If A ( - 1, 6), B( - 3, - 9) and C(5, - 8) are the vertices of a ΔABC

Question:

If $A(-1,6), B(-3,-9)$ and $C(5,-8)$ are the vertices of a $\triangle A B C$, find the equations of its medians.

Solution:

Construction: - Draw median from vertices $A, B$ and $C$ on lines $B C, A C$ and

AC respectively .Let the mid - points of lines BC,AC and AB be L,M and N respectively.

Now find the coordinate of L, M and N using mid - point theorem

$(\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)$

coordinates of $L=\left(\frac{-3+5}{2}, \frac{-9+(-8)}{2}\right) \Rightarrow\left(1, \frac{-17}{2}\right)$

coordinates of $M=\left(\frac{-1+5}{2}, \frac{6+(-8)}{2}\right) \Rightarrow(2,-1)$

coordinates of $\mathrm{N}=\left(\frac{-1+(-3)}{2}, \frac{6+(-9)}{2}\right) \Rightarrow\left(-2, \frac{-3}{2}\right)$

Now equation of medians AL, BM and CN using two point form

For median AL

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

$y-6=\frac{\frac{-17}{2}-6}{1-(-1)}(x-(-1))$

$y-6=\frac{\frac{-17-12}{2}}{2}(x+1) \Rightarrow y-6=\frac{-29}{4}(x+1)$

$4(y-6)=-29(x+1)$

$4 y-24+29 x+29=0$

$29 x+y+5=0$

For median BM,

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

$y-(-9)=\frac{-1-(-9)}{2-(-3)}(x-(-3))$

$y+9=\frac{8}{5}(x+3) \Rightarrow 5(y+9)=8(x+3)$

$5 y+45=8 x+24$

$8 x-5 y+24-45=0$

$8 x-5 y-21=0$

For median CN,

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

$y-(-8)=\frac{\frac{-3}{2}-(-8)}{-2-5}(x-5)$

$y+8=\frac{\frac{-3+16}{2}}{2}(x-5) \Rightarrow y+8=\frac{13}{4}(x-5)$

$4(y+8)=13(x-5)$

$4 y+32=13 x-65$

$13 x-4 y-65-32=0$

$13 x-4 y-97=0$

So, the required line of equations for medians are for AL: 29x + y + 5 = 0

For BM: $8 x-5 y-21=0$

For CN: $13 x-4 y-97=0$