Question

$A(3,7);B(5,11),C(-2,8)$ are the vertices of $△ABC$. $AD$ is one of the medians of the triangle. The equation of the median $AD$ is

• A. $5x+3y-36=0$
• B. $5x+3y+36=0$
• C. $3y-5x-36=0$
• D. $3y-5x+36=0$

Answer 1

The correct option is A $5x+3y-36=0$

The given vertices of the $\Delta ABC$ are

$A(x_1,y_1)=(3,7),B(x_2,y_2)=(5,11)$ and $C(x_3,y_3)=(-2,8)$.

The median $AD$ will meet $BD$ at the mid point $M_{BD}$ of $BD$.

So, the equation of $BD$ will be the line passing through $A$ and $M_{BD}$.

The mid piont of $BD$, by the section formula, is

$M_{BD}(x_4,y_4)=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})=(\frac{5-2}{2},\frac{11+8}{2})=(\frac{3}{2},\frac{19}{2})$.

We know that the equation of a line passing through two points $A(x_1,y_1)$ and $M_{BD}(x_4,y_4)$ is $\frac{y-y_1}{y_4-y_1}=\frac{x-x_1}{x_4-x_1}$.

$\therefore$ line $AD\equiv \frac{y-\frac{19}{2}}{7-\frac{19}{2}}=\frac{x-\frac{3}{2}}{3-\frac{3}{2}}$

$\Rightarrow 6y-57=-10x+15$

$\Rightarrow 3y+5x-36=0$