Question

Find the equations of the medians of a triangle, the equations of whose sides are :
$3x+2y+6=0,2x-5y+4=0$ and $x-3y-6=0.$

Answer 1

The equations of the sides of a triangle are given by the following equations

$3x+2y+6=0⟶\left(1\right)$

$2x-5y+4=0⟶\left(2\right)$

$x-3y-6=0⟶\left(3\right)$

Solving the equations taking two at a time we can find the vertices of the triangle whose sides are given by $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$

Solving $\left(1\right)$ (2)$

$3x+2y+6=0$ and $2x-5y+4=0$

we get, $x=-2$ and $y=0$

Solving $\left(2\right)$ & $\left(3\right)$

$x-3y-6=0$

$2x-5y+4=0$ } $\Rightarrow$ $x=-42$ and $y=-16$

Solving $\left(3\right)$ & $\left(1\right)$

$x-3y-6=0$

$3x+2y+6=0$ } $\Rightarrow$ $x=\frac{-6}{11}$ and $y=\frac{-24}{11}$

So, the vertices are

$A(-2,0)$ $B=(-42,-16)$ $C=(\frac{-6}{11},\frac{-24}{11})$

let the midpoint of the side $AB$ be $P$, which has coordinates $(-22,-8)$

let the midpoint of the side $BA$ be $Q$, which has coordinates $(\frac{-468}{11},\frac{-200}{11})$

let the midpoint of the side $CA$ be $R$, which has coordinates $(\frac{-28}{11},\frac{-24}{11})$

Now the equation of a line in the two-point form is given by

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

So the equation of the median $PC$ is given by : $y+\frac{24}{11}=\frac{-8+\frac{24}{11}}{-22+\frac{6}{11}}(x+\frac{6}{11})$

$\Rightarrow 11y+24=\frac{64}{236}(11x+6)$

$\Rightarrow 11y+24=\frac{16}{59}(11x+6)$

$\Rightarrow 176x-649y-1032=0$

the equation of the median $RB$ is given by

$y+\frac{24}{11}=\frac{-16+24/11}{-42+28/11}(x+\frac{28}{11})\Rightarrow 2387y+5208=836x+2128$

$\Rightarrow 836x-2387y-3080=0$

the equation of the median $AQ$ is given by

$y-0=\frac{-200/11}{\frac{-468}{11}+2}(x+2)$

$\Rightarrow 100x-223y+200=0$