Recall from Intermediate Algebra that two non-vertical lines are perpendicular if and only if they have negative reciprocal slopes. That is to say, if one line has slopem1and the other has slopem2thenm1 · m2 = −1.Please note that a horizontal line is perpendicular to a vertical line and vice versa, so we assumem1 ≠ 0andm2 ≠ 0.Below, you are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-10-07T07:51:06+02:00

Complete Question

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Answer:

The line is [tex]y = -4 x[/tex]

Step-by-step explanation:

From the question we are told that

The line is [tex]y = \frac{1}{4} x + 8[/tex]

The point is P(0,0 )

Generally the equation for a line is [tex]yn = mx + c[/tex]

Now comparing this with the given equation we see that

[tex]m =\frac{1}{4} = slope[/tex]

[tex]C =8 = intercept [/tex]

Generally the slope of the perpendicular to the given line is mathematically evaluated as

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

=> [tex]m_1 = - \frac{1}{ \frac{1}{4}}[/tex]

=> [tex]m_1 = -4 [/tex]

Generally from the univesal equation of a line and the given point we have that

y = mx + c at p(0,0) and [tex]m_1 = -4[/tex] is

0 = -4 (0 ) + c

=> c = 0

Hence the equation perpendicular to the line given at the point given is

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