Find the condition that the diagonals of a parallelogram formed by $ax+by+c=0$. Find the condition that the.diagonals of a parallelogram formed by $ax+by+c=0$, $ax+by+c'=0$, $a'x+b'y+c=0$ and $a'x+b'y+c'=0$ are at right angles.
My Attempt:
The equation of diagonal passing through the point of intersection of $ax+by+c=0$ and $a'x+b'y+c=0$ is 
$$(ax+by+c)+ K(a'x+b'y+c)=0$$
Where $K$ is any arbitrary constant.
Again, The equation of the diagonal passing through the point of intersection of $ax+by+c=0$ and $a'x+b'y+c'=0$ is
$$(ax+by+c)+L(a'x+b'y+c')=0$$
Where $L$ is any arbitrary constant.
How do I complete the rest?
 A: For two parallel lines in the $xy$-plane given by the equations
\begin{align*}
Ax+By+C_1 = 0\\[4pt]
Ax+By+C_2 = 0\\[4pt]
\end{align*}
the distance between them is given by the formula
$$\frac{\left|C_2 - C_1\right|}{\sqrt{A^2+B^2}}$$
(see https://en.wikipedia.org/wiki/Distance_between_two_straight_lines)
But given a parallelogram,
$$\text{the diagonals are perpendicular}$$
$$\text{if and only if}$$ 
$$\text{the parallelogram is a rhombus}$$
$$\text{if and only if}$$ 
$$\text{the distance between the pairs of opposite sides are equal}$$
Applying the above to the lines specified for the edges of your parallelogram, you get
$$
\frac{\left|c - c'\right|}{\sqrt{a^2+b^2}} 
= 
\frac{\left|c - c'\right|}{\sqrt{(a')^2+(b')^2}}$$
which yields the condition
$$a^2+b^2 = (a')^2+(b')^2$$
A: $ax+by+c=0$ ...(1)
$a'x+b'y+c=0$ ...(2)
$ax+by+c'= 0$ ...(3)
$a'x+b'y+c'=0 $...(4)
Let P be the intersection of (1) and (2)
Equation of a line passing through P is given by
$ ax+by+c +K(a'x+b'y+c) = 0$ ...(5) 
Let R be the intersection of (3) and (4) for which $ ax+by=-c'$ and $a'x+b'y=-c'$
Substituting in (5) $(c-c')+K(c-c') = 0$  or, $K = -1 $
Hence, from (5), $(a-a')x+(b-b')y=0$ ...(6). 
This is the equation of the diagonal PR.
Let S be the intersection of (1) and (4)
Equation of a line passing through S is given by
$ax+by+c + L(a'x+b'y+c') = 0 $ …….(7) 
Let Q be the intersection of (2) and (3) for which $ax+by = -c'$ and $a'x+b'y = -c$
Substituting these in (5) $c-c'+L(c'-c) = 0$ or, $L=1 $
Substituting in (7), $(a+a')x+(b+b')y+c+c'=0$ ...(8). This is the equation of the diagonal QS.
Now slope of PR  is $$\frac{-(a-a')}{(b-b')}$$
slope of QS   is $$\frac{-(a+a')}{(b+b')}$$
Since the diagonals are perpendicular,
Product of slopes=$$\frac{-(a-a')}{(b-b')}*\frac{-(a+a')}{(b+b')} = \frac{a^{2}-a'^{2}}{b^{2}-b'^{2}}=-1$$
Hence ${a^{2}+b^{2}}={a'^{2}+b'^{2}}$
