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Write an equation of the line perpendicular to the line –4x + 3y = –15 and passes through the point (–8, –13)

1 Answer

5 votes
Answer:

4y = -3x - 76

Explanations:

The given equation is:

-4x + 3y = -15

Make y the subject of the formula to express the equation in the form

y = mx + c


\begin{gathered} -4x\text{ + 3y = -15} \\ 3y\text{ = 4x - 15} \\ y\text{ = }(4)/(3)x\text{ - }(15)/(3) \\ y\text{ = }(4)/(3)x\text{ - 5} \end{gathered}

Comparing the equation with y = mx + c

the slope, m = 4/3

the y-intercept, c = -5

The equation perpendicular to the equation y = mx + c is:


y-y_1\text{ = }(-1)/(m)(x-x_1)

The line passes through the point (-8, -13). That is, x₁ = -8, y₁ = -13

Substitute m = 4/3, x₁ = -8, y₁ = -13 into the equation above


\begin{gathered} y\text{ - (-13) = }(-1)/((4)/(3))(x\text{ - (-8))} \\ y\text{ + 13 = }(-3)/(4)(x\text{ + 8)} \\ y\text{ + 13 = }(-3)/(4)x\text{ - 6} \\ y\text{ = }(-3)/(4)x\text{ - 6 - 13} \\ y\text{ = }(-3)/(4)x\text{ - 19} \\ 4y\text{ = -3x - }76 \end{gathered}

User Michal Bernhard
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