Question

If the coordinates of the vertices of the triangle $\mathrm{ABC}$ be $\left(-1,6\right),\left(-3,-9\right),$ and $\left(5,-8\right)$ respectively, then the equation of the median through $\mathrm{C}$ is:

• A. $13x-14y-47=0$
• B. $13x-14y+47=0$
• C. $13x+14y+47=0$
• D. $13x+14y-47=0$

Answer 1

The correct option is C

$13x+14y+47=0$

Explanation for the correct option.

Step 1: Draw a figure

Given the vertices of the triangle are $A(-1,6),B(-3,-9)$ and $C(5,-8)$.

Step 2: Find the coordinate of median

Let the vertices of the triangle $A(-1,6),B(-3,-9)$ and $C(5,-8)$.

Let, $CD$ be the median.

Then the coordinates of $D$ are:

$\begin{array}{rcl}D& =& \left[\frac{-1+\left(-3\right)}{2},\frac{-9+6}{2}\right]\\ & =& \left(-2,\frac{-3}{2}\right)\end{array}$

Step 3: Find the equation of the median

We know that line passing through $\left(x_1,y_1\right),\left(x_2,y_2\right)$ is given by

$\mathit{y}\mathbf{-}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$.

Equation of line passing through $C\left(5,-8\right)$ and $D\left(-2,-\frac{3}{2}\right)$ is :

$\begin{array}{rcl}\therefore y-\left(-8\right)& =& \frac{{\displaystyle \frac{-3}{2}}-\left(-8\right)}{-2-5}\left(x-5\right)\\ & \Rightarrow & y+8=-\frac{13}{14}\left(x-5\right)\\ & \Rightarrow & 14y+112=-13x+65\\ & \Rightarrow & 13x+14y+47=0\end{array}$

Hence, option C is correct.