Question

Consider two points $A(1,a)$ and $B(5,b)$. If the equation of the line bisecting the line segment $AB$ perpendicularly is $x-3y=0$ then $|ab|$ is

Answer 1

Given points $A(1,a)$ and $B(5,b)$
Midpoint of line segment $AB$
$=(3,\frac{a+b}{2})$
This point lies on the line $x-3y=0$, so
$3-\frac{3(a+b)}{2}=0\Rightarrow a+b=2\cdots \left(1\right)$
Slope of the line segment $AB$
$m=\frac{b-a}{4}$
Slope of given line is
$=\frac{1}{3}$
So,
$m=-3\Rightarrow \frac{b-a}{4}=-3\Rightarrow a-b=12$
from equation $\left(1\right)$
$a=7,b=-5$
$\therefore |ab|=35$