Question

$ABCD$ is a parallelogram where $A(x,y),B(5,8),C(4,7)$ and $D(2,-4)$ . Find
(i) the coordinates of $A$.
(ii) the equation of the diagonal $BD$.

Answer 1

i)Let $(a,b)$ be the mid-point of line segment joining $B$ and $D$

$\Rightarrow a=\frac{5+2}{2}=\frac{7}{2}.........\left(1\right)$

$\Rightarrow b=\frac{8-4}{2}=\mathrm{2..........}\left(2\right)$

It is known that diagonals of parallelogram bisect each other. Midpoint of line segment joining $A$ and $C$ will also be the same.

$a=\frac{x+4}{2}........\left(3\right)$

$b=\frac{y+7}{2}.........\left(4\right)$

Equating $\left(1\right)$ and $\left(3\right)$

$\frac{7}{2}=\frac{x+4}{2}$

$7=x+4$

$x=3$

Equating $\left(2\right)$ and $\left(4\right)$

$2=\frac{y+7}{2}$

$4=y+7$

$y=-3$

Hence $A\equiv (3,-3)$

ii)Equation of the diagonal BD

$\frac{y-8}{x-5}=\frac{-4-8}{2-5}$

$\frac{y-8}{x-5}=\frac{-12}{-3}$

$\frac{y-8}{x-5}=4$

$y-8=4x-20$

Equation of $BD$: $4x-y-12=0$