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Which equation represents a line which is perpendicular to the line y = -x + 8?

Sagot :

Two lines are perpendicular between each other if their slopes fulfills the following property

[tex]m_1m_2=-1[/tex]

where m1 and m2 represents the slopes of line 1 an 2, respectively.

To find the slope of a line we can write it in the form slope-intercept form

[tex]y=mx+b[/tex]

Our original line is

[tex]y=-\frac{1}{8}x+8[/tex]

Then its slope is

[tex]m_1=-\frac{1}{8}[/tex]

Now we have to find the slope of the second line. Using the first property,

[tex]\begin{gathered} m_1m_2=-1_{} \\ -\frac{1}{8}m_2=-1_{} \\ m_2=(-1)(-8) \\ m_2=8 \end{gathered}[/tex]

Then the second line has to have a slope of 8.

The options given to us are:

[tex]\begin{gathered} x+8y=8 \\ x-8y=-56 \\ 8x+y=5 \\ y-8x=4 \end{gathered}[/tex]

Then we have to determine which of these options have a slope of 8. To do that we write them in the slope-intercept form:

[tex]\begin{gathered} x+8y=8\rightarrow y=-\frac{1}{8}x+1 \\ x-8y=-56\rightarrow y=\frac{1}{8}x+7 \\ 8x+y=5\rightarrow y=-8x+5 \\ y-8x=4\rightarrow y=8x+4 \end{gathered}[/tex]

Once we have the options in the right form, we note that the only one of them that has a slope of 8 is the last one.

Then the line perpendicular to the original one is

[tex]y-8x=4[/tex]