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 of intercepts 5,6 on the axes.

Answer 1

The required line is

$(x+y-4)+\lambda (2x-3y-1)=0$

or $x(1+2\lambda )+y(1-3\lambda )-4-\lambda =0$

And it is perpendicular to $\frac{x}{5}+\frac{y}{6}=1$

$\therefore$( slope of required line) $\times$

$(slopeof\frac{x}{5}+\frac{y}{6}=1)=-1$

$\Rightarrow -\left(\frac{1+2\lambda }{1-3\lambda }\right)\times \frac{-6}{5}=-1$

$\Rightarrow \frac{1+2\lambda }{1-3\lambda }=\frac{-5}{6}$

$\Rightarrow 6+12\lambda =-5+15\lambda$

$\Rightarrow 11=3\lambda or\lambda =\frac{11}{3}$

$\therefore$ The required line is

($x+y-4)+\frac{11}{3}(2x-3y-1)=0$

$2x+3y-12+22x-3y-11=0$

$25x-30y-23=0$