Question

If $\left(2,1\right)\left(-1,2\right)\left(3,3\right)$ are the mid points of the sides $BC,CA.ABof\Delta ABC$, then the equation of AB is.

Answer 1

$△,E$ and $F$ are midpoint of $AB,BC$ and $CA$

let equation of $AB$ be $y=mk+c$

$(3,3)$ is on $AB\Rightarrow 3=3m+c---\left(1\right)$

$(3,3)$ is mid point of $(x_1,y_1)$ and $(x_2,y_2)$

$\left(\frac{x_1+x_2}{2}\right)=3\frac{y_1+y_2}{2}=3$

$(x_1+x_2)=6,(y_1+y_2)=6---\left(2\right)$

Similarly,

$(x_2+x_3)=4,(y_2+y_3)=2---\left(3\right)$

$(x_3+x_1)=-2,(y_3+y_1)=4---\left(4\right)$

From $\left(2\right)\left(3\right)$ and $\left(4\right)$

$2(x_1+x_2+x_3)=8,2(y_1+y_2+y_3)=12$

$x_1+x_2+x_3=4$

$x_2+x_3=4$

$\Rightarrow x_1=0$

$y_1+y_2+y_3=6$

$y_2+y_3=2$

$y_1=4$

$\therefore (0,4)$ is point on $y=3x+c$

$\Rightarrow 4=m\left(0\right)+c\Rightarrow c=4---\left(5\right)$

from $\left(5\right)$ ans $\left(1\right)m=-1/3$

$\therefore$ equation of $AB:y=\frac{-x}{3}+4$