Question

In triangle $ABC$, the coordinates of vertices $A$, $B$ and $C$ are $(4,7)$, $(-2,3)$ and $(0,1)$ respectively. Find the equations of the medians passing through the vertices $A$, $B$ and $C$.

Answer 1

Given, $A(4,7),B(-2,3),C(0,1)$ are the vertices of $\Delta ABC$.

Let $AD,BE$ and $CF$ are the medians through point $A,B,C$ respectively.

$AD$ is the median.

$D$ is the midpoint of $BC$.

By midpoint formula,

Coordinate of $D=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{-2+0}{2},\frac{3+1}{2})$

$=(\frac{-2}{2},\frac{4}{2})$

$D=(-1,2)$

The equation of median $AD$ where $A=(4,7)$ and $D=(-1,2)$

$x_1{y}_1$ $x_2{y}_2$

The equation of the median is given by,

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

$\frac{x-4}{4+1}=\frac{y-7}{7-2}$

$\frac{x-4}{5}=\frac{y-7}{5}$

$\therefore x-4=y-7$

$x-y=-7+4$

$\therefore x-y=-3$

$\therefore x-y+3=0$

$x-y+3$ is the equation of median AD.

Again, $BE$ is the median.

$E$ is the midpoint of $AC$.

By midpoint formula,

Coordinate of $E=(\frac{4+0}{2},\frac{7+1}{2})$

$=(2,4)$

The equation of median BE is given by,

$\frac{x+2}{-2-2}=\frac{y-3}{3-4}$

$\frac{x+2}{-4}=\frac{y-3}{-1}$

$-x-2=-4y+12$

$-x-2+4y-12=0$

$-x+4y-14=0$

$x-4y+14=0$

$\therefore x-4y+14=0$ is the equation of median $BE$.

Again, $CF$ is the median.

$F$ is the midpoint of $AB$.

By midpoint formula,

Coordinate of $F=(\frac{4-2}{2}+\frac{7+3}{2})$

$=(1,5)$

The equation of median $CF$ is given by,

$\frac{x-0}{0-1}=\frac{y-1}{1-5}$

$\frac{x}{-1}=\frac{y-1}{-4}$

$-4x=-y+1$

$4x-y+1=0$

$\therefore 4x-y+1=0$ is the equation of median $CF$.