# Coordinate Geometry

## Distance Formula

### 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 know 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 know 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 know 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