Question

The quadratic equation whose roots are the $x$ and $y$ intercepts of the line passing through $(1,1)$ and making a triangle of area $A$ with the co-ordinate axes is

• A. $x^2+Ax+2A=0$
• B. $x^2-2Ax+2A=0$
• C. $x^2-Ax+2A=0$
• D. $(x-A)(x+A)=0$

Answer 1

The correct option is B $x^2-2Ax+2A=0$

Line passing through $(1,1)$ is
$y-1=m(x-1)$
$y=mx+1-m$
$x$ intercept : $x=\frac{m-1}{m}$ ---(1)
$y=0$
$y$ intercept : $y=1-m$ ---(2)
$x=0$
Given area of $OAB=A=\frac{1}{2}\times \left(OB\right)\left(OA\right)$
$=\frac{1}{2}\times (1-m)\frac{(m-1)}{m}$
$A=\frac{-(m^2-1)}{2m}$ ---(3)
Quadratic equation whose roots are $x,y$ is
$x^2-(x+y)x_1+xy=0$
$x_1^2-\left(\frac{2m-m^2-1}{m}\right)x_1-\frac{(m-1{)}^2}{m}=0$
$x_1^2+\left(\frac{m^2-2m+1}{m}\right)x_1-\frac{(m-1{)}^2}{m}=0$
From 3, we get
$x_1^2-2A{x}_1+2A=0$