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Line p has an equation of y=-8x+6. Line q, which is perpendicular to line p, includes the point (2,–2). What is the equation of line q?

Write the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

Sagot :

Leora03

Answer:

[tex]y = \frac{1}{8} x - 2 \frac{1}{4} [/tex]

Step-by-step explanation:

Slope-intercept form

y= mx +c, where m is the slope and c is the y-intercept

Line p: y= -8x +6

slope= -8

The product of the slopes of perpendicular lines is -1. Let the slope of line q be m.

m(-8)= -1

m= -1 ÷(-8)

m= ⅛

Substitute m= ⅛ into the equation:

y= ⅛x +c

To find the value of c, substitute a pair of coordinates that the line passes through into the equation.

When x= 2, y= -2,

-2= ⅛(2) +c

[tex] - 2 = \frac{1}{4} + c[/tex]

[tex]c = - 2 - \frac{1}{4} [/tex]

[tex]c = - 2 \frac{1}{4} [/tex]

Thus, the equation of line q is [tex]y = \frac{1}{8} x - 2 \frac{1}{4} [/tex].

Answer:

y=1/8x-9/4

Step-by-step explanation:

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