NCERT Solutions
Class 10 Maths
Coordinate Geometry
Ex. 7.1 Q.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)
(i) Distance between two points (x1, y1) and (x2, y2) is given as:
√{(x2 – x1)2 + (y2 – y1)2}
Therefore, the distance between two points (2, 3) and (4, 1) is
D = √{(4 – 2)2 + (1 - 3)2}
=> D = √{22 + (-2)2}
=> D = √{4 + 4}
=> D = √8
=> D = 2√2
(ii) Distance between two points (x1, y1) and (x2, y2) is given as:
√{(x2 – x1)2 + (y2 – y1)2}
Therefore, the distance between two points (2, 3) and (4, 1) is
D = √{(-1 + 5)2 + (3 - 7)2}
=> D = √{42 + (-4)2}
=> D = √{16 + 16}
=> D = √32
=> D = 4√2
(iii) Distance between two points (x1, y1) and (x2, y2) is given as:
√{(x2 – x1)2 + (y2 – y1)2}
Therefore, the distance between two points (2, 3) and (4, 1) is
D = √{(-a - a)2 + (-b - b)2}
=> D = √{(-2a)2 + (-2b)2}
=> D = √{4a2 + 4b2}
=> D = 2√(a2 + b2)