Question

Find the equation of the straight line perpendicular to

2y + x = 3

and passing

through (3, −1).

Answer 1

Answer:

The answer is $2x-y=7$

Step-by-step explanation:

Slope of the line $2y+x=3$ is

           $=-\dfrac{\text{coefficent of x}}{\text{coefficent of y}}=-\dfrac{1}{2}$

Thus slope of line perpendicular to above line is

                  $=-\dfrac{1}{-\frac{1}{2}}=2$  

(Since product of slopes of perpendicular lines is -1)

Thus required line through (3,-1) is given by

            $y-(-1)=2(x-3)$, using point slope form

          $\Rightarrow y+1=2x-6\\\Rightarrow 2x-y=7$