Question

The equation of the medians of a triangle formed by the lines $x+y-6=0,x-3y-2=0$ and $5x-3y+2=0$, is

• A. $x=2,x+9y-14=0$ and $7x-9y-2=0$
• B. $x=2,x-9y-14=0$ and $7x-9y+2=0$
• C. $x=2,x+9y+14=0$ and $x-9y-2=0$
• D. $x=2,x-9y-14=0$ and $7x+9y+2=0$

Answer 1

The correct option is A $x=2,x+9y-14=0$ and $7x-9y-2=0$

Let the given equation form a triangle $\Delta ABC$ whose sides equations are:
$AB:x+y-6=0...\left(i\right)$
$BC:x-3y-2=0...\left(ii\right)$
$CA:5x-3y+2=0...\left(iii\right)$
So intersection point of $AB,BC$ is $B(5,1)$
similarly for $BC,CA$ and $CA,AB$ we get $C(-1,-1)$ and $A(2,4)$ respectively
Let $G$ be the centriod, then all the medians will pass through $G$ only and $G\equiv (2,\frac{4}{3})$
So equation of the median passing through $A$ will be
$x=2$
equation of the median passing through $B$ will be
$y-\frac{4}{3}=\frac{4/3-1}{2-5}(x-2)\Rightarrow x+9y=14$
equation of the median passing through $C$ will be
$y+1=\frac{4/3+1}{2+1}(x+1)\Rightarrow 7x-9y=2$