Class 10 Mathematics

Coordinate Geometry

NCERT Exercise 7.1-Part 1

Question: 1 - Find the distance between the following pairs of points:

(i) (2,3), (4,1)

Solution: Let (x1, y1) = (2, 3) and (x2, y2) = (4, 1)

We now that distance formula

`=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((4-2)^2+(1-3)^2)`

`=sqrt(2^2+-2^2)`

`=sqrt(4+4)`

`=sqrt8=2sqrt2` units


(ii) (-5, 7), (-1, 3)

Solution: Let (x1, y1) = (-5, 7) and (x2, y2) = (-1, 3)

We now that distance formula

`=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((-1+5)^2+(3-7)^2)`

`=sqrt((-4)^2+(-4)^2)`

`=sqrt(16+16)`

`=sqrt(32)=4sqrt2` units


(iii) (a, b), (-a, -b)

Solution: Let (x1, y1) = (a, b) and (x2, y2) = (-a, -b)

We now that distance formula

`=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt((-a-a)^2+(-b-b)^2)`

`=sqrt((-2a)^2+(-2b)^2)`

`=sqrt(4a^2+4b^2)`

`=sqrt(4(a^2+b^2)`

`=2sqrt(a^2+b^2)` units