Find the condition that the diagonals of a parallelogram formed by $ax+by+c=0$.

Question

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?

Answer 1

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

$$a^2+b^2 = (a')^2+(b')^2$$

Answer 2

$ax+by+c=0$ ...(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 ${a^{2}+b^{2}}={a'^{2}+b'^{2}}$