# GSEB Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

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

## Gujarat Board Textbook Solutions Class 10 Maths 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 the given points be A(2, 3) and B(4, 1). We know that the distance between two points A(x1, y1) and B(x2, y2) is given bý
AB = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
AB = $$\sqrt{(4-2)^{2}+(1-3)^{2}}$$
= $$\sqrt{4+4}$$ = $$\sqrt{8}$$ = 2$$\sqrt{4}$$

(ii) Let the given points be A(-5, 7) and B(-1,3) We know that the distance between two points A(-5, 7) and B(-1, 3) is given by
AB = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
AB = $$\sqrt{(-1-(-5))^{2}+(3-(-1))^{2}}$$
= $$\sqrt{16 + 16}$$ = $$\sqrt{32}$$ = 4$$\sqrt{2}$$

(iii) Let the given points be A(a, b) and B(-a, -b). We know that the distance between two points A(a, b) and B(-a, -b) is given by
AB = $$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
AB = $$\sqrt{(-a-a)^{2}+(-b-b)^{2}}$$
= $$\sqrt{(-2 a)^{2}+(-2 b)^{2}}$$
= $$\sqrt{(-2 a)^{2}+(-2 b)^{2}}$$
= $$\sqrt{4 a^{2}+4 b^{2}}$$
= $$\sqrt{4\left(a^{2}+b^{2}\right)}$$
= $$2 \sqrt{a^{2}+b^{2}}$$ 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 the given points be A(0, 0) and B(36, 15), then
AB = $$\sqrt{(36-0)^{2}+(15-0)^{2}}$$
= $$\sqrt{(36)^{2}+(15)^{2}}$$
= $$\sqrt{1296+225}$$
= $$\sqrt{1521}$$ = 39
Yes, we can find the distance between the two towns A and B discussed in section 7.2 and this distance = 39 km.

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are coimear.
Solution:
Let the given points be A(1, 5), B(2, 3) and C(- 2, -11), then
AB = $$\sqrt{(2-1)^{2}+(3-5)^{2}}$$
⇒ AB = $$\sqrt{(1)^{2}+(-2)^{2}}$$
⇒ AB = $$\sqrt{1+4}$$ = $$\sqrt{5}$$
AC = $$\sqrt{(-2-1)^{2}+(-11-5)^{2}}$$
⇒ AC = $$\sqrt{(-3)^{2}+(-16)^{2}}$$
⇒ AC = $$\sqrt{9+256}$$
⇒ AC = $$\sqrt{265}$$
and BC = $$\sqrt{(-2-2)^{2}+(-11-3)^{2}}$$
⇒ BC = $$\sqrt{(-4)^{2}+(-14)^{2}}$$
⇒ BC = $$\sqrt{16+196}$$
⇒ BC = $$\sqrt{212}$$
Here, we see that AB + BC  AC, BC + AC AB and AB + AC BC.
Hence, the points A, B and C are not coimear. Question 4.
Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Solution:
Let the given points be A(5, -2), B(6, 4) and C(7, -2), then
AB = $$\sqrt{(6-5)^{2}+[4-(-2)]^{2}}$$
= $$\sqrt{(1)^{2}+(6)^{2}}$$ = $$\sqrt{1+36}$$
= $$\sqrt{37}$$
BC = $$\sqrt{(7-6)^{2}+(-2-4)^{2}}$$
= $$\sqrt{(1)^{2}+(-6)^{2}}$$
= $$\sqrt{1+36}$$ = $$\sqrt{37}$$
Since AB = BC
Therefore, ABC is an isosceles triangle.

Question 5.
In a classroom, 4 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. Solution:
Let A(3, 4), B(6, 7), C(9, 4) and D(6, 1) be the given points. Then,
AB = $$\sqrt{(6-3)^{2}+(7-4)^{2}}$$
= $$\sqrt{(3)^{2}+(3)^{3}}$$ = $$\sqrt{9+9}$$
= $$\sqrt{18}$$ = 3$$\sqrt{2}$$

BC = $$\sqrt{(9-6)^{2}+(4-7)^{2}}$$
= .$$\sqrt{(3)^{2}+(-3)^{2}}$$ + $$\sqrt{9+9}$$
=$$\sqrt{18}$$ = 3$$\sqrt{2}$$

CD = $$\sqrt{(6-9)^{2}+(1-4)^{2}}$$
= $$\sqrt{(-3)^{2}+(-3)^{2}}$$
= $$\sqrt{9+9}$$ = $$\sqrt{18}$$ = 3$$\sqrt{2}$$

DA = $$\sqrt{(3-6)^{2}+(4-1)^{2}}$$
= $$\sqrt{(-3)^{2}+(3)^{2}}$$ $$\sqrt{9+9}$$
= $$\sqrt{18}$$ = 3$$\sqrt{2}$$

AC = $$\sqrt{(9-3)^{2}+(4-4)^{2}}$$ = 6
BD = $$\sqrt{(6-6)^{2}+(1-7)^{2}}$$ = 6
We see that,
AB = BC = CD = DA
and AC = BD = 6
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:
(j) (-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 the given points be A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0).
Then,
AB = $$\sqrt{[1-(-1)]^{2}-[0-(-2)]^{2}}$$
= $$\sqrt{(1+1)^{2}+(-2)^{2}}$$
= $$\sqrt{(2)^{2}+(-2)^{2}}$$
= $$\sqrt{4+4}$$ = $$\sqrt{8}$$

BC = $$\sqrt{(-1-1)^{2}+(2-0)^{2}}$$
= $$\sqrt{(-2)^{2}+(2)^{2}}$$
= $$\sqrt{4+4}$$ = $$\sqrt{8}$$

CD = $$\sqrt{[3-(-1)]^{2}+(0-2)^{2}}$$
= $$\sqrt{(-3+1)^{2}+(-2)^{2}}$$
= $$\sqrt{(-2)^{2}+(-2)^{2}}$$
= .$$\sqrt{4+4}$$ = $$\sqrt{8}$$

and DA = $$\sqrt{[-1-(-3)]^{2}+(-2-0)^{2}}$$
= $$\sqrt{(-1+3)^{2}+(-2)^{2}}$$
= $$\sqrt{(2)^{2}+(-2)^{2}}$$
= $$\sqrt{4+4}$$ = $$\sqrt{8}$$
Thus, we have
AB = BC = CD = DA

⇒ All sides are equal Also,
AC = $$\sqrt{[-1-(-1)]^{2}+[2-(-2)]^{2}}$$
= $$\sqrt{(-1+1)^{2}+(2+2)^{2}}$$
= $$\sqrt{0+(4)^{2}}$$ = $$\sqrt{16}$$ = 4
and BD = $$\sqrt{(-3-1)^{2}+0}$$
= $$\sqrt{(-4)^{2}}$$ = $$\sqrt{16}$$ = 4
Since, all four sides of the quadrilateral are equal and diagonals are also equal, sothe given points form a square.

(ii) Let the given points be A(-3, 5), B(3, 1), C(0, 3) and D(-1, -4).
Then, AB = $$\sqrt{[3-(-3)]^{2}+(1-5)^{2}}$$
= $$\sqrt{(3+3)^{2}+(1-5)^{2}}$$
= $$\sqrt{(6)^{2}+(-4)^{2}}$$
= $$\sqrt{36+16}$$ = $$\sqrt{52}$$ = 2$$\sqrt{13}$$ BC= $$\sqrt{(0-3)^{2}+(3-1)^{2}}$$
= $$\sqrt{(-3)^{2}+(2)^{2}}$$
= $$\sqrt{9+4}$$ = $$\sqrt{13}$$

AB = $$\sqrt{(-3)^{2}+4}$$ = $$\sqrt{13}$$
∴ AB = BC + AC
∴ A, B and C are collin ear.
So, quadrilateral ABCD is not formed.

(iii) Let the given points be A(4, 5), B(7, 6), C(4, 3) and D(1, 2)
Then, AB = $$\sqrt{(7-4)^{2}+(6-5)^{2}}$$
⇒ AB = $$\sqrt{(3)^{2}+(1)^{2}}$$
= $$\sqrt{9+1}$$ = $$\sqrt{10}$$

BC = $$\sqrt{(4-7)^{2}+(3-6)^{2}}$$
= $$\sqrt{(-3)^{2}+(-3)^{2}}$$
= $$\sqrt{9+9}$$ = $$\sqrt{18}$$

CD = $$\sqrt{[-1-(-3)]^{2}+(-2-0)^{2}}$$
= $$\sqrt{(-3)^{2}+(-1)^{2}}$$
= $$\sqrt{9+1}$$ = $$\sqrt{10}$$

and DA = $$\sqrt{(4-1)^{2}+(5-2)^{2}}$$
= $$\sqrt{(3)^{2}+(3)^{2}}$$
= $$\sqrt{9+9}$$ = $$\sqrt{18}$$
Here, we have
AB = CD = $$\sqrt{10}$$
And, BC = DA = $$\sqrt{18}$$ ⇒ Opposite sides of quadrilateral are equaL
Also, AC = $$\sqrt{(4-4)^{2}+(3-5)^{2}}$$
= $$\sqrt{0+(-2)^{2}}$$
= $$\sqrt{0+4}$$ = $$\sqrt{4}$$ = 2

and BD = $$\sqrt{(1-7)^{2}+(2-6)^{2}}$$
= $$\sqrt{(-6)^{2}+(-4)^{2}}$$
= $$\sqrt{36 + 16}$$ = $$\sqrt{52}$$
= AC ≠ BD
Here, sides AB = DC and BC = AD.
Therefore, the quadrilateral formed by given points is a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Solution:
Let the required point be ‘P’ which is on the x-axis, so its ordinate = 0 and abscissa (say) = x.
Therefore, coordinates of the point P are (x, 0).
Let the given points be A(2, -5) and B(-2, 9).
It is given that:
AP = BP
$$\sqrt{(2-x)^{2}+(-5-0)^{2}}$$ = $$\sqrt{[x-(-2)]^{2}+(0-9)^{2}}$$
= $$\sqrt{(2-x)^{2}+(-5)^{2}}$$ = $$\sqrt{(x+2)^{2}+(-9)^{2}}$$
= $$\sqrt{(2)^{2}+(x)^{2}-2(2)(x)+25}$$ = $$\sqrt{(x)^{2}+(2)^{2}+2(x)(2)+81}$$
= $$\sqrt{4+x^{2}-4 x+25}$$ = $$\sqrt{x^{2}+4+4 x+81}$$
= $$\sqrt{x^{2}-4 x+29}$$ = $$\sqrt{x^{2}+4 x+85}$$
Squaring both sides, we get
x2 – 4x + 29 = x2 + 4x + 85
-4x -4x = 85 – 29
-8x = 56
= x = -7
Therefore, the required point 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:
We have, P(2, -3), Q( 10, y) and PQ = 10 units.
Now, PQ2 = (10)2 = 100
(10 – 2)2 + [(y – (-3)]2= 100
(8)2 + (y + 3)2 = 100
64+y2 + 6y + 9 = 100
y2 + 6y – 27 = 0
y2 + 9y – 3y – 27 = o
y(y + 9) -3 (y + 9) = 0
(y + 9)(y – 3) = 0
y + 9 = 0
or y – 3 = 0
y = -9
or y = 3
y = -9, 3
Hence, the required 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:
We have, P(5, 3) and R(x, 6) and Q(0, 1). It is also given that,
PQ = RQ
PQ2 = RQ2
(0 – 5)2 + [1 – (3)]2
= (0 – x)2 + (1 – 6)2
25 + 16 = x2 + 25
x2 = 16
x = ±4
Therefore, co-ordinates of R are R(±4, 6).
Now, QR = $$\sqrt{(0 \pm 4)^{2}+(1-6)^{2}}$$
= $$\sqrt{41}$$
PR = $$\sqrt{(\pm 4-5)^{2}+[6-(-3)]^{2}}$$
= $$\sqrt{(4-5)^{2}+81}$$ or $$\sqrt{(-4-5)^{2}+81}$$
= $$\sqrt{82}$$ or $$\sqrt{162}$$ = $$\sqrt{82}$$ or 9$$\sqrt{2}$$

Question 10.
Find the relation between x andy such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Solution:
Let P(x, y), A(3, 6) and B(-3, 4) be the given points.
It is given that,
AP = BP
= $$\sqrt{(x-3)^{2}+(y-6)^{2}}$$
= $$\sqrt{[x-(-3)]^{2}+(y-4)^{2}}$$
= $$\sqrt{(x-3)^{2}+(y-6)^{2}}$$
= $$\sqrt{(x+3)^{2}+(y-4)^{2}}$$
x2 – 6x + 9 + y2 – 12y + 36
= x2 + 6x + 9 + y2 – 8y + 16 (Squaring both sides)
– 6x -6x -12y + 8y + 36 – 16 = 0
-12x – 4y + 20 = 0
-4(3x + y – 5) = 0
3x + y – 5 = 0