Question

Write the equation of a line in slope-intercept form that is perpendicular to the given line and passes through the point (-9,-17).
Y=9x-4

Answer 1

Answer:

y = -1/9x -18

Step-by-step explanation:

Y = 9x-4

This is in slope intercept form (y=mx+b where m is the slope)

The slope is 9

We want a line that is perpendicular

Perpendicular lines have slopes that multiply to -1

Let m be the slope of the perpendicular line

m*9 = -1

Divide each side by 9

m = -1/9

The slope of the perpendicular line is -1/9

Using the slope intercept form

y = mx+b

y = -1/9 x+b

Substitute the point into the equation

-17 = -1/9(-9) +b

-17 = 1+b

Subtract 1 from each side

-17-1 =1-1+b

b= -18

The equation is

y = -1/9x -18

Answer 2

Answer:

y = -1/9x - 18

Step-by-step explanation:

Using the equation of line

y - y_1 = m(x - x_1)

y = mx + C

First first the slope of the equation

y = 9x - 4

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

Slope = 9, but because it is perpendicular

m = -1/9

Using the equation

y - y_1 = m(x - x_1)

With the point (-9,-17)

x_1 = -9

y_1 = -17

y - (-17) = -1/9(x - (-9)

y + 17 = -1/9(x + 9)

y = -1/9(x + 9) - 17

Open the bracket.

y = (-x -9)/9 - 17

Lcm is 9

y = -x -9 - 153 / 9

y = -x - 162 / 9

Let's separate

y = -x/9 - 162/9

y = -x/9 - 18

y = -1/9x - 18

Therefore, the equation of the line is y = -1/9x - 18