What is the slope of any line perpendicular to the line passing through #(-8,2)# and #(-12,-20)#?

Question

Genesis Allen · Accepted Answer

#-2/11# The slope of the line passing through (-8,2)and (-12.-20)is #(-20-2)/(-12+8)=-22/-4=11/2# The slope of the line perpendicular to the line passing through (-8,2)and (-12.-20) is the ngative inverse of #11/2 # i.e.#=-2/11#

Kennedy Adams · Answer

undefined To find the slope of the line passing through (-8, 2) and (-12, -20), first, calculate the slope using the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$,

where $(x_1, y_1) = (-8, 2)$ and $(x_2, y_2) = (-12, -20)$.

$m = \frac{-20 - 2}{-12 - (-8)} = \frac{-22}{-4} = 5.5$.

The slope of the line passing through these two points is 5.5.

To find the slope of a line perpendicular to this line, take the negative reciprocal of the slope.

The negative reciprocal of 5.5 is $-\frac{1}{5.5} = -\frac{2}{11}$.

So, the slope of any line perpendicular to the line passing through (-8, 2) and (-12, -20) is $-\frac{2}{11}$.

Jace Anderson · Answer

undefined First, calculate the slope of the line passing through the points (-8,2) and (-12,-20) using the formula:

slope = (y2 - y1) / (x2 - x1)

Substitute the coordinates into the formula:

slope = (-20 - 2) / (-12 - (-8))
      = (-22) / (-4)
      = 5.5

Since perpendicular lines have slopes that are negative reciprocals of each other, the slope of any line perpendicular to the given line is the negative reciprocal of 5.5, which is -1/5.5, or approximately -0.1818.