Math Practice

Parallel and Perpendicular Lines Practice Problems

Q: What makes two lines parallel?

Equal slopes and different y-intercepts. If the intercepts match too, the equations describe the same line and are called coincident rather than parallel.

Q: What is the perpendicular slope rule?

The slopes are negative reciprocals, so their product is -1. Flip the fraction **and** change the sign — doing only one of the two is the classic error.

Q: How do I find a perpendicular slope for a whole number?

Write it as a fraction first. A slope of 6 is 6/1, so the perpendicular slope is -1/6, not -6.

Q: Is there a shortcut in standard form?

Yes. For a parallel line keep A and B and recompute C from the new point. For a perpendicular line swap A and B and change one sign, then recompute C.

Fifteen hand-built problems on the two slope relationships, each fully worked. They cover finding a parallel or perpendicular slope, classifying a pair of equations as parallel, perpendicular, coincident or neither, writing a new line through a given point, the standard-form shortcuts, and perpendicular bisectors.

Questions

1.
What slope is parallel to \(y = 4x - 1\)?
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Parallel lines have equal slopes, so \(m = 4\).
2.
What slope is perpendicular to \(y = 4x - 1\)?
Show full solution
The negative reciprocal of \(4 = \dfrac{4}{1}\) is \(-\dfrac{1}{4}\). Check: \(4 \times -\dfrac{1}{4} = -1\).
3.
What slope is perpendicular to a line of slope \(\dfrac{5}{8}\)?
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Flip the fraction and change the sign: \(-\dfrac{8}{5}\).
4.
What slope is perpendicular to a line of slope \(-3\)?
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\(-3 = -\dfrac{3}{1}\); flipping gives \(-\dfrac{1}{3}\) and changing the sign gives \(\dfrac{1}{3}\).
5.
Are \(y = 2x + 1\) and \(y = -\dfrac{1}{2}x + 4\) perpendicular? Answer yes or no.
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\(2 \times -\dfrac{1}{2} = -1\), so yes.
6.
Are \(y = \dfrac{3}{4}x\) and \(y = \dfrac{4}{3}x\) perpendicular? Answer yes or no.
Show full solution
\(\dfrac{3}{4} \times \dfrac{4}{3} = 1\), not \(-1\). The fraction was flipped but the sign was never changed.
7.
Write the line parallel to \(y = \dfrac{2}{3}x + 1\) through \((3,\ 4)\), in slope-intercept form. Give only the right-hand side.
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Keep the slope \(\dfrac{2}{3}\): \(y - 4 = \dfrac{2}{3}(x - 3)\), so \(y = \dfrac{2}{3}x + 2\).
8.
Write the line perpendicular to \(y = -\dfrac{2}{3}x + 1\) through \((4,\ -1)\), in slope-intercept form. Give only the right-hand side.
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The perpendicular slope is \(\dfrac{3}{2}\). Then \(y + 1 = \dfrac{3}{2}(x - 4)\), so \(y = \dfrac{3}{2}x - 7\).
9.
Are \(3x + 4y = 8\) and \(6x + 8y = 9\) parallel, coincident, or intersecting? Answer with one word.
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Doubling the first gives \(6x + 8y = 16\), which is not \(9\). Same slope, different constant — parallel and distinct.
10.
Are \(x - 5y = 10\) and \(3x - 15y = 30\) parallel, coincident, or intersecting? Answer with one word.
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Dividing the second by 3 gives \(x - 5y = 10\) — exactly the first. They are the same line, so the system has infinitely many solutions.
11.
A line parallel to \(2x - 7y = 1\) passes through \((7,\ 0)\). Written as \(2x - 7y = C\), what is \(C\)?
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Keep \(A\) and \(B\), then substitute the point: \(2(7) - 7(0) = 14\). So \(2x - 7y = 14\).
12.
A line perpendicular to \(2x + 7y = 5\) passes through \((3,\ 2)\). Written as \(7x - 2y = C\), what is \(C\)?
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Swap \(A\) and \(B\) and change one sign to get the perpendicular family \(7x - 2y = C\). Substituting the point: \(21 - 4 = 17\).
13.
Find the perpendicular bisector of the segment from \((0,\ 0)\) to \((6,\ 4)\), in slope-intercept form. Give only the right-hand side.
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Midpoint \((3,\ 2)\); the segment has slope \(\dfrac{2}{3}\), so the bisector has slope \(-\dfrac{3}{2}\). Then \(y - 2 = -\dfrac{3}{2}(x - 3)\), giving \(y = -\dfrac{3}{2}x + \dfrac{13}{2}\).
14.
How far apart are the parallel lines \(y = 3x + 4\) and \(y = 3x - 6\)? Give the distance to three decimal places.
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\(d = \dfrac{|b_1 - b_2|}{\sqrt{m^2 + 1}} = \dfrac{10}{\sqrt{10}} = \sqrt{10} \approx 3.162\). It is less than the vertical gap of 10 because the distance is measured perpendicular to the lines.
15.
The triangle \(A(0,\ 0)\), \(B(4,\ 2)\), \(C(1,\ -2)\) has a right angle. At which vertex? Answer A, B or C.
Show full solution
\(AB\) has slope \(\dfrac{2}{4} = \dfrac{1}{2}\) and \(AC\) has slope \(\dfrac{-2}{1} = -2\). Their product is \(-1\), so the right angle is at \(A\).

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Mathovia Team

Editorial Team

Parallel and Perpendicular Lines Practice Problems

Two lessons feed this set: Parallel Lines and Perpendicular Lines. They are practised together here because almost every exam question that asks for one also asks for the other, and the two rules are only distinguishable if you have drilled them side by side.

How to use this set

Everything reduces to a slope comparison, so get both slopes first — rearranging into or using — and only then decide.

The classification problems want one word: parallel, coincident, or intersecting. The construction problems want the right-hand side of the equation.

Common mistakes to watch for

Flipping without changing the sign. Perpendicular to is , not .

Calling coincident lines parallel. Reduce both equations fully before deciding — if they match exactly, they are one line.

Changing the slope when writing a parallel line. The slope is the one thing that stays.

Using the vertical gap as the distance. The distance between parallel lines is measured perpendicular to them, so it is shorter.

Where to go next

Slope is the number both rules act on, and Point-Slope Form is how you turn a new slope and a given point into an equation.

Frequently Asked Questions

What makes two lines parallel?+

Equal slopes and different -intercepts. If the intercepts match too, the equations describe the same line and are called coincident rather than parallel.

What is the perpendicular slope rule?+

The slopes are negative reciprocals, so their product is . Flip the fraction **and** change the sign — doing only one of the two is the classic error.

How do I find a perpendicular slope for a whole number?+

Write it as a fraction first. A slope of is , so the perpendicular slope is , not .

Is there a shortcut in standard form?+

Yes. For a parallel line keep and and recompute from the new point. For a perpendicular line swap and and change one sign, then recompute .

Are a horizontal and a vertical line perpendicular?+

Geometrically yes, but the slope rule cannot express it — one slope is and the other is undefined, and there is no product to take. This pair is the standard exception.

How do I check a quick sign?+

A perpendicular pair always has one positive and one negative slope, unless one line is horizontal or vertical. If your two slopes share a sign, something has gone wrong.

∥ ⊥Parallel and Perpendicular Lines Practice Problems

Parallel and Perpendicular Lines Practice Problems

How to use this set

Common mistakes to watch for

Where to go next

Frequently Asked Questions

More practice problems

Parallel and Perpendicular Lines Practice Problems