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  • Let  be a straight line passing through origin and <img alt="L_{2}" src="https://entrancecorner.onc

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

Option: 1

x + y = 0 


Option: 2

x – y = 0
 


Option: 3

x + 7y = 0


Option: 4

x – 7y = 0


Answers (1)

The intersection of \mathrm{L_2 \equiv x+y-1=0} and circle gives

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

\mathrm{\begin{array}{ll} \text { Or } & x^2-3 x+2=0 \\ \text { Or } & (x-2)(x-1)=0 \quad \Rightarrow \quad x=1,2 \end{array}}

\therefore Points of intersection are A(1, 0), B(2, –1)

\mathrm{ \therefore \quad A B=\sqrt{2}=\text { intercept on } L_2}

Let L_1 be y = mx. Its intersection with circle gives

\mathrm{\begin{aligned} & x^2+(m x)^2-x+3(m x)=0 \\ & \Rightarrow \quad x=0, \frac{1-3 m}{1+m^2} \end{aligned}}

The points of intersection are 0(0, 0) and \mathrm{D\left(\frac{1-3 m}{1+m^2}, \frac{m(1-3 m)}{1+m^2}\right)}

\mathrm{\begin{aligned} & \text { (Intercept } O D)^2=\left(\frac{1-3 m}{1+m^2}\right)^2\left(1+m^2\right) \\ & \therefore O D \quad=\frac{1-3 m}{\sqrt{1+m^2}} \\ & A B=O D \Rightarrow(1-3 m)^2=2\left(1+m^2\right) \\ & (7 m+1)(m-1)=0 \\ & \therefore m=-\frac{1}{7}, 1 \end{aligned}}

Hence, \mathrm{L_1} is either y = x or \mathrm{y=-\frac{1}{7} x}

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Kshitij

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