Question

If the coordinates of a point on the y-axis which is equidistant from the points $(13,2)$ and $(12,-3)$ is $(p,q)$, then find the value of $p+q$.

Answer 1

Given Co-ordinates of a point on y -axis equidistant from the points (13,2) and (12, -3) is (p,q)

To find P+q

Sol -$\because$ the point lies on y-axis $\therefore$ i.e. us of the

form (0,y)

$\Rightarrow p=0$

(0,q) is equidistant from (13,2) and (12,-3)

$\sqrt{(0-13{)}^2+(q-2{)}^2}=\sqrt{(0-12{)}^2+(q+3{)}^2}$

$(0-13{)}^2+(q-2{)}^2=(0-12{)}^2+(q+3{)}^2$

$169+q^2+4-4q=144+q^2+9+6q$

$173-4q=153+6q$

$173-153=6q+44q$

$20=10qv$

$2=q$

$\therefore p+q=0+2=2$