Question

Statement 1: If the diagonals of the quadrilateral formed by the lines $px+qy+r=0,p^{\prime }x+q^{\prime }y+r=0$, $px+qy+r^{\prime }=0,p^{\prime }x+q^{\prime }y+r^{\prime }=0$ are at right angles, then $p^2+q^2=p^{\prime 2}+q^{\prime 2}$

Statement 2 : Diagonals of a rhombus are bisected and perpendicular to each other.

Answer 1

The quadrilateral is obviously a parallelogram and if the diagonal are at right angles, it must be a rhombus.

Hence, the distance between the pairs of opposite sides must be the same, i.e.
$\frac{|r-r^{\prime }|}{\sqrt{p^2+q^2}}=\frac{|r-r^{\prime }|}{\sqrt{p^{\prime 2}+q^{\prime 2}}}\Rightarrow p^2+q^2=p^{\prime 2}+q^{\prime 2}$

Hence, Statement 1 is true, Statement 2 is true.
Also, Statement 2 is correct explanation for Statement 1.