Question

The equations of perpendiculars of the sides $AB$ and $AC$ of triangle $ABC$ are $x-y-4=0$ and $2x-y-5=0$ respectively. If the vertex $A$ is $(-2,3)$ and point of intersection of perpendiculars bisectors is $(\frac{3}{2},\frac{5}{2})$, find the equation of medians to the sides $AB$ & $AC$ respectively.

Answer 1

Equation of ${}^{\prime }A{B}^{\prime }$ which is ${\perp }_{ar}$ to $x-y-4=0$

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

$\Rightarrow x+y=1$-----(i)

Equation of ${}^{\prime }A{C}^{\prime }$

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

$\Rightarrow 2y-6=-x-2$

$\Rightarrow x+2y=4$-----(ii)

Equation of line $EF$

$y-\frac{5}{2}=2(x-\frac{3}{2})$

$\Rightarrow 4x-2y=1$-----(iii)

Equation of $DF$ is $2x-2y+2=0$------(iv)

Solving (ii) & (iii) $E(1,\frac{3}{2})$

Solving (i) & (iv) $D(0,1)$

$\therefore B(2,-1),C(4,0)$

median of ${}^{\prime }A{B}^{\prime }$ is $x+4y=4$

median of ${}^{\prime }A{C}^{\prime }$ is $5x+2y=8$

Hence, this is the answer.