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Exercises Equations and inequalities

Example 16 from Algebra and Trigonometry, 2.2 Linear Equations in One Variable

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to $5 x - 3 y + 4 = 0$ and passing through the point $(- 4, 1) .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
The first step is to write the equation in slope-intercept form.

$\begin{matrix} 5 x - 3 y + 4 & = & 0 \\ −3 y & = & −5 x - 4 \\ y & = & \frac{5}{3} x + \frac{4}{3} \end{matrix}$

We see that the slope is $m = \frac{5}{3} .$ This means that the slope of the line perpendicular to the given line is the negative reciprocal, or $- \frac{3}{5} .$ Next, we use the point-slope formula with this new slope and the given point.

$\begin{matrix} y - 1 & = & - \frac{3}{5} (x - (−4)) \\ y - 1 & = & - \frac{3}{5} x - \frac{12}{5} \\ y & = & - \frac{3}{5} x - \frac{12}{5} + \frac{5}{5} \\ y & = & - \frac{3}{5} x - \frac{7}{5} \end{matrix}$

How did it go?

PreviousWriting the Equation of a Line Parallel to a Given Line Passing Through a Given Point Next Modeling a Linear Equation to Solve an Unknown Number Problem

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 2.2, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.