Question

Find the equation of the straight line passing through the point of intersection of intersection of $2x+y-1=0$ and $x+3y-2=0$ and making with the coordinate axes a triangle of area $3/8sq.units.$

Answer 1

The equation of straight line passing through the point of intersection of $2x+y-1=0$ and $x+3y-2=0$ is

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

$(2+\lambda )x+(1+3\lambda )y-1-2\lambda =\mathrm{0......................................}\left(i\right)$

$\frac{x}{(1+2\lambda )/(2+\lambda )}+\frac{y}{(1+2\lambda )/(1+3\lambda )}=1$

So the point of intersection of this line with coordinate axes are

$(\frac{1+2\lambda }{2+\lambda },0)$ and $(0,\frac{1+2\lambda }{1+3\lambda })$

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

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

$4(1+2\lambda {)}^2=3(2+\lambda \left)\right(1+3\lambda )$

$4(1+4\lambda +4{\lambda }^2)=3(2+7\lambda +3{\lambda }^2)$

$7{\lambda }^2-5\lambda -2=0$

$\lambda =1,-\frac{2}{7}.....................\left(ii\right)$

Hence the equations of the required lines from eq (i) and (ii) are:

$3x+4y-1-2=0$ and $(2-\frac{2}{7})x+(1-\frac{6}{7})y-1+\frac{4}{7}=0$

$3x+4y-3=0$ and $12x+y-3=0$