Question

The line $\frac{x}{a}+\frac{y}{b}=1$ cust the axes in $A$ and $B$. Another variable line cuts at the axes in $A^{\prime }$ and $B^{\prime }$ such that $OA+OB=O{A}^{\prime }+O{B}^{\prime }$ then prove that the locus of the point of intersection of the lines $A{B}^{\prime }$ and $A^{\prime }B$ is the line $x+y=a+b$

Answer 1

Let the other variable line be $\frac{x}{a^{\prime }}+\frac{y}{b^{\prime }}=1$, where
$a+b=a^{\prime }+b^{\prime }.....\left(1\right)$
$A{B}^{\prime }$ is $\frac{x}{a}+\frac{y}{b^{\prime }}=\mathrm{1....}\left(2\right)$
$A^{\prime }B$ is $\frac{x}{a^{\prime }}+\frac{y}{b}=\mathrm{1....}\left(3\right)$
Subtracting $x(\frac{1}{a}-\frac{1}{a^{\prime }})+y(\frac{1}{b^{\prime }}-\frac{1}{b})=0$
or $x\frac{(a^{\prime }-a)}{a{a}^{\prime }}+y\frac{(b-b)}{b{b}^{\prime }}=0$
or $\frac{x}{a{a}^{\prime }}+\frac{y}{b{b}^{\prime }}=\mathrm{0...}\left(4\right)$
In order to find the locus of their point of intersection, we have to eliminate the variables $a^{\prime },b^{\prime }$ ($a,b$ being fixed) from their equations
Putting for $a^{\prime }$ and $b^{\prime }$ from (2) and (3) in (4)
$\frac{x}{a}(1-\frac{y}{b}).\frac{1}{2}+\frac{y}{b}(1-\frac{x}{a}).\frac{1}{y}=0$
or $\frac{x+y}{ab}=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$
$x+y=a+b$ is the required locus