Question

Find the equation of the straight line passing through the point of intersection of 2x + 3y + 1 = 0 and 3x − 5y − 5 = 0 and equally inclined to the axes.

Answer 1

The equation of the straight line passing through the points of intersection of 2x + 3y + 1 = 0 and 3x − 5y − 5 = 0 is

2x + 3y + 1 + λ(3x − 5y − 5) = 0

$\Rightarrow$(2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0
$\Rightarrow y=-\left(\frac{2+3\lambda }{3-5\lambda }\right)-\left(\frac{1-5\lambda }{3-5\lambda }\right)$

The required line is equally inclined to the axes. So, the slope of the required line is either 1 or −1.

$$\begin{gathered} \therefore -\frac{2+3\lambda }{3-5\lambda }=1\mathrm{and}-\frac{2+3\lambda }{3-5\lambda }=-1 \\ \Rightarrow -2-3\lambda =3-5\lambda \mathrm{and}2+3\lambda =3-5\lambda \\ \Rightarrow \lambda =\frac{5}{2}\mathrm{and}\frac{1}{8} \end{gathered}$$

Substituting the values of λ in (2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0, we get the equations of the required lines.

$$\begin{gathered} \left(2+\frac{15}{2}\right)x+\left(3-\frac{25}{2}\right)y+1-\frac{25}{2}=0\mathrm{and}\left(2+\frac{3}{8}\right)x+\left(3-\frac{5}{8}\right)y+1-\frac{5}{8}=0 \\ \Rightarrow 19x-19y-23=0\mathrm{and}19x+19y+3=0 \end{gathered}$$