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Fine the equation of the line that is perpendicular to the given line and passes through the given point. Enter the right side of the equation as a single fraction. y=7x-1/4;(5,5)
The equation is y=
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Fine The Equation Of The Line That Is Perpendicular To The Given Line And Passes Through The Given Point Enter The Right Side Of The Equation As A Single Fracti class=

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Answer:

[tex]y=-\frac{4}{7} x+\frac{55}{7}[/tex]

Step-by-step explanation:

Change the given equation to slope-intercept to get [tex]y=\frac{7}{4}x-\frac{1}{4}[/tex]. When you multiply the slopes of perpendicular lines, you get -1. -1 divided by [tex]\frac{7}{4}[/tex] is [tex]-\frac{4}{7}[/tex]. The slope of the new line is then [tex]-\frac{4}{7}[/tex]. The perpendicular line passes through (5, 5) so you can have the equation[tex]5=-\frac{4}{7} (5)+b[/tex]. Simplifying gets [tex]b = \frac{55}{7}[/tex] . So now the final equation for the perpendicular line is [tex]y=-\frac{4}{7} x+\frac{55}{7}[/tex]

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