The equation of the line through the point of intersection of the lines \(3 x-4 y+1=0\) and \(5 x+y-1=0\)…
The equation of the line through the point of intersection of the lines \(3 x-4 y+1=0\) and \(5 x+y-1=0\) and making equal non-zero intercepts on the coordinate axes is
\(2 x+2 y=3\)
\(23 x+23 y=6\)
\(23 x+23 y=11\)
\(2 x+2 y=7\)
Solution
We have, equation of the line through the point of intersection of the lines \(3 x-4 y+1=0\) and \(5 x+y-1=0\) is
\(\begin{array}{lc} & (3 x-4 y+1)+\lambda(5 x+y-1)=0 \\ \Rightarrow & (3+5 \lambda) x+(\lambda-4) y=(\lambda-1) \\ \Rightarrow \quad & \frac{x}{\frac{\lambda-1}{3+5 \lambda}}+\frac{y}{\frac{\lambda-1}{\lambda-4}}=1\end{array}\)
According to given information
\(\begin{array}{rlrl}
\frac{\lambda-1}{3+5 \lambda} & =\frac{\lambda-1}{\lambda-4} \text { and } \lambda \neq 1 \\
\Rightarrow & \lambda-4 =5 \lambda+3 \\
\Rightarrow & 4 \lambda =-7 \Rightarrow \lambda=-\frac{7}{4}
\end{array}\)
So, equation of required line is
\(\begin{array}{rlrlrl}
& \left(3-\frac{35}{4}\right) x+\left(-\frac{7}{4}-4\right) y & =\left(-\frac{7}{4}-1\right) \\
& \Rightarrow -\frac{23}{4} x-\frac{23}{4} y =-\frac{11}{4} \\
& \Rightarrow 23 x+23 y =11
\end{array}\)
Hence, option (3) is correct.