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  • Let  be a straight line passing through the origin and <img alt="\mathrm{ L_2}" src="http

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

Option: 1

x+y=0


Option: 2

2x-y=0


Option: 3

x+7 y=0


Option: 4

x-7y=0


Answers (1)

best_answer

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

\mathrm{\begin{aligned} & x^2+y^2-x+3 y=0 \\ \Rightarrow & x^2+m^2 x^2-x+3 m x=0 \\ \Rightarrow & x\left[x\left(1+m^2\right)-(1-3 m)\right]=0 \\ \therefore & x=0, x=\frac{1-3 m}{1+m^2} \text { and } y=0, y=m\left(\frac{1-3 m}{1+m^2}\right) \\ \therefore & \text { Intercept }=\sqrt{\left(\frac{1-3 m}{1+m^2}\right)^2+m^2\left(\frac{1-3 m}{1+m^2}\right)^2} \\ = & \frac{1-3 m}{1+m^2} \sqrt{1+m^2}=\frac{1-3 m}{\sqrt{1+m^2}} \end{aligned}}

Similarly, for the line x + y = 1 or y = 1 – x

\mathrm{\begin{aligned} & x^2+(1-x)^2-x+3(1-x)=0 \\ & \Rightarrow \quad x^2+x^2-2 x+1-x+3-3 x=0 \\ & \Rightarrow \quad 2 x^2-6 x+4=0 \\ & \Rightarrow \quad x^2-3 x+2=0 \Rightarrow(x-1)(x-2)=0 \\ & \therefore \quad x=1,2 \text { and } y=0,-1 \\ & \therefore \quad \text { Intercept }=\sqrt{(1-2)^2+(0+1)^2}=\sqrt{2} \\ & \text { Given that, } \sqrt{2}=\frac{1-3 m}{\sqrt{1+m^2}} \end{aligned}}

\mathrm{\begin{aligned} & \Rightarrow \quad 2+2 m^2=9 m^2-6 m+1 \Rightarrow 7 m^2-6 m-1=0 \\ & \Rightarrow \quad(7 m+1)(m-1)=0 \\ & \Rightarrow m=1,-\frac{1}{7} \\ & \therefore \quad \text { Lines are } y=x \text { or } y=-\frac{1}{7} x \\ & \Rightarrow x-y=0 \text { or } x+7 y=0 \end{aligned}}

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