Question

given that $A(5,4),B(-3,-2)$ and $C(1,-8)$ are the vertices of a $△ABC$
Find: (i) the slope of median $AD$
(ii) the slope of altitude $BM$.

Answer 1

$\left(i\right)$ Given that $AD$ is the median

We know that Median is the line drawn from a vertex to the midpoint of opposite side.

So, $D$ is the midpoint of $B,C$

The midpoint of $(x_1,y_1)and(x_2,y_2)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Therefore, $D=(\frac{-3+1}{2},\frac{-2-8}{2})$

$\Rightarrow D=(\frac{-2}{2},\frac{-10}{2})$

$\Rightarrow D=(-1,-5)$

So, Now we need the slope of $AD$

The slope of $(x_1,y_1)and(x_2,y_2)=\frac{y_2-y_1}{x_2-x_1}$

Therefore Slope of $AD=\frac{-5-4}{-1-5}=\frac{-9}{-6}$

$\Rightarrow$Slope of median $AD=\frac{3}{2}$

$\left(ii\right)$Slope of altitude $BM$

$\Rightarrow BM\perp AC$

So, $\left(SlopeofBM\right)\times \left(SlopeofAC\right)=-1$

$\Rightarrow \left(SlopeofBM\right)=\frac{-1}{\left(SlopeofAC\right)}$ .........$\left(1\right)$

So, we first find the slope of $AC$

The slope of $(x_1,y_1)and(x_2,y_2)=\frac{y_2-y_1}{x_2-x_1}$

Therefore Slope of $AC=\frac{-8-4}{1-5}=\frac{-12}{-4}$

$\Rightarrow$Slope of $AC=3$

Substituting this in $\left(1\right)$, we get,

$\Rightarrow SlopeofBM=\frac{-1}{3}$

$\Rightarrow$Slope of altitude $BM=\frac{-1}{3}$