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

$L_1+\lambda l_2=0$ is the equation of line passing through two lines,$L_1$ and $L_2$

$therefore(2x+y-1)+\lambda (x+3y-2)=0$ is the required equation . ...(i)

or$x(2+\lambda )+y(1+3\lambda )-1-2\lambda =0$

$\frac{x}{\frac{1+2\lambda }{2+\lambda }}+\frac{4}{\frac{1+2\lambda }{1+3\lambda }}=1$

Area of $Delta=\frac{1}{2}\times OB\times OA$

$\frac{8}{3}=\frac{1}{2}\times$ (y intercept) $\times$ (x intercept)

$\frac{8}{3}=\frac{1}{2}\times \left(\frac{1+2\lambda }{1+3\lambda }\right)\times \left(\frac{1+2\lambda }{2+\lambda }\right)$

$\frac{16}{3}=\frac{1+4{\lambda }^2+4\lambda }{2+3{\lambda }^2+7\lambda }$

$30+48{\lambda }^2+112\lambda =-3-12{\lambda }^2-12\lambda$

$60{\lambda }^2+12\lambda +35=0$

$\lambda =\frac{-124\pm \sqrt{(124{)}^2-4\times 60\times 35}}{2\times 60}$

$=\frac{-124\pm \sqrt{15376-8400}}{120}$

$Approximately=1$

$\therefore$ Substituting in (i) $\Rightarrow 2x+4y-3=0,$

$12x+y-3=0$