Haryana State Board HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 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).

Solution:

Let required ratio be k : 1 and line 2x + y – 4 divides the line segment.

Join A(2, – 2) and B(3, 7) at the point P(x, y).

Then co-ordinates of P are.

â‡’ x = \(\frac{3 k+2}{k+1}\)

y = \(\frac{7 k-2}{k+1}\)

Thus, the co-ordinates of P are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\).

Since, point P lies on the given line 2x + y – 4 = 0.

Therefore, 2 Ã— \(\left(\frac{3 k+2}{k+1}\right)+\frac{7 k-2}{k+1}\) – 4 = 0

\(\frac{6 k+4+7 k-2-4 k-4}{k+1}\) = 0

\(\frac{9 k-2}{k+1}\) = 0

9k – 2 = 0

k = \(\frac{2}{9}\)

k : 1 = 2 : 9

Hence, required ratio is 2 : 9.

Question 2.

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

Solution:

Let A(x, y), B(1, 2)and C(7, 0) are given points

Since, the given points are coimear.

Therefore,Area of âˆ†ABC = 0

â‡’ \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

â‡’ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2)] = 0

â‡’ \(\frac{1}{2}\) [2x – y + 7y – 14] = 0

â‡’ \(\frac{1}{2}\) [2x + 6y – 14] = 0

â‡’ 2x + 6y – 14 = 0

â‡’ x + 3y – 7 = 0

Hence, required relation is x + 3y – 7 = 0.

Question 3.

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

Solution :

Let P(x, y) be the centre of a circle which passes through A(6, – 6), B(3, – 7) and C(3, 3)

âˆ´ PA = PB = PC (equal radii of circle)

Now PA = PB

â‡’ PA^{2} = PB^{2}

â‡’ (6 – x)^{2} + (- 6 – y)^{2} = (3 – x)^{2} + (- 7 – y)^{2}

â‡’ 36 + x^{2} – 12x + 36 + y^{2} + 12y = 9 + x^{2} – 6x + 49 + y^{2} + 14y

â‡’ x^{2} + y^{2} – 12x + 12y + 72 = x^{2} + y^{2} – 6x + 14y + 58

â‡’ x^{2} + y^{2} – 12x + 12y – x^{2} – y^{2} + 6x – 14y = 58 – 72

â‡’ – 6x – 2y = – 14

â‡’ 3x + y = 7 ……………..(1)

and PB = PC

â‡’ PB^{2} = PC^{2}

â‡’ (3 – x)^{2} + (- 7 – y)^{2} = (3 – x)^{2} + (3 – y)^{2}

â‡’ (3 – x)^{2} + (- 7 – y)^{2} – (3 – x)^{2} = (3 – y)^{2}

â‡’ (- 7 – y)^{2} = (3 – y)^{2}

â‡’ 49 + y^{2} + 14y = 9 + y^{2} – 6y

â‡’ y^{2} + 14y – y^{2} + 6y = 9 – 49

20y = – 40

y = – \(\frac{40}{20}\)

y = – 2

Substituting the value ofy in equation (1), we get

3x – 2 = 7

3x = 7 + 2 = 9

x = \(\frac{9}{3}\)

x = 3

Hence, centre of the circle is (3, – 2).

Question 4.

The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the co-ordinates of the other two vertices.

Solution :

Let the opposite vertices of a square be A(- 1, 2) and C(3, 2) and unknown vertex be B(x, y) then

AB = BC (sides of square)

â‡’ AB^{2} = BC^{2}

â‡’ (x + 1)^{2} + (y – 2)^{2} = (3 – x)^{2} + (2 – y)^{2}

â‡’ x^{2} + 1 + 2x + y^{2} + 4 – 4y = 9 + x^{2} – 6x + 4 + y^{2} – 4y

â‡’ x^{2} + 2x – 4y + y^{2} + 5 = x^{2} + y^{2} – 6x – 4y + 13

x^{2} + 2x – 4y + y^{2} – y^{2} + 6x + 4y = 13 – 5

8x = 8

x = \(\frac{8}{8}\)

x = 1 ……………….(1)

In right âˆ†ABC, we get

AC^{2} = AB^{2} + BC^{2} [By Pythagoras theorem]

â‡’ (3 + 1)^{2} + (2 – 2)^{2} = (x + 1)^{2} + (y – 2)^{2} + (3 – x)^{2} + (2 – y)^{2}

â‡’ 16 = x^{2} + 1 + 2x + y^{2} + 4 – 4y + 9 + x^{2} – 6x + 4 + y^{2} – 4y

â‡’ 16 = 2x^{2} + 2y^{2} – 4x – 8y + 18

â‡’ 2x^{2} + 2y^{2} – 4x – 8y = 16 – 18 = – 2

â‡’ x^{2} + y^{2} – 2x – 4y= – 1

â‡’ (1)^{2} + y^{2} – 2 Ã— 1 – 4y = – 1 [âˆµ From (1) x = 1]

â‡’ 1 + y^{2} – 4y – 2 = – 1

y^{2} – 4y = – 1 – 1 + 2

â‡’ y(y – 4) = 0

y= 0 or y -4 = 0

y = 0 or y = 4

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

Question 5.

The class X students of a secondary school in Krishinagar have been alloted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m 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 remaininq area of the plot.

(i) Taking A as origin, find the co-ordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of âˆ† PQR if C is the origin ? Also, calculate the area-s of the triangles in these cases. What do you observe?

Solution :

(i) Taking A as origin the coordinates of the vertices of P, Q and R are (4, 6), (3, 2) and (6, 5) respectively.

(ii) Taking C as the origin, co-ordinates of P, Q and R are (12, 2), (13, 6) and (10, 3).

In 1st condition area of âˆ†PQR = \(\frac{1}{2}\) [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)]

= \(\frac{1}{2}\) [4 Ã— (- 3) + 3 Ã— (- 1) + 6 Ã— 4]

= \(\frac{1}{2}\) (- 12 – 3 + 24)

= \(\frac{1}{2}\) Ã— 9

= \(\frac{9}{2}\) square units.

In 2nd condition area of âˆ†PQR = \(\frac{1}{2}\) [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)]

= \(\frac{1}{2}\) [12 Ã— 3 + 13 Ã— 1 + 10 Ã— (- 4)]

= \(\frac{1}{2}\) (36 + 13 – 40)

= \(\frac{1}{2}\) Ã— 9

= \(\frac{9}{2}\) square units

Hence, coordinates of âˆ†PQR.

(i) Taking A as origin and taking AD as x-axis, AB as y-axis are (4, 6), (3, 2) and (6, 5).

(ii) Taking C as origin and taking CB x-axis, CD as y-axes are (12, 2), (13, 6) and (10, 3)

Areas in two cases are \(\frac{9}{2}\) square units.

Question 6.

The vertices of a âˆ†ABC 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 \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\). Calculate the area of the âˆ†ADE and compare it with the area of âˆ†ABC. (Recall theorem 6.2 and theorem 6.6).

Solution:

We have,

\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{4}\)

â‡’ \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}=4\)

â‡’ \(\frac{\mathrm{AB}}{\mathrm{AD}}-1=\frac{\mathrm{AC}}{\mathrm{AE}}-1\) = 4- 1

â‡’ \(\frac{\mathrm{AB}-\mathrm{AD}}{\mathrm{AD}}=\frac{\mathrm{AC}-\mathrm{AE}}{\mathrm{AE}}\) = 3

â‡’ \(\frac{\mathrm{DB}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{AE}}\) = 3

â‡’ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}=\frac{1}{3}\)

â‡’ AD : DB = AE : EC = 1 : 3.

Point D divides the AB in the ratio 1 : 3.

Co-ordinates of D

Point E divides the AC in the ratio 1 : 3.

Co-ordinates of E

Area of âˆ†ABC

= \(\frac{1}{2}\) [x_{1} (y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\) [4(5 – 2) + 1(2 – 6) + 7(6 – 5)]

= \(\frac{1}{2}\) (4 Ã— 3 + 1 Ã— – 4 + 7 Ã— 1)

= \(\frac{1}{2}\) (12 – 4 + 7)

= \(\frac{1}{2}\) Ã— 15

= \(\frac{15}{2}\) square units

\(\frac{\text { Area of } \triangle \mathrm{ADE}}{\text { Area of } \triangle \mathrm{ABC}}=\frac{\frac{15}{32}}{\frac{15}{2}}\)

\(\frac{\text { Area of } \triangle \mathrm{ADE}}{\text { Area of } \triangle \mathrm{ABC}}=\frac{15}{32} \times \frac{2}{15}=\frac{1}{16}\)

Hence, Area of âˆ†ADE = \(\frac{15}{32} \) square units

and Area of ADE : Area of âˆ†ABC = 1 : 16.

Question 7.

Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of âˆ†ABC.

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

(ii) Find the co-ordinates of the points P on AD such that AP : PD = 2 : 1.

(iii) Find the co-ordinates 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(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are the vertices of âˆ†ABC, find the co-ordinates of the centroid of the triangle.

Solution :

(i) D is the mid-point of BC.

Co-ordinates of point D are \(\left(\frac{6+1}{2}, \frac{5+4}{2}\right)\) i.e., \(\left(\frac{7}{2}, \frac{9}{2}\right)\).

(ii) P divides the AD in the ratio 2 : 1.

Co-ordinates of point P = \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)

= \(\left(\frac{2 \times \frac{7}{2}+1 \times 4}{2+1}, \frac{2 \times \frac{9}{2}+1 \times 2}{2+1}\right)\)

= \(\left(\frac{7+4}{3}, \frac{9+2}{3}\right)=\left(\frac{11}{3}, \frac{11}{3}\right)\)

(iii) E is the mid-point of AC.

Co-ordinates of point E = \(\left(\frac{4+1}{2}, \frac{2+4}{2}\right)=\left(\frac{5}{2}, 3\right)\)

Q divides BE in the.ratio 2 : 1.

Co-ordinates of point Q.

= \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1^{\prime} y_2+m_2 y_1}{m_1+m_2}\right)\)

= \(\left(\frac{2 \times \frac{5}{2}+1 \times 6}{2+1}, \frac{2 \times 3+1 \times 5}{2+1}\right)\)

= \(\left(\frac{5+6}{3}, \frac{6+5}{3}\right)=\left(\frac{11}{3}, \frac{11}{3}\right)\)

F is the mid-point of AB.

Co-ordinates of point F = \(\left(\frac{4+6}{2}, \frac{2+5}{2}\right)=\left(5, \frac{7}{2}\right)\)

Point R divides the CF in the ratio 2 : 1.

Co-ordinates of point R = \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)

= \(\left(\frac{2 \times 5+1 \times 1}{2+1}, \frac{2 \times \frac{7}{2}+1 \times 4}{2+1}\right)\)

= \(\left(\frac{10+1}{3}, \frac{7+4}{3}\right)=\left(\frac{11}{3}, \frac{11}{3}\right)\)

(iv) We observe that the co-ordinates of point P, Q, R, are same. Therefore, the point P, Q, R, are same point.

(v) See solution in Basic concepts; co-ordinates of the centroid.

Question 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.

Solution :

Since P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.

âˆ´ Co-ordinates of point P are \(\left[\frac{(-1-1)}{2}, \frac{(-1+4)}{2}\right]\) i.e., (- 1, \(\frac{3}{2}\))

Co-ordinates of point Q are \(\left[\left(\frac{-1+5}{2}\right),\left(\frac{4+4}{2}\right)\right]\) i.e., (2, 4)

Co-ordinates of point R are \(\left[\left(\frac{5+5}{2}\right),\left(\frac{4-1}{2}\right)\right]\) i.e., (5, \(\frac{3}{2}\))

Co-ordinates of point R are \(\left[\left(\frac{5-1}{2}\right),\left(\frac{-1-1}{2}\right)\right]\) i.e., (2, – 1).

Now,

Thus PQ = QR = RS = SP, but diagonal PR â‰ diagonal SQ.

Hence quadrilateral PQRS is a rhombus.