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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?

pi-π
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  • Firstly, $K,L$ are unknown constants, not "arbitrary" constants. Putting that minor nitpick aside, since the diagonals are perpendicular, the normal vectors for those two lines need to be perpendicular. That should clinch it. – quasi Apr 18 '17 at 12:53
  • @quasi, where does vector emerge for co-ordinates solution? – pi-π Apr 18 '17 at 12:56
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    A line with equation of the form $Ax + By + C = 0$ has normal vector ${\langle}A,B{\rangle}$. – quasi Apr 18 '17 at 12:57
  • @quasi, I have never dealt cordinates using vectors. Could you please elaborate? – pi-π Apr 18 '17 at 12:58
  • It's should be in your textbook. But this may not be the right approach since it only gives you one equation. – quasi Apr 18 '17 at 13:00
  • @quasi, Under which topic should I search for it in my textbook? – pi-π Apr 18 '17 at 13:02
  • Which textbook are you using (author, title, edition)? – quasi Apr 18 '17 at 13:03
  • This tedious problem must belong to SL loney. –  Apr 18 '17 at 14:14

2 Answers2

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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

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

quasi
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  • Why are the distance between the pairs of opp. sides equal? – pi-π Apr 18 '17 at 15:53
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    In a rhombus, all sides are equal. The area is equal to the product of any side and the altitude to that side, hence, since the sides are equal, so are the altitudes. – quasi Apr 18 '17 at 15:55
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$ax+by+c=0$ ...(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 ${a^{2}+b^{2}}={a'^{2}+b'^{2}}$

Saurav Bhattacharyya
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