Write an equation of a line that is perpendicular to y = 3x + 3 and passes through (−6, 3).

y equals negative one-third times x plus 1
y equals negative one-third times x minus 5
y = 3x + 21
y = 3x − 15

Respuesta :

Slope Intercept Form of a line is y = mx + b, where m = slope, b = y - intercept, and x and y are variables.

To write the equation of a line in slope intercept form, we need to find m and b first.

Parallel lines have equal slopes.

Perpendicular lines have slopes that are negative reciprocals.

y = 3x + 5 is a line with a slope of m = 3.

A line perpendicular to that line will have a slope that is the negative reciprocal of 3.

The reciprocal of 3 is 1/3. So the negative reciprocal of 3 is -1/3.

Therefore, we want to write the equation of a line with slope, m = -1/3, and passes through the point (6, - 8) = (x, y).

To write the equation in slope intercept form, y = mx + b, we also need to find the y - intercept, which is b.

y = mx + b

-8 = (-1/3)(6) + b      (we've set up the equation with only one unknown, b, that we can now solve for)

-8 = -2 + b

b = -6

With a slope, m = -1/3, and a y-intercept, b = -6, the equation of our line relating x and y is:

y = (-1/3)x - 6

Write an equation of a line that is perpendicular to y = 3x + 3 and passes through (−6, 3).

y equals negative one-third times x plus 1

y equals negative one-third times x minus 5

y = 3x + 21

y = 3x − 15

y=(-1/3)x-6