Question

Find the equation of a straight line passing through the point of intersection of x+2y+3=0 and 3x+4y+7=0 and perpenicular to the straight line x-y+0=0

Answer 1

The equation of the equired line is

$(x+2y+3)+\lambda (3x+4y+7)=0$

or $x(1+3\lambda )+y(2+4\lambda )+3+7\lambda =0$

$m_1=slopeoftheline=-\left(\frac{1+3\lambda }{2+4\lambda }\right)$

The line is perpendiccular to x-y+9=0

whose slope $(m_2=1)$

$\therefore m_1\times m_2=-1$

$\Rightarrow =-\left(\frac{1+3\lambda }{2+4\lambda }\right)\times 1=-1$

$\Rightarrow 1+3\lambda =2+4\lambda$

$\Rightarrow \lambda =-1$

$\therefore$ The required line is

$x+2y+3+\lambda (2x+4y+7)=0$

Put the value of $\lambda =-1$

$x+2y+3-(3x+4y+7)=0$

$-2x-2y-4=0$

or,$x+y+2=0$