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What is the equation of a line that passes through the point (8,-2) and is parallel to the line whose equations is 3x+4y=15

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For this case we have:


When two lines are parallel, their slopes are equal.


Be a line of the form
y = mx + b

Where:


m is the slope


b is the cut point


If we have:
3x + 4y = 15

We can rewrite it as:



4y = 15-3x


y = - \frac {3} {4} x + \frac {15} {4}

Thus, the slope of that line is given by
m = - \frac {3} {4}

Since that line is parallel to the one we want to find, then
m = - \frac {3} {4}is the same for both lines.


The equation of the line that we want to find follows the form:



y_(2) = m_(2)x_(2) + b_(2)

Where
m_(2)= - \frac {3} {4}

So, we have:



y_(2) = - \frac {3} {4} x_(2) + b_(2)

We have as data the point
(x_(2), y_(2)) = (8, -2) that passes through the line we want to find. Substituting the points we find the cut point
b_(2):



-2 = - \frac {3} {4} (8) + b_(2)



-2 = -6 + b_(2)\\b_(2) = -2 + 6\\b_(2) = 4

Thus, the equation of the requested line is given by:



y_(2) = - \frac {3} {4} x_(2) + 4

Answer:



y_(2) = - \frac {3} {4} x_(2) + 4

User Takuto
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