Question

Find the equation of line which is parallel. $3x-2y+9=0$ and passes through the $(5,-6)$.

Answer 1

Given that the line is parallel to $3x-2y+9=0$ and passes through the point $(5,-6)$-

As we know that if two lines are parallel then their slope is equal.

Given line-

$3x-2y+9=0$

$\Rightarrow 2y=3x+9$

$\Rightarrow y=\frac{3}{2}x+\frac{9}{2}$

Slope of given line $=\frac{3}{2}$

$\therefore$ Slope of required line $=$ slope of given line $=\frac{3}{2}$

As the line is passing therough $(5,-6)$.

Therefore, equation of required line is-

$(y-(-6))=\frac{3}{2}(x-5)$

$2(y+6)=3(x-5)=0$

$2y+12=3x-15$

$3x-2y-15-12=0$

$3x-2y-27=0$

Hence the equation of required line is $3x-2y-27=0$.