Question

One side of a rectangle lies along the line $4x+7y+5=0$. Two of its vertices are $(-3,1)$ and $(1,1)$. Then find the equations of other sides.

Answer 1

Since, the side $AB$ is perpendicular to $AD$

$\therefore$ Its equation if of the form $7x-4y+\lambda =0$

Since, it passes through $(-3,1)$

$\therefore 7(-3)-4\left(1\right)+\lambda =0\Rightarrow \lambda =25$

$\therefore$ Equation of $AB$ is $7x-4y+25=0$

Now, $BC$ is parallel to $AD$

Therefore, its equation is $4x+7y+\lambda =0$

Since, it passes through $(1,1)$.

$\therefore 4\left(1\right)+7\left(1\right)+\lambda =0\Rightarrow \lambda =-11$

$\therefore$ equation of $BC$ is $4x+7y-11=0$

Now,equation of $DC$ is $7x-4y+\lambda =0$

$\Rightarrow 7\left(1\right)-4\left(1\right)+\lambda =0\Rightarrow \lambda =-3$

$\therefore 7x-4y-3=0$