Question

If the quadrilateral formed by the lines $ax+by+c=0$, $a^{\prime }x+b^{\prime }y+c=0,ax+by+c^{\prime }=0,a^{\prime }x+b^{\prime }y+c^{\prime }=0$ have perpendicular diagonals, then

• A. $b^2+c^2=b^{\prime 2}+c^{\prime 2}$
• B. $c^2+a^2=c^{\prime 2}+a^{\prime 2}$
• C. $a^2+b^2=a^{\prime 2}+b^{\prime 2}$
• D. None of these

Answer 1

The correct option is C $a^2+b^2=a^{\prime 2}+b^{\prime 2}$

Since the diagonals are perpendicular, so the given quadrilateral is a rhombus.

So, the distances between two pairs of parallel sides are equal.

Hence,
$\left|\frac{c-c^{\prime }}{\sqrt{a^2+b^2}}\right|=\left|\frac{c-c^{\prime }}{\sqrt{{a^{\prime }}^2+{b^{\prime }}^2}}\right|$

Since the distance between two parallel lines
$ax+by+c_1=0$ and $ax+by+c_2=0$ is $\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|$

Thus $a^2+b^2={a^{\prime }}^2+{b^{\prime }}^2.$