Question

Find the equation of a straight line through the point of intersection of the lines 4x-3y=0 and 2x-5y+3=0 and parallel to 4x+5y+6=0.

Answer 1

$\text{LIne through the intersection of}4x-35=0and2x-5y+3=0is$

$(4x-3y)+\lambda (2x-5y+3)=0...\left(i\right)$

$or,x(4x+2\lambda )-y(3+5\lambda )+3\lambda =0$

$Andtherequiredlineisparallelto$ $4x+5y+6$

$\therefore slopeofrequired=slopeof4x+5y+6=\frac{-4}{3}$

$\therefore \frac{-(4+2\lambda )}{-(3+5\lambda )}=\frac{-4}{3}$

$\Rightarrow 5(4+2\lambda )=-4(3+5\lambda )$

$\Rightarrow 20+10\lambda =-12-20\lambda$

$\Rightarrow 30\lambda =-32$

$\Rightarrow \lambda =\frac{-16}{15}$

$Putting\lambda inequation\left(i\right)$

$(4x-3y)-\frac{16}{15}(2x-5y+3)=0$

$\Rightarrow 60x-5y-32+40y-48=0$

$\Rightarrow 28x+5y-48=0$

$Istherequiredline.$