 18

⇒ a = 16

∴ From equation (i),

$$16=\frac{10+k}{2}$$

⇒ 32 = 10 + k

⇒ k = 22

∴ A ≡ (10, – 6), B ≡ (22, 4)

$$\text{AB}=\sqrt{(22-10)^{2}+(4+6)^{2}}\\=\sqrt{(12)^{2}+(10)^{2}}\\\sqrt{144+100}\\\sqrt{244}=2\sqrt{61}$$

$$\textbf{7. If D}\space\bigg(-\frac{\textbf{1}}{\textbf{2}},-\frac{\textbf{5}}{\textbf{2}}\bigg),\textbf{E(7,3) and F}\bigg(\frac{\textbf{7}}{\textbf{2}},\frac{\textbf{7}}{\textbf{2}}\bigg)$$

are the mid-points of sides of ΔABC, find the area of the ΔABC.

Sol. Let A ≡ (x1, y1), B ≡ (x2, y2) and C ≡ (x3, y3)

$$\text{Since,}\space\text{D}\bigg(-\frac{1}{2},\frac{5}{2}\bigg),\text{E}(7,3)\space\text{and F}\bigg(\frac{7}{2},\frac{7}{2}\bigg)$$

be the mid-points of the sides BC, CA and AB, respectively.

$$\text{Since},\text{D}\bigg(-\frac{1}{2},-\frac{5}{2}\bigg)\space\text{is the mid-point of BC.}\\\therefore\space\frac{x_2+x_3}{2}=\frac{1}{2}\\\text{and}\space\frac{y_2+y_3}{2}=\frac{5}{2}$$

⇒ x2 + x3 = – 1 ...(i)

and y2 + y3 = 5 ...(ii)

As E(7, 3) is the mid-point of CA.

$$\therefore\space\frac{x_3+x_1}{2}=\frac{1}{2}\\\text{and}\space\frac{y_2+y_3}{2}=\frac{5}{2}$$

⇒ x2 + x3 = – 1 ...(i)

and y2 + y3 = 5 ...(ii)

As E(7, 3) is the mid-point of CA.

$$\therefore\space\frac{x_3+x_1}{2}=7\\\text{and}\space\frac{y_3+y_1}{2}=3$$

⇒ x3 + x1 = 14 ...(iii)

and y3 + y1 = 6 ...(iv)

$$\text{Also, F}\bigg(\frac{7}{2},\frac{7}{2}\bigg)\space\text{is the mid-point of AB.}\\\therefore\space\frac{x_1+x_2}{2}=\frac{7}{2}\\\text{and}\space\frac{y_1+y_2}{2}=\frac{7}{2}$$

⇒ x1 + x2 = 7 ...(v)

and y1 + y2 = 7 ...(vi)

Adding Eqs. (i), (iii) and (v), we get

2(x1 + x2 + x3) = 20

⇒ x1 + x2 + x3 = 10 ...(vii)

Subtracting Eqs. (i), (ii) and (v) from Eq. (vii) respectively, we get

x1 = 11, x2 = – 4, x3 = 3

Adding Eqs. (ii), (iv), and (vi), we get

2(y1 + y2 + y3) = 18

⇒ y1 + y2 + y3 = 9 ...(viii)

Subtracting Eqs. (ii), (iv) and (vi) from Eq. (viii) respectivgely, we get

y1 = 4, y2 = 3, y3 = 2

Hence, the vertices of ΔABC are A(11, 4), B(– 4, 3) and C(3, 2).

$$\therefore\space\text{Area of ΔABC =}\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\=\frac{1}{2}[11(3-2)+(\normalsize-4)(2-4)+3(4-3)]\\=\frac{1}{2}[11×1+(\normalsize-4)(2-4)+3(4-3)]\\=\frac{1}{2}[11×1+(\normalsize-4)(\normalsize-2)+3(1)]\\=\frac{1}{2}[11+8+3]\\=\frac{22}{2}=11$$

∴ Area of ΔABC = 11

8. Find the values of k, if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

Sol. Since, the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

∴ Area of ΔABC = 0

$$\Rarr\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0\\\Rarr\frac{1}{2}[(k+1)(2k+3-5k)+3k(5k-2k)+(5k-1)(2k-(2k-3))]=0\\\Rarr\frac{1}{2}[(k+1)(-3k+3)+3k(3k)+(5k+1)(2k-2k-3)]=0\\\Rarr\frac{1}{2}[-3k^{2}+3k-3k+3+9k^{2}-15k+3]=0\\\Rarr\space\frac{1}{2}[6k^{2}-15k+6]=0\space(\text{(Multiply by 2)})$$

⇒ 6k2 – 15k + 6 = 0

⇒ 2k2 – 5k + 2 = 0 (Divide by 3)

⇒ 2k2 – 4k – k + 2 = 0

⇒ 2k(k – 2) – 1(k – 2) = 0

⇒ (k – 2)(2k – 1) = 0

If k – 2 = 0 ⇒ k = 2

$$\text{If 2k – 1 = 0 ⇒ k =}\frac{1}{2}\\\therefore\space k=2,\frac{1}{2}$$