# Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Question 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)
Solution:
(i) Let P(2, 3) and Q(4, 1) be the given points.
Here, x1 = 2, y1 = 3 and x2 = 4, y2 = 1.
∴ PQ = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
So, PQ = $$\sqrt{(4-2)^{2}+(1-3)^{2}}$$ = $$\sqrt{(2)^{2}+(-2)^{2}}$$
or PQ = $$\sqrt{4+4}$$ = $$\sqrt{8}$$ = 2$$\sqrt{2}$$

(ii) Let P(- 5, 7) and Q(- 1, 3) be the given points.
Here, x1 = -5, y1 = 7 and x2 = -1, y2 = 3.
∴ PQ = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
So, PQ = $$\sqrt{(-1+5)^{2}+(3-7)^{2}}$$ = $$\sqrt{(4)^{2}+(-4)^{2}}$$
or PQ = $$\sqrt{16+16}$$ = $$\sqrt{32}$$ = $$\sqrt{16 \times 2}$$ = 4$$\sqrt{2}$$

(iii) Let P(a, b) and Q(- a, – b) be the given points.
Here, x1 = a, y1 = b and x2 = a, y2 = – b.
∴ PQ = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
So, PQ = $$\sqrt{(-a-a)^{2}+(-b-b)^{2}}$$
= $$\sqrt{(-2 a)^{2}+(-2 b)^{2}}$$
or PQ = $$\sqrt{4 a^{2}+4 b^{2}}$$
= 2$$\sqrt{a^{2}+b^{2}}$$

Question 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 discussed in Section 7.2?
Solution:
Let P(0, 0) and Q(36, 15) be the given points.
Here, x1 = 0, y1 = 0 and x2 = 36, y2 = 15.
∴ PQ = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
So, PQ = $$\sqrt{(36-0)^{2}+(15-0)^{2}}$$ = $$\sqrt{1296+225}$$
= $$\sqrt{1521}$$ = 39
In fact, the positions of towns A and B are given by (0, 0) and (36, 15) respectively and so the distance between them = 39 km as calculated above.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Let A(1, 5), B(2, 3) and C(- 2, – 11) be the given points. Then, we have :

Clearly, BC ≠ AB + AC, AB ≠ BC + AC and AC ≠ BC + AB.
Hence, A, B and C are not collinear.

Question 4.
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Solution:
Let A(5, – 2), B(6, 4) and C(7, – 2) are the given points. Then,
AB = $$\sqrt{(6-5)^{2}+(4+2)^{2}}$$
= $$\sqrt{1+36}$$
= $$\sqrt{37}$$
BC = $$\sqrt{(7-6)^{2}+(-2-4)^{2}}$$
= $$\sqrt{1+36}$$
= $$\sqrt{37}$$
Clearly, AB = BC
∴ ∆ ABC is an isosceles triangle.

Question 5.
In a classroom, four friends are Seated at the points A, B, C and D as 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, and why?
Solution:
Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).
By using distance formula, we get

Thus, the four sides are equal and the diagonals are also equal. Therefore, ABCD is a square.
Hence, Champa is correct.

Question 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)
Solution:
(i) Let A(- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0) be the given points. Then,

Clearly, four sides AB, BC, CD and DA are equal.
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 points. Plot these points as shown.

Clearly, the points A, C and B are collinear.
So, no quadrilateral is formed by these points.

Aliter :
AB = $$\sqrt{(3+3)^{2}+(1-5)^{2}}$$
= $$\sqrt{36+16}$$
= $$\sqrt{52}$$
= 2$$\sqrt{13}$$

BC = $$\sqrt{(3-0)^{2}+(1-3)^{2}}$$
and AC = $$\sqrt{(0+3)^{2}+(3-5)^{2}}$$ = $$\sqrt{13}$$
So, AC = AB + BC.
Hence, there, will not be any quadrilateral.

(iii) Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2) be the .given points. Then,

Clearly, AB = CD, BC = DA and AC = BD.
∴ The quadrilateral ABCD is a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Since the point on x-axis have its ordinate – 0, so P(x, 0) is any point on the x-axis.
Since P(x, 0) is equidistant from A(2, – 5) and P (- 2, 9), therefore PA = PB gives. PA² = PB²
or (x – 2)² + (0 + 5)² = (x + 2)² + (0 – 9)²
or x² – 4x + 4 + 25 = x² + 4x + 4 + 81
or – 4x – 4x = 81 – 25
– 8x = 56
or x = $$\frac { 56 }{ -8 }$$ = – 7
∴ The point equidistant from given points on the x-axis is (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Solution:
P(2, – 3) and Q(10, y) are given points such that PQ = 10 units.

Thus, the possible value of y is – 9 or 3.

Question 9.
If Q(0, 1) is equidistant from P(5, – 3) and R(x, 6), find the values of x. Also, find the distances QR and PR.
Solution:
Since the point Q(0, 1) is equidistant from P(5, – 3) and R(x, 6), therefore
QP = QR gives QP² – QR²
or (5 – 0)² + (- 3 – 1)² = (x – 0)² + (6 – 1)²
or 25 + 16 = x² + 25
or x² = 16 So, x = ± 4
Thus, R is (4, 6) or (- 4, 6).
Now, QR = Distance between Q(0, 1) and R(4, 6)
$$\sqrt{(4-0)^{2}+(6-1)^{2}}$$ = $$\sqrt{16+25}$$ = $$\sqrt{41}$$
Also, QR = Distance between Q(0, 1) and R(- 4, 6)
$$\sqrt{(-4-0)^{2}+(6-1)^{2}}$$ = $$\sqrt{16+25}$$ = $$\sqrt{41}$$
and PR = Distance between P(5, – 3) and R(4, 6)
$$\sqrt{(4-5)^{2}+(6+3)^{2}}$$ = $$\sqrt{1+81}$$ = $$\sqrt{82}$$
and PR = Distance between P(5, – 3) and R(-4, 6)
$$\sqrt{(-4-5)^{2}+(6+3)^{2}}$$ = $$\sqrt{81+81}$$ = 9$$\sqrt{2}$$

Question 10.
Find a relation between x and y such that the point Or, y) is equidistant from the points (3, 6) and (-3,4).
Solution:
Let the point P(x, y) be equidistant from the points A(3, 6) and B(- 3, 4).
i.e., PA = PB gives PA² = PB²
or (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
or x²- 6x + 9 + y² – 12y + 36 = x²+ 6x + 9 + y² – 8y + 16
or 6x – 6x – 12y + 8y + 36 – 16 0
or – 12x – 4y + 20 = 0
or 3x + y – 5 = 0,
which is the required relation.