Question

The coordinate of a and b are ( -4,8) and (x,3) respectively. If ab=13,then the possible value of x is

Answer 1

Given :

AB=13cm

A(-4,8)

B(x,3)

By distance formula:

$\sqrt{(x2-x1)^2+(y2-y1)^2}=13$

$\sqrt{(x-(-4))^2+(8-3)^2}=13$

Squaring both sides we get:

$\implies(x+4)^2+5^2=13^2$

$\implies (x+4)^2=169-25$

$\implies (x+4)^2=144$

Two cases arise

$\implies x+4=12$

$\implies x=12-4=8$

It can also be -12

$\implies x+4=-12$

$\implies x=-12-4$

$\implies x=-16$

So the values of x are 8 and -16.

Hope it helps.