What if the original line is horizontal?

A horizontal line has a slope of 0 and an equation of the form $y = c$. The perpendicular line is vertical. Its equation is $x = x_1$, where $x_1$ is the x-coordinate of the given point.

What if the original line is vertical?

A vertical line has an undefined slope and an equation of the form $x = c$. The perpendicular line is horizontal. Its equation is $y = y_1$, where $y_1$ is the y-coordinate of the given point.

Do perpendicular lines always intersect?

Yes. In two-dimensional Euclidean geometry, any two lines with slopes that are negative reciprocals of each other are not parallel and will always intersect at exactly one point at a 90-degree angle.

How to find the equation of a line perpendicular to another

To find the equation of a line perpendicular to a given line, you must determine the negative reciprocal of the original line's slope. This method applies whenever you are given the equation of an initial line and a specific coordinate point through which the new perpendicular line must pass.

The setup

Identify the equation of the given line and the coordinates of the point $(x_1, y_1)$ the new line passes through. Convert the given equation to slope-intercept form $y = mx + b$ if it is not already in that format.

The steps

Extract the slope $m$ from the given line's equation. 2. Calculate the perpendicular slope $m_{\perp} = -1/m$ . 3. Substitute $m_{\perp}$ and the given point $(x_1, y_1)$ into the point-slope formula $y - y_1 = m_{\perp}(x - x_1)$ . 4. Rearrange the resulting equation into slope-intercept form $y = m_{\perp}x + b_{ ext{new}}$ or standard form as required by the problem.

Checking the result

Multiply the slope of your new line by the slope of the original line. The product must be exactly $-1$ . Verify that substituting the given point $(x_1, y_1)$ into your new equation yields a true statement.

Common errors

A frequent mistake is failing to change the sign when finding the reciprocal slope, resulting in just $1/m$ instead of $-1/m$ . Another common error is incorrectly reusing the $y$ -intercept from the original line instead of solving for the new line's intercept.

Worked example

Find the equation of the line perpendicular to $2x + 3y = 6$ that passes through the point $(4, -1)$ .

Convert the original line to slope-intercept form: $3y = -2x + 6$ becomes $y = -\frac{2}{3}x + 2$ . The original slope is $m = -\frac{2}{3}$ . The perpendicular slope is the negative reciprocal: $m_{\perp} = \frac{3}{2}$ . Use the point-slope form with the point $(4, -1)$ : $y - (-1) = \frac{3}{2}(x - 4)$ . Simplify the equation: $y + 1 = \frac{3}{2}x - 6$ . Subtract 1 from both sides to get the final equation in slope-intercept form: $y = \frac{3}{2}x - 7$ .

FAQ

Run your own problem

Answer form

Drag an image or PDF of the problem here, or click to upload

Ctrl + V to paste

References: OpenStax Intermediate Algebra, Chapter 2 · Khan Academy: Analytic Geometry · Stewart Calculus, Appendix B