Find the equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 and making with the coordinate axes a triangle of area \[\frac{3}{8}\] sq. units.

#### Solution

The equation of the straight line passing through the point of intersection of 2x + y − 1 = 0 and x + 3y − 2 = 0 is given below:

2x + y − 1 + λ (x + 3y − 2) = 0

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

\[\Rightarrow \frac{x}{\frac{1 + 2\lambda}{2 + \lambda}} + \frac{y}{\frac{1 + 2\lambda}{1 + 3\lambda}} = 1\]

So, the points of intersection of this line with the coordinate axes are \[\left( \frac{1 + 2\lambda}{2 + \lambda}, 0 \right) \text { and } \left( 0, \frac{1 + 2\lambda}{1 + 3\lambda} \right)\].

It is given that the required line makes an area of \[\frac{3}{8}\] square units with the coordinate axes.

\[\frac{1}{2}\left| \frac{1 + 2\lambda}{2 + \lambda} \times \frac{1 + 2\lambda}{1 + 3\lambda} \right| = \frac{3}{8}\]

\[ \Rightarrow 3\left| 3 \lambda^2 + 7\lambda + 2 \right| = 4\left| 4 \lambda^2 + 4\lambda + 1 \right|\]

\[ \Rightarrow 9 \lambda^2 + 21\lambda + 6 = 16 \lambda^2 + 16\lambda + 4\]

\[ \Rightarrow 7 \lambda^2 - 5\lambda - 2 = 0\]

\[ \Rightarrow \lambda = 1, - \frac{2}{7}\]

\[3x + 4y - 1 - 2 = 0 \text { and } \left( 2 - \frac{2}{7} \right)x + \left( 1 - \frac{6}{7} \right)y - 1 + \frac{4}{7} = 0\]

\[ \Rightarrow 3x + 4y - 3 = 0 \text { and } 12x + y - 3 = 0\]