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 given equations are as follows:

3x + 2y + 6 = 0 ... (1)

2x − 5y + 4 = 0 ... (2)

x − 3y − 6 = 0 ... (3)

In triangle ABC, let equations (1), (2) and (3) represent the sides AB, BC and CA, respectively.

Solving (1) and (2):
x = −2, y = 0

Thus, AB and BC intersect at B (−2, 0).

Solving (1) and (3):
x = $-\frac{6}{11}$, y = $-\frac{24}{11}$

Thus, AB and CA intersect at $A\left(-\frac{6}{11},-\frac{24}{11}\right)$.

Similarly, solving (2) and (3):
x = −42, y = −16

Thus, BC and CA intersect at C (−42, −16).

Let D, E and F be the midpoints the sides BC, CA and AB, respectively.Then,
Then, we have:

$D=\left(\frac{-{\displaystyle 2}-42}{2},\frac{{\displaystyle 0}-16}{2}\right)=\left(-22,-8\right)$

$E=\left(\frac{-{\displaystyle \frac{6}{11}}-42}{2},\frac{-{\displaystyle \frac{24}{11}}-16}{2}\right)=\left(\frac{-{\displaystyle 234}}{11},-\frac{100}{11}\right)$

$F=\left(\frac{-{\displaystyle \frac{6}{11}}-2}{2},\frac{-{\displaystyle \frac{24}{11}}+0}{2}\right)=\left(-\frac{14}{11},-\frac{12}{11}\right)$



Now, the equation of median AD is

$$\begin{gathered} y+\frac{24}{11}=\frac{-8+{\displaystyle \frac{24}{11}}}{-22+{\displaystyle \frac{6}{11}}}\left(x+\frac{6}{11}\right) \\ \Rightarrow 16x-59y-120=0 \end{gathered}$$

The equation of median BE is

$$\begin{gathered} y-0=\frac{-{\displaystyle \frac{100}{11}-0}}{-{\displaystyle \frac{234}{11}+2}}\left(x+2\right) \\ \Rightarrow 25x-53y+50=0 \end{gathered}$$

And, the equation of median CF is

$$\begin{gathered} y+16=\frac{-{\displaystyle \frac{12}{11}+16}}{-{\displaystyle \frac{14}{11}+42}}\left(x+42\right) \\ \Rightarrow 41x-112y-70=0 \end{gathered}$$