The equation of the line through the intersection of $3 x-4 y+1$ and $5 x+y-1=0$ which cuts off equal intercepts on the axes is given by

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Aicharya · Accepted Answer

Equation of family of lines passes through the intersection of lines $3 x-4 y+1=0$ and $5 x+y-1=0$ is $\begin{aligned} & (3 x-4 y+1)+\lambda(5 x+y-1) =0 \ \Rightarrow & (3+5 \lambda) x+(\lambda-4) y+(1-\lambda) =0 \ \Rightarrow & \frac{x}{\frac{\lambda-1}{5 \lambda+3}}+\frac{y}{\lambda-1} =1, \quad\{\lambda \neq 1\} \end{aligned}$ $\because$ The obtain lines makes equal intercepts with the axes, so $\begin{aligned} & \left|\frac{\lambda-1}{5 \lambda+3}\right| =\left|\frac{\lambda-1}{\lambda-4}\right| \ \Rightarrow & |\lambda-4| =|5 \lambda+3| \quad \because \{\lambda \neq 1\} \ \Rightarrow & \lambda =-\frac{7}{4}, \frac{1}{6} \end{aligned}$ $\therefore$ Equation of required lines are $-23 x-23 y+11=0$ and $23 x-23 y+5=0$ or $23 x+23 y-11=0$ and $23 x-23 y+5=0$ Hence, option (a) is correct.

The equation of the line through the intersection of \(3 x-4 y+1\) and \(5 x+y-1=0\) which cuts off equal…

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