Question

Find the equation of the straight line drawn through the point of intersection of the lines x + y = 4 and 2x − 3y = 1 and perpendicular to the line cutting off intercepts 5, 6 on the axes.

Answer 1

The equation of the straight line passing through the point of intersection of x + y = 4 and 2x − 3y = 1 is
x + y − 4 + λ(2x − 3y − 1) = 0

\[\Rightarrow\] (1 + 2λ)x + (1 − 3λ)y − 4 − λ = 0   ... (1)

\[\Rightarrow y = - \left( \frac{1 + 2\lambda}{1 - 3\lambda} \right)x + \frac{4 + \lambda}{1 - 3\lambda}\] 

The equation of the line with intercepts 5 and 6 on the axis is

\[\frac{x}{5} + \frac{y}{6} = 1\]              ... (2)

The slope of this line is \[- \frac{6}{5}\].

The lines (1) and (2) are perpendicular.

\[\therefore - \frac{6}{5} \times \left( - \frac{1 + 2\lambda}{1 - 3\lambda} \right) = - 1\]

\[ \Rightarrow \lambda = \frac{11}{3}\]

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

\[\Rightarrow \left( 1 + \frac{22}{3} \right)x + \left( 1 - 11 \right)y - 4 - \frac{11}{3} = 0\]

\[ \Rightarrow 25x - 30y - 23 = 0\].