Question

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight line $x+y=1$. If the intercepts made by the circle $x^2+y^2-x+3y=0$ on $L_1$ and $L_2$ are equal, then which of the following equations can represent $L_1$

• A. $x+y=0$
• B. $x-y=0$
• C. $x+7y=0$
• D. $x-7y=0$

Answer 1

Answer: (B), (C)

$x-y=0$
$x+7y=0$

Let the equation of the line passing through origin be $y=mx$. Therefore

$x^2+y^2-x+3y=0\Rightarrow x^2+m^2{x}^2-x+3mx=0$

$\Rightarrow x[x(1+m^2)-(1-3m)]=0$

$\therefore x=0,x=\frac{1-3m}{1+m^2}$ and

$y=0,y=m\frac{1-3m}{1+m^2}$

$\therefore$ Intercept $=\sqrt{{\left(\frac{1-3m}{1+m^2}\right)}^2+m^2{\left(\frac{1-3m}{1+m^2}\right)}^2}=\frac{1-3m}{1+m^2}\sqrt{1+m^2}=\frac{1-3m}{\sqrt{1+m^2}}$

Similarly for the line $x+y=1\Rightarrow y=1-x$

$x^2+{(1-x)}^2-x+3(1-x)=0\Rightarrow x^2+x^2-2x+1-x+3-3x=0$

$\therefore x=1,2$ and $y=0,-1$

$\therefore$ Intercept $\sqrt{{(1-2)}^2+{(0+1)}^2}=\sqrt{2}$

Given that $2+2m^2=9m^2-6m+1\Rightarrow 7m^2-6m-1=0$

$\Rightarrow (7m+1)(m-1)=0\Rightarrow m=1,-\frac{1}{7}$

$\therefore$ Lines are $y=x,y=-\frac{1}{7}x$ or $x-y=0,x+7y=0$