Parallel and Perpendicular Lines on the SAT
Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals: you flip the fraction and change the sign, so $\dfrac{2}{3}$ becomes $-\dfrac{3}{2}$. Multiply the two slopes and you always get $-1$.
So start by finding the slope of the line you are given. If it is written as $Ax + By = C$, solve for $y$, or use $m = -\dfrac{A}{B}$.
These questions are usually short. Students lose points in two places: they forget to change the sign when finding a perpendicular slope, or they read the slope straight off an equation that is not in slope-intercept form.
The blue lines $y = 2x + 1$ and $y = 2x - 4$ have the same slope, so they are parallel. The red line $y = -\dfrac{1}{2}x + 1$ has slope $-\dfrac{1}{2}$, the negative reciprocal of $2$, so it is perpendicular to both.
What you will see
| You see | Do this |
|---|---|
| "Parallel to $y = 3x - 4$" | Same slope: $3$. |
| "Perpendicular to $y = \dfrac{2}{3}x + 1$" | Negative reciprocal: $-\dfrac{3}{2}$. |
| A line in the form $Ax + By = C$ | Slope is $-\dfrac{A}{B}$. For $2x + 5y = 10$, the slope is $-\dfrac{2}{5}$. |
| "Passes through the point $(3, 1)$" | Use the new slope and the point: $y - 1 = m(x - 3)$, then simplify. |
| "No solution" for two linear equations | Same slope, different $y$-intercepts: parallel lines that never meet. |
| Two points on each line | Find each slope with $\dfrac{y_2 - y_1}{x_2 - x_1}$, then compare. |
Examples, step by step
Example 1: a parallel line through a point
Find the line parallel to $y = -2x + 5$ that passes through $(3, 1)$.
- Parallel means the same slope: $m = -2$.
- Use the point: $y - 1 = -2(x - 3)$, so $y - 1 = -2x + 6$.
- $y = -2x + 7$. Check: $-2(3) + 7 = 1$.
Example 2: a perpendicular line from standard form
Find the line through the origin that is perpendicular to $2x + 5y = 10$.
- Slope of the given line: $-\dfrac{A}{B} = -\dfrac{2}{5}$.
- Negative reciprocal: $\dfrac{5}{2}$.
- Through the origin, the $y$-intercept is $0$: $y = \dfrac{5}{2}x$.
Example 3: finding a constant
The lines $3x + ky = 7$ and $6x - 2y = 5$ are parallel. Find $k$.
- Slope of the second line: $-\dfrac{6}{-2} = 3$.
- Slope of the first line: $-\dfrac{3}{k}$.
- Set them equal: $-\dfrac{3}{k} = 3$, so $k = -1$.
Where students slip
- Flipping without changing the sign. Perpendicular to slope $\dfrac{2}{3}$ is $-\dfrac{3}{2}$, not $\dfrac{3}{2}$.
- Changing the sign without flipping. $-\dfrac{2}{3}$ is not perpendicular to $\dfrac{2}{3}$. You have to flip the fraction and change the sign.
- Reading the slope from standard form. In $2x + 5y = 10$ the slope is not $2$. It is $-\dfrac{2}{5}$.
- Calling identical lines parallel. Same slope and same intercept is one line with infinitely many solutions, not two parallel lines.
- Sign slips with the point. Through $(8, -1)$, the point-slope form is $y + 1 = m(x - 8)$.
Practice questions
Find the slope of the given line first, every time. Then open the solution to check.
Easy
Line $k$ is parallel to the line $y = 3x - 4$. What is the slope of line $k$?
- A$-3$
- B$3$
- C$-\dfrac{1}{3}$
- D$-4$
Show solution
Answer: B. Parallel lines have the same slope, so the slope is $3$. Choice C is the slope of a perpendicular line.
EasyStudent-produced response
Line $p$ is perpendicular to the line $y = \dfrac{2}{3}x + 1$. What is the slope of line $p$?
Enter your answer.
Show solution
Answer: $-\dfrac{3}{2}$ (or $-1.5$). Flip $\dfrac{2}{3}$ to get $\dfrac{3}{2}$ and change the sign: $-\dfrac{3}{2}$. Check: $\dfrac{2}{3} \cdot \left(-\dfrac{3}{2}\right) = -1$.
Medium
Which equation represents the line that is parallel to $y = -2x + 5$ and passes through the point $(3, 1)$?
- A$y = \dfrac{1}{2}x - \dfrac{1}{2}$
- B$y = -2x - 5$
- C$y = -2x + 7$
- D$y = -2x + 1$
Show solution
Answer: C. The slope is $-2$. $y - 1 = -2(x - 3)$ gives $y = -2x + 7$. Choice A also passes through $(3, 1)$, but it is perpendicular, not parallel.
Medium
Which equation represents a line that passes through the origin and is perpendicular to the line $2x + 5y = 10$?
- A$y = -\dfrac{2}{5}x$
- B$y = \dfrac{2}{5}x$
- C$y = -\dfrac{5}{2}x$
- D$y = \dfrac{5}{2}x$
Show solution
Answer: D. The given line has slope $-\dfrac{2}{5}$. The negative reciprocal is $\dfrac{5}{2}$, and through the origin the line is $y = \dfrac{5}{2}x$. Choice A is parallel to the given line.
MediumStudent-produced response
In the $xy$-plane, the lines $y = kx + 3$ and $y = 4x - 1$ are parallel, where $k$ is a constant. What is the value of $k$?
Enter your answer.
Show solution
Answer: $4$. Parallel lines have equal slopes, so $k = 4$. The different intercepts, $3$ and $-1$, keep them from being the same line.
MediumStudent-produced response
The lines $3x + ky = 7$ and $6x - 2y = 5$ are parallel, where $k$ is a constant. What is the value of $k$?
Enter your answer.
Show solution
Answer: $-1$. The second line has slope $-\dfrac{6}{-2} = 3$. The first has slope $-\dfrac{3}{k}$. So $-\dfrac{3}{k} = 3$ and $k = -1$.
HardStudent-produced response
Line $\ell$ passes through $(1, 2)$ and $(4, 8)$. Line $n$ passes through $(0, 5)$ and $(6, a)$, and line $n$ is perpendicular to line $\ell$. What is the value of $a$?
Enter your answer.
Show solution
Answer: $2$. Line $\ell$ has slope $\dfrac{8 - 2}{4 - 1} = 2$, so line $n$ has slope $-\dfrac{1}{2}$. Then $\dfrac{a - 5}{6 - 0} = -\dfrac{1}{2}$, so $a - 5 = -3$ and $a = 2$.
Hard
Line $m$ is perpendicular to the line $y = -\dfrac{3}{4}x + 2$ and passes through the point $(8, -1)$. What is the $y$-intercept of line $m$?
- A$-\dfrac{35}{3}$
- B$\dfrac{29}{3}$
- C$-1$
- D$\dfrac{35}{3}$
Show solution
Answer: A. The slope of line $m$ is $\dfrac{4}{3}$. $y + 1 = \dfrac{4}{3}(x - 8)$. At $x = 0$: $y = -1 - \dfrac{32}{3} = -\dfrac{35}{3}$. Choice B adds $\dfrac{32}{3}$ instead of subtracting it.
Hard
Each system below includes the equation $y = 2x + 3$. Which system has no solution?
- A$y = 2x + 3$ and $4x - 2y = -6$
- B$y = 2x + 3$ and $2x + 4y = 6$
- C$y = 2x + 3$ and $4x - 2y = 6$
- D$y = 2x + 3$ and $y = 3x + 2$
Show solution
Answer: C. $4x - 2y = 6$ becomes $y = 2x - 3$: same slope, different intercept, so the lines are parallel and never meet. Choice A becomes $y = 2x + 3$, the same line, so it has infinitely many solutions.
Hard
Which line is perpendicular to the line $y = \dfrac{3}{4}x + 1$?
- A$3x - 4y = 8$
- B$3x + 4y = 12$
- C$4x - 3y = 12$
- D$4x + 3y = 12$
Show solution
Answer: D. Perpendicular needs slope $-\dfrac{4}{3}$. For $4x + 3y = 12$, the slope is $-\dfrac{4}{3}$. Choice A has slope $\dfrac{3}{4}$ (parallel), and choice C has slope $\dfrac{4}{3}$ (sign not changed).
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