Question

Three vertices of a parallelogram $ABCD$ taken in order are $A(3,6),B(5,10)$ and $C(3,2)$. Find the equation of the side $AD$ of the parallelogram $ABCD$.

Answer 1

Given three coordinates of parallelogram-

$A(3,6)$

$B(5,10)$

$C(3,2)$

Now,

Mid-point of $AC$ will be equal to mid-point of $BD$.

Therefore,

Mid-point of $AC=(\frac{3+3}{2},\frac{6+2}{2})=(3,4)$

Mid-point of $BD=(\frac{x+5}{2},\frac{y+10}{2})$

Now,

$\frac{x+5}{2}=3\Rightarrow x=1$

$\frac{y+10}{2}=4\Rightarrow y=-2$

Hence the coordinate of fourth point will be $(1,-2)$.

Therefore, equation of line $AD$ will be-

$(y-6)=\frac{-2-6}{1-3}(x-3)$

$y-6=4(x-3)$

$y-6=4x-12$

$4x-y=6$

Hence the equation of line $AD$ will be-

$4x-y=6$