Discussion Forum

The equation  of the straight  line passing  through the point of intersection of $5x-6y-1$ , $3x+2y+5=0$, and perpendicular to the line $3x-5y+11=0$ is 

Answer:

Explanation:

We know that point of intersection of two lines 

i.e,   $ax_{1}+by_{1}+c_{1}=0$ and $ax_{2}+by_{2}+c_{2}=0$  is 

    $\left(\frac{b_{1}c-b_{2}c}{a_{1}b_{2}-a_{2}b_{1}},\frac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}\right)$

$\therefore$ point of intersection of two lines  i.e,  

5x-6y-1=0  and 3x+2y+5=0 is 

     $\left(\frac{-6\times5-2\times-1}{5\times2-3\times-6},\frac{-1\times3-5\times5}{5\times2-3\times-6}\right)$

     =  $\left(\frac{-30+2}{10+18},\frac{-3-25}{10+18}\right)$

  = $\left(\frac{-28}{28},\frac{-28}{28}\right)=\left(-1,-1\right)$

 and slope of line 3x-5+11=0 is m= $\frac{3}{5}$

Equation of the line passing through (-1,-1)  and perpendicular to 3x-5y+11=0 is 

  (y+1)= $-\frac{5}{3} (x+1)$

 $\Rightarrow$    $3(y+1)=-\frac{5}{3} (x+1)$

 $\Rightarrow$    $3y+3=-5x-5$

 $\Rightarrow$   $5x+3y+3+3+5=0$

 $\Rightarrow$    $5x+3y+8=0$