Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1
Gujarat Board Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1
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
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,
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).
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
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