Question

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

Answer 1

Given: Points A(4, 7), B($-$2, 3) and C(0, 1) are the vertices of $△$ABC and
AP, BQ and CR are the medians of the triangle, i.e., P, Q and R are the midpoints of sides BC, AC and AB, respectively.

∴ x-coordinate of P = $\frac{0-2}{2}=\frac{-2}{2}=-1$

Also, y-coordinate of P = $\frac{3+1}{2}=\frac{4}{2}=2$

The coordinates of point P are ($-$1, 2).
Now, slope of the median AP is given below:

$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{7-2}{4-(-1)} \\ =\frac{5}{5} \\ =1 \end{gathered}$$

So, equation of the median AP is given below:
y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$ 2= {x $-$ ($-$1)}
⇒ y $-$2= x +1
⇒ x $-$y +3 = 0

Now, x-coordinate of Q = $\frac{4+0}{2}=2$

Also, y-coordinate of Q = $\frac{1+7}{2}=4$

The coordinates of point Q are (2, 4).
Now, slope of the median BQ is given below:

$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{3-4}{-2-2} \\ =\frac{-1}{-4} \\ =\frac{1}{4} \end{gathered}$$

So, equation of the median CR is given below:
y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$3= $\frac{1}{4}$ {x $-$ ($-$2)}
⇒ 4y $-$12= x +2
⇒ x$-$4y+14=0

Now, x-coordinate of R= $\frac{4-2}{2}=1$

Also, y-coordinate of R = $\frac{7+3}{2}=5$

The coordinates of point R are (1,5).
Now, slope of the median CR is given below:

$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{1-5}{0-1} \\ =4 \end{gathered}$$

So, equation of the median CR is given below:
y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$1 = 4 (x $-$ 0)
⇒ y $-$1= 4x
⇒ 4x$-$y+1=0