Question

If $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{c}+\frac{y}{d}=1$ intersect the axes at four concyclic points and $a^2+c^2=b^2+d^2$, then these lines can intersect at $(a,b,c,d>0)$

• A. $(1,1)$
• B. $(1,-1)$
• C. $(2,-2)$
• D. $(3,3)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(1,1)$
B $(1,-1)$
C $(2,-2)$
D $(3,3)$

$A,B,C$ and $D$ are concyclic then $ac=bd$
Intersection of $\frac{x}{a}+\frac{y}{b}=1$ ...(1) and

$\frac{x}{c}+\frac{y}{d}=1$ ...(2) is

$(\frac{ac(d-b)}{ad-bc},\frac{bd(a-c)}{ad-bc})$
Given:

$a^2+c^2=b^2+d^2$ $⟹a^2+c^2-2ac=b^2+d^2-2bd$
$⟹(a-c{)}^2=(b-d{)}^2$ $⟹(a-c)=\pm (b-d)$
$\therefore$ Points of intersection can lie on the these lines $x=y$ or $x=-y$

Points can be $(1,1),(1,-1),(2,-2)$ and $(3,3)$

Hence, options 'A', 'B', 'C' and 'D' are correct.