Question

Find the equation of the sides of the median of the triangle formed by the joining of points (2,4),(4,6),(-6,-10).

Answer 1

Since, median divides the hint segment into two parts.

Let AD be the median on BC,

BF be the median on AC and EC be the median on AB

coordinates of D,E and F are :

$D=\equiv (\frac{4-6}{2},\frac{6-10}{2}),E\equiv (\frac{2+4}{2},\frac{4+6}{2})$

$F\equiv (\frac{2+(-6)}{2},\frac{4+(-10)}{2})$

$D\equiv (-1,-2),E\equiv (3,5),F\equiv (-2,-3)$

Equation of line : $(y-y_1)=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$

Equation of $AD$:

$A\underset{x_1,y_1}{(2,4)},D\underset{x_2,y_2}{(-1,-2)}$

$(y-4)=\left(\frac{-2-4}{-1-2}\right)(x-2)$

$(y-4)=\left(\frac{-6}{-3}\right)(x-2)$

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

$\Rightarrow y-4=2x-4$

$AD\Rightarrow 2x-y=0$

Equation of $BF:$

$B(4,6),F(-2,-3)$

$(y-6)=\left(\frac{-3-6}{-2-4}\right)(x-4)$

$\Rightarrow (y-6)=\frac{-9}{-6}(x-4)$

$\Rightarrow 2(y-6)=3(x-4)\Rightarrow 2y-12=3x-12$

$BF\Rightarrow 3x-2y=0$

Equation of $EC:$

$(3,5),C(-6,-10)$

$(y-5)=\left(\frac{-10-5}{-6-3}\right)(x-3)$

$\Rightarrow 3y-15=5x-15$

$\Rightarrow (y-5)=\frac{-15}{-9}(x-3)$

$EC\Rightarrow 5x-3y=0$