Question

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.

Answer 1

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

$\Rightarrow \frac{x}{{\displaystyle \frac{1+2\lambda }{2+\lambda }}}+\frac{y}{{\displaystyle \frac{1+2\lambda }{1+3\lambda }}}=1$

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

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

$$\begin{gathered} \frac{1}{2}\left|\frac{1+2\lambda }{2+\lambda }\times \frac{1+2\lambda }{1+3\lambda }\right|=\frac{3}{8} \\ \Rightarrow 3\left|3{\lambda }^2+7\lambda +2\right|=4\left|4{\lambda }^2+4\lambda +1\right| \\ \Rightarrow 9{\lambda }^2+21\lambda +6=16{\lambda }^2+16\lambda +4 \\ \Rightarrow 7{\lambda }^2-5\lambda -2=0 \\ \Rightarrow \lambda =1,-\frac{2}{7} \end{gathered}$$

Hence, the equations of the required lines are

$$\begin{gathered} 3x+4y-1-2=0\mathrm{and}\left(2-\frac{2}{7}\right)x+\left(1-\frac{6}{7}\right)y-1+\frac{4}{7}=0 \\ \Rightarrow 3x+4y-3=0\mathrm{and}12x+y-3=0 \end{gathered}$$