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{4left(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