Question

Let the line $\frac{x}{a}+\frac{y}{b}=1$ cuts the $x$ and $y$ axes at $A$ and $B$ respectively. Now a line parallel to the given line cuts the coordinate axis at $P$ and $Q$ and points $P$ and $Q$ are joined to $B$ and $A$ respectively. The locus of intersection of the joining lines is

• A. $\frac{x}{a}-\frac{y}{b}=0$
• B. $\frac{x}{a}+\frac{y}{b}=0$
• C. $\frac{x}{b}-\frac{y}{a}=0$
• D. $\frac{x}{b}+\frac{y}{a}=0$

Answer 1

The correct option is A $\frac{x}{a}-\frac{y}{b}=0$

$\frac{x}{a}+\frac{y}{b}=1$

It meets $x$ axis at $A(9,a)$ and $y$ axis at $B(0,b)$

$\frac{x}{a}+\frac{y}{b}=1$

$\to bx+ay=ab$

equation perpendicular to above line is

$ax-by=k$

$\frac{ax}{k}\frac{by}{k}=1$

$\frac{x}{k/a}+\frac{y}{\left(\frac{-k}{b}\right)}=1$

It meets $x$ axis at $p(\frac{k}{a},0)$ and $Q(0,\frac{-k}{b})$

Hence equation of $AQ$ is

$\frac{x}{a}+\frac{y}{\left(\frac{-k}{b}\right)}=1$

$\frac{x}{a}-\frac{by}{k}=1.......\left(i\right)$

$\to$ equation of $PB$ is

$\frac{x}{\left(\frac{k}{a}\right)}+\frac{y}{b}=1$

$\frac{ax}{k}+\frac{y}{b}=1.......\left(ii\right)$

$\to$Let $(x,2)$ be point of intersection of $AQ$

$\frac{x}{a}-\frac{bv}{k}=1....\left(iii\right)$

$\frac{ax}{k}+\frac{v}{b}=1$

$\to \frac{ay}{k}=1-\frac{v}{b}\to k=\frac{ab.x}{b-v}$

Put this in equation $\left(iii\right)$

$\frac{x}{a}-\frac{bv}{\left(\frac{abx}{b-v}\right)}=1$

$\frac{x}{a}-\frac{bv(b-v)}{abx}=1$

$\frac{x}{a}-\frac{v(b-v)}{ax}=1\to x^{2}vb+v^2=ax$

$x^2+v^2-ax+b=0$

Hence equation of locus

$x^2+y^2-ax-by=0$