Question

Find the equation of the straight line which passes through the point of intersection of the lines 3x − y = 5 and x + 3y = 1 and makes equal and positive intercepts on the axes.

Answer 1

The equation of the straight line passing through the point of intersection of 3x − y = 5 and x + 3y = 1 is

3x − y − 5 + λ(x + 3y − 1) = 0

$\Rightarrow$(3 + λ)x + (−1 + 3λ)y − 5 − λ = 0 ... (1)
$\Rightarrow y=-\left(\frac{3+\lambda }{-1+\lambda }\right)x+\frac{5+\lambda }{-1+\lambda }$

The slope of the line that makes equal and positive intercepts on the axis is −1.

From equation (1), we have:

$$\begin{gathered} -\frac{3+\lambda }{-1+3\lambda }=-1 \\ \Rightarrow \lambda =2 \end{gathered}$$

Substituting the value of λ in (1), we get the equation of the required line.

$$\begin{gathered} \Rightarrow \left(3+2\right)x+\left(-1+6\right)y-5-2=0 \\ \Rightarrow 5x+5y-7=0 \end{gathered}$$