Question

If the coordinates of the vertices of $\Delta ABC$ be (-1, 6), (-3, -9) and (5, -8) respectively then the equation of the median through C is:
a). 13x – 14y - 47 = 0
b). 13x – 14y + 47 = 0
c). 13x + 14y + 47 = 0
d). 13x + 14y – 47 = 0

Answer 1

Hint: First of all find the centroid of the triangle because the median of the triangle from any point passes through the centroid. As we have to find an equation of a median passing through point C so we have a point C and the centroid. So, we can write the equation of a line if two points are given.

Complete step-by-step solution -
The triangle ABC, the median passing through C and the centroid E are shown below:

Firstly, we are going to find the centroid of the $\Delta ABC$. So, the formula of the centroid of $\Delta ABC$ is shown below:
$E(x,y)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$
Here $ x_1, x_2, x_3, y_1, y_2, y_3 $ are the x and y coordinates of A, B and C respectively. $ A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) $.
Moving on we are putting the values of A (-1, 6), B (-3, -9) and C (5, -8) in the formula of centroid.
$\begin{align}
  & E\left( x,y \right)=\left( \dfrac{-1-3+5}{3},\dfrac{6-9-8}{3} \right) \\
 & \Rightarrow E\left( x,y \right)=\left( \dfrac{1}{3},-\dfrac{11}{3} \right) \\
\end{align}$
Now, we have got the centroid of $\Delta ABC$ and the point C is already given so we can write the equation of a straight line passing through point C.
We have point C (5, -8) and centroid $E\left( \dfrac{1}{3},-\dfrac{11}{3} \right)$. So, for finding the equation of a line passing through these points we need the slope of the line and a point lies on that line.
The formula for slope of the line passing through two points $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ is as follows:
$m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
The important point in the above equation is that these $ x_1, x_2, y_1, y_2 $ are not the ones that we have described above for the centroid. Now, taking E as P and C as Q, the slope of the median is:
$\begin{align}
  & m=\dfrac{-8+\dfrac{11}{3}}{5-\dfrac{1}{3}} \\
 & \Rightarrow m=\dfrac{-13}{14} \\
\end{align}$
From the above calculation, we have found the slope of the equation of the median. Now, we need a point and it is given that median is passing through point C so we have the point C (5, -8) and to write an equation of a straight line we need a slope and a point that we have shown below:
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
In the above equation, the value of $m=-\dfrac{13}{14}$ and $ (x_1, y_1) $ is the point C (5, -8).
$\begin{align}
  & y+8=-\dfrac{13}{14}\left( x-5 \right) \\
 & \Rightarrow 14\left( y+8 \right)=-13x+65 \\
 & \Rightarrow 13x+14y=65-112 \\
 & \Rightarrow 13x+14y+47=0 \\
\end{align}$
Hence, the equation of the median passing through point C is 13x + 14y + 47 = 0.
Hence, the correct option is (c).

Note: You can also verify the answer that you are getting is correct or not by putting the x and y of the coordinates of point C and the centroid of the triangle in the equation of a straight line that you have obtained. The point C and centroid will always satisfy the equation of a straight line that we have derived.