Question

A variable line passes through a fixed point $(a,b)$ and meets the coordinates axes in $A$ and $B.$ The locus of the point of intersection of lines through A, B parallel to coordinate axes is-

• A. $\frac{x}{a}+\frac{y}{b}=2$
• B. $\frac{a}{x}+\frac{b}{y}=1$
• C. $\frac{x}{a}+\frac{y}{b}=1$
• D. $\frac{x}{a}+\frac{y}{b}=3$

Answer 1

The correct option is B $\frac{a}{x}+\frac{b}{y}=1$

$Letslopeoflinebemthenequationofliney-b=m(x-a)atpointAy=0\therefore -b=m(x-a)orx=\left(\frac{-b}{m}\right)+aatpointBx=0\therefore y-b=m(0-a)y=-ma+b\therefore h=\left(\frac{-b}{m}\right)+a\to \left(1\right)andK=-ma+borm=\left(\frac{b-K}{a}\right)usevalueofminequation\left(1\right),wegeth=\left(\frac{-b}{\left(\frac{b-K}{a}\right)}\right)+ah=\left(\frac{-ab}{b-K}\right)+ah=\left(\frac{-ab+a(b-k)}{b-k}\right)hb-hK=-ab+ab-aK\left(\frac{hb-hK}{hK}\right)=\left(\frac{-aK}{hK}\right)or\left(\frac{b}{K}\right)-1=\left(\frac{-a}{h}\right)or\left(\frac{a}{h}\right)+\left(\frac{b}{K}\right)=1or\left(\frac{a}{x}\right)+\left(\frac{b}{y}\right)=1$