Question

Write an equation of a line perpendicular to y=3x+5 that contains the points (-3,-4)

Answer 1

The equation of line is:

$y = - \frac{1}{3}x-5$

Further explanation:

Given equation of line is:

y = 3x+5

The standard form of point-slope form is:

y = mx+b

Comparing the given equation with the standard form is:

m = 3

We know that product of slopes of two perpendicular lines is -1

Let m2 be the slope of line perpendicular to y = 3x+5

Then

$3 * m_2 = -1\\m_2 = -\frac{1}{3}$

Putting the value of slope in standard form

$y = -\frac{1}{3}x + b$

To find the value of b, putting (-3,-4) in equation

$-4 = -\frac{1}{3}(-3) + b\\-4 = 1 + b\\b = -4-1\\b = -5$

Putting the values of slope and b in equation

$y = - \frac{1}{3}x-5$

Keywords: Slope, Point-intercept form

Learn more about perpendicular lines at:

• brainly.com/question/1332667
• brainly.com/question/13096001

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