Haryana State Board HBSE 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 1.

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

Solution :

Let P(x, y) be the required point By section formula co-ordinates of P are

⇒ x = 1, y = 3

Hence, the co-ordiantes of required point are (1, 3).

Question 2.

Find the co-ordinates of the points

Solution :

Let P and Q be the points of trisection of AB.

i.e., AP = PQ = QB

Let AP = PQ = QB = K

PB = PQ + QB = 2K

AP + PQ = 2K

∴ AP : PB = K : 2K = 1 : 2

and, AQ : QB = 2K : K = 2 : 1.

Therefore, P divides AB in the ratio 1 : 2.

By section formula the co-ordinates of P are

So, the co-ordinates of p are (2, – \(\frac{5}{3}\))

Now, Q divides AB in the ratio 2 : 1.

By section formula, co-ordinates of Q are

So, the co-ordinates of Q are (0, – \(\frac{7}{3}\))

Hence, co.-ordmates of trisection are (2, \(\frac{5}{3}\)) and (0, – \(\frac{7}{3}\))).

Question 3.

To conduct sports day tivities 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 th the distance of AD on the 2nd line and posts a green flag. Preet runs 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 half way between the line segment joining the two flags, where should she post her flag?

Solution:

From the figure taking A as origin (0, 0), x-axis along AB and y-axis along AD, we will determine the co-ordinates of the green flag and red flag.

Now, the green flag in 2nd line and its distance parallel to AD.

AD = \(\frac{1}{4}\) × 100 = 25 m.

So, the co-ordinates of green flag are (2, 25), we mark this position as G.

Similarly, co-ordinates of red flag = (8, 20), we mark this position as R.

Now,the distance between G and R = \(\sqrt{(8-2)^2+(20-25)^2}\)

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

= \(\sqrt{36+25}\)

= \(\sqrt{61}]\)

Position of the blue flag is the mid point of GR.

Co-ordinates of blue flag = \(=\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)

= \(\left(\frac{10}{2}, \frac{45}{2}\right)\)

= (5, 22.5)

It means that blue flag is in the 5th line and at a distance of 22-5 m along the direction parallel to AD.

Hence, distance between G and R = \(\sqrt{61}]\) m, and blue flag in 5th line at a distance of 22.5 m.

Question 4.

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

Solution :

Let the required ratio be m_{1} : m_{2}

By section formula, we have

Now, \(\frac{6 m_1-3 m_2}{m_1+m_2}\) = – 1

⇒ 6m_{1} – 3m_{2} = – m_{1} – m_{2}

⇒ 6m_{1} + m_{1} = 3 m_{2} – m_{2}

⇒ 7m_{1} = 2m_{2}

⇒ \(\frac{m_1}{m_2}=\frac{2}{7}\)

⇒ m_{1} : m_{2} = 2 : 7

Hence, the required ratio is 2 : 7.

Question 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.

Solution :

Let the required ratio be k : 1.

By section formula, we have

Since, point lies on x-axis, so its ordinate is zero.

0 = \(\frac{5 k-5}{k+1}\)

5k – 5 = 0

k = \(\frac{5}{5}\) = 1

So, the required ratio is 1 : 1.

Putting the value of k we get co-ordinates of the point of division.

y = 0

Hence, required ratio = 1: 1 and co-ordinates of the point of division = (- \(\frac{3}{2}\), 0).

Question 6.

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

Solution :

Let given vertices of a parallelogram be A(1, 2), B(4, y), C(x, 6) and D(3, 5). Join AC and BD.

We know that diagonals of a parallelogram bisect each other

So, O is the mid point of AC as well as that of BD.

Co-ordinates of mid point of AC = Co-ordinates of mid point of BD.

1 + x = \(\frac{7 \times 2}{2}\) and y + 5 = 8

x = 7 – 1 and y = 8 – 5

x = 6 and y = 3

Hence, x = 6 and y = 3

Question 7.

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

Solution :

Let the co-ordinates of point A be (x, y).

∵ O is the mid point of AB

Co-ordinates of O are 2 = \(\frac{x+1}{2}\) and – 3 = \(\frac{y+4}{2}\)

⇒ 4 = x + 1 and – 6 = y + 4

⇒ 3 = x and – 10 = y

⇒ x = 3, and y = – 10

Hence, co-ordinates of point A are (3, – 10).

Question 8.

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

Solution :

Let the co-ordinates of P be (x, y), we have

AP = \(\frac{3}{7}\) AB

AP + PB = AB

By section formula, the co-ordinates of point P are

Hence, co-ordinates of P are \(\left(-\frac{2}{7},-\frac{20}{7}\right)\).

Question 9.

Find the co-ordinates of the points which divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts.

Solution:

Let P, Q, R be three points that divides the line segment joining A(- 2, 2) and B(2, 8) in four equal parts.

Since Q is the mid point of AB.

Therefore, co-ordinates of point Q are \(\left(-\frac{2+2}{2}, \frac{2+8}{2}\right)\) i.e. (0, 5)

Similarly, P is the mid point of AQ.

Therefore, co-ordinates of P are \(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)\) i.e., (- 1, \(\frac{7}{2}\))

and R is the mid point of QB.

Therefore, co-ordinates of R are \(\left(\frac{0+2}{2}, \frac{5+8}{2}\right)\) i.e., (1, \(\frac{13}{2}\))

Hence, the required co-ordinates are (- 1, \(\frac{7}{2}\)), (0, 5) and (1, \(\frac{13}{2}\)).

Question 10.

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

Solution :

Let coordinates of vertices of a rhombus be A(3, 0), B(4, 5), C(- 1, 4) and D(- 2, – 1), then

Now, area of rhombus = \(\frac{1}{2}\) (Product of diagonals)

= \(\frac{1}{2}\) × AC × BD

= \(\frac{1}{2}\) × 4√2 × 6√2

= 24 square units

Hence, area of rhombus = 24 square units.