Question

find the equation of the line perpendicular to 2x+3y-6=0 and passing through (-1,1)

Answer 1

$\textbf{Concept:}$

$\text{Equation of any line perpendicular to ax+by+c=0 and}$

$\text{passing through the point (x_1,y_1) is of the form}\\\\\boxed{\bf\,bx-ay=bx_1-ay_1}$

$\text{Given line is 2x+3y-6=0}$

$\text{The equation of the line perpendicular to 2x+3y-6=0 can be put in the form}\\\\3x-2y=3x_1-2y_1$

$\text{Here,}(x_1,y_1)=(-1,1)$

$\implies\;3x-2y=3(-1)-2(1)$

$\implies\;3x-2y=-3-2$

$\implies\;3x-2y=-5$

$\implies\boxed{\bf{3x-2y+5=0}}$

Answer 2

Step-by-step explanation:

the answer of these question 3x-2y+5=0