Question

find the equation of the straight line passing through the point (0,-1) which is perpendicular to the line y= -3/4 x-3

Answer 1

Answer:

$4x - 3y -3=0$

Step-by-step explanation:

Given equation ,

$\implies y =\dfrac{-3}{4}x - 3$

Compare it to slope intercept form to find the slope which is $y=mx+c$ , we have ,

• $m =\dfrac{-3}{4}$

Now we know that the product of slope of two perpendicular lines is-1 . Hence the slope of the perpendicular line is ,

• $m_{perp}= \dfrac{4}{3}$

The given point is (0,-1) , on using Point slope form of the line we have,

$\implies y-y_1 = m(x - x_1) \\\\\implies y -(-1) = \dfrac{4}{3}(x -0 ) \\\\\implies y + 1 =\dfrac{4}{3}x \\\\\implies \underline{\boxed{\red{ 4x - 3y -3 = 0 }}}$