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What is the slope m of a line that is perpendicular to the line represented by x = 8? (If an answer is undefined, enter UNDEFINED.)

Answer :

Answer:

slope = 0

Step-by-step explanation:

x = 8 is a vertical line parallel to the y- axis

A line perpendicular to it is a horizontal line parallel to the x- axis.

The x- axis has a slope of zero thus the line perpendicular to x = 8 is zero

MrRoyal

The slope of a line is simply the ratio of a change in the vertical axis to the horizontal axis. The line perpendicular to [tex]x = 8[/tex] has a slope of 0

Given that

[tex]x = 8[/tex]

See attachment for the line of [tex]x = 8[/tex]

First, we determine the slope of [tex]x = 8[/tex]

The slope of a line (m) is calculated as:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]

[tex]x = 8[/tex] means that, the value of x is always 8; irrespective of the y-value.

So:

[tex]x_2 = x_1 = 8[/tex]

[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex] becomes

[tex]m = \frac{y_2 - y_1}{8-8}[/tex]

[tex]m = \frac{y_2 - y_1}{0}[/tex]

Assume the slope of the line perpendicular to [tex]x = 8[/tex] is [tex]m_2[/tex].

[tex]m_2[/tex] is calculated as follows:

[tex]m_2 = -\frac{1}{m}[/tex]

Substitute [tex]m = \frac{y_2 - y_1}{0}[/tex]

[tex]m_2 = -\frac{1}{(y_2 - y_1)/0}[/tex]

This gives

[tex]m_2 = -\frac{1\times 0}{y_2 - y_1}[/tex]

[tex]m_2 = -\frac{0}{y_2 - y_1}[/tex]

[tex]m_2 = -0[/tex]

[tex]m_2 = 0[/tex]

The slope of the line perpendicular to [tex]x = 8[/tex] is 0.

Read more about slopes at

https://brainly.com/question/3605446

${teks-lihat-gambar} MrRoyal

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