Question

Find the equation of the line that is perpendicular to y=1/3x-3 and passes through the point (3,-2)

Answer 1

Answer:

y = -3x + 7

Step-by-step explanation:

Using the equation of line

y = mx + C

y - y_1 = m(x - x_1)

First find the slope

With eqn y = 1/3x - 3

Note, if two lines are perpendicular, their slope will be negative reciprocal

slope = m = -3/1

Using y - y_1 = m(x - x_1)

With point ( 3, -2)

x_1 = 3

y_1 = -2

y - (-2) = -3/1(x - 3)

y + 2 = -3/1(x -3)

Open bracket

y + 2 = -3(x - 3)/1

y + 2 = (-3x +9)/1

y = (-3x + 9 )/1 - 2

LCM is 1

y = -3x + 9 - 2

y = -3x + 7

The equation of the line is

y = -3x + 7