Question

Find the equation of a right bisector of the line segment joining the points $(2,-3)$ and $(-1,5)$.

Answer 1

End point of the given line are $(2,-3)$ and $(-1,5)$.

Therefore,

Mid-point of line $=(\frac{2+(-1)}{2},\frac{(-3)+5}{2})=(\frac{1}{2},1)$

Now,

Slope of the given line $=\frac{5-(-3)}{-1-2}=-\frac{8}{3}$

As we know that the product of slope of two perpendicular lines is $-1$

Therefore,

Slope of required line $=\frac{3}{8}$

Therefore,

Equation of line with slope $\frac{3}{8}$ and passing through $(\frac{1}{2},1)$-

$(y-1)=\frac{3}{8}(x-\frac{1}{2})$

$16(y-1)=3(2x-1)$

$16y-16=6x-3$

$\Rightarrow 6x-16y+13=0$

Hence the equation of required line is $6x-16y+13=0$.