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Parallel and Perpendicular Lines: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Three directions

Line A has slope $-\frac52$ . Line B is perpendicular to A, and line C is perpendicular to B. Find the slope of C.
1. Hint 1
  Each right-angle change requires the negative reciprocal.
2. Hint 2
  Find the slope of B before applying the relationship to C.
Answer

$-\frac52$ .

Full solution

The slope of B is $2/5$ , since
$-\frac52\cdot\frac25=-1$
Taking its negative reciprocal gives
$m_C=-\frac52$
Checking $\frac25(-\frac52)=-1$ confirms the second right angle.

Answer

$-\frac52$ .

Key idea

Applying the negative reciprocal twice returns the original nonzero slope.
Problem 1: Mark complete
Problem 2 Footpath across a road

A park map is drawn on a coordinate grid measured in meters, with east as the positive $x$ direction and north as the positive $y$ direction. A straight road on the map has slope $-\frac34$ , and a straight footpath crosses the road at a right angle. Walking along the footpath, Maya moves 6 m north while also moving east. How far east does she move?
1. Hint 1
  The footpath's direction on the map is fixed by its right angle with the road.
2. Hint 2
  Find the footpath's slope, then treat the 6 m north as a rise and solve for the run east.
Answer

$\frac92$ m, or 4.5 m.

Full solution

The footpath is perpendicular to the road, so its slope is the negative reciprocal of $-\frac34$ , which is $\frac43$ .
On the map, a move north is a rise and a move east is a run.
If the run is $r$ meters, then
$\frac6r=\frac43$
The run is positive, so multiplying both sides by $3r$ is valid:
$18=4r$ $r=\frac92$
Thus Maya moves $\frac92$ m, or 4.5 m, east.
Her move has slope $6/(9/2)=4/3$ , the footpath's slope, as required.

Answer

$\frac92$ m, or 4.5 m.

Key idea

A known perpendicular slope fixes the ratio of rise to run, so one known leg of a move determines the other.
Problem 2: Mark complete
Problem 3 Two constants

Choose a real constant $c$ so that $y=-2x+c$ is parallel to each of $y=-2x+1$ and $y=-2x-4$ , and is distinct from both. Give one possible equation.
1. Hint 1
  The slope is already shared; distinctness depends on the intercepts.
2. Hint 2
  Choose an intercept height different from both listed heights.
Answer

Any $y=-2x+c$ with $c\ne1$ and $c\ne-4$ ; for example, $y=-2x$ .

Full solution

Take $c=0$ , giving
$y=-2x$
All three slopes are $-2$ , while the intercept heights are $0$ , $1$ , and $-4$ .
Therefore the new line is distinct from, and parallel to, both given lines.
Any real $c$ other than 1 and $-4$ also works.

Answer

Any $y=-2x+c$ with $c\ne1$ and $c\ne-4$ ; for example, $y=-2x$ .

Key idea

A shared slope with different intercepts produces distinct parallel lines.
Problem 3: Mark complete
Problem 4 A parallel through $C$

The figure shows segments $AB$ and $BC$ . Write the equation of the line through $C$ parallel to $AB$ , in point-slope form using $C$ . Determine whether this new line is perpendicular to $BC$ .

Segments $AB$ and $BC$ on a unit grid.

Text description of this figure

A square coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative three to four and the vertical y-axis from negative two to five, with gridlines, tick marks and number labels at every whole number, arrowheads on both axes, and the origin labeled 0. Three points are plotted and labeled with their letters only: point A at negative two, negative one; point B at two, one; and point C at one, three. A segment joins A to B, and a second segment joins B to C. No coordinate pairs, equations or other lines are shown.
1. Hint 1
  Use the two endpoints of each segment to find its direction.
2. Hint 2
  The new line keeps the slope of $AB$ ; compare that slope with the slope of $BC$ .
Answer

$y-3=\frac12(x-1)$ ; yes, it is perpendicular to $BC$ .

Full solution

The points read from the grid are $A(-2,-1)$ , $B(2,1)$ , and $C(1,3)$ .
The slope of $AB$ is
$m_{AB}=\frac{1-(-1)}{2-(-2)}$ $m_{AB}=\frac12$
Using $C$ gives
$y-3=\frac12(x-1)$

The slope of $BC$ is
$(3-1)/(1-2)=-2.$
Since $\frac12(-2)=-1$ , the new line is perpendicular to $BC$ .
The new equation holds at $C$ , and its intercept $5/2$ differs from the intercept 0 of $AB$ .

Answer

$y-3=\frac12(x-1)$ ; yes, it is perpendicular to $BC$ .

Key idea

A new line can inherit one segment direction while meeting another at a right angle.
Problem 4: Mark complete
Problem 5 Shared meeting point

Lines $2x-y=-1$ and $x+y=7$ meet at $P$ . Write the equation of the line through $P$ perpendicular to the first line, in slope-intercept form.
1. Hint 1
  The meeting point supplies the location required for the new line.
2. Hint 2
  Solve each equation for $y$ , so that each slope can be read and the two heights can be set equal.
3. Hint 3
  Find the common point, then combine it with the negative reciprocal of the first slope.
Answer

$y=-\frac12x+6$ .

Full solution

Subtracting $2x$ from both sides of $2x-y=-1$ and then multiplying by $-1$ gives
$y=2x+1$
Subtracting $x$ from both sides of $x+y=7$ gives
$y=-x+7$

At the meeting point the two heights agree, so
$2x+1=-x+7$
Adding $x$ and subtracting 1 gives
$3x=6$ $x=2$
Either line then gives $y=5$ , so $P=(2,5)$ .

The first line has slope 2, so the new slope is $-1/2$ .
Point-slope form gives
$y-5=-\frac12(x-2)$ $y=-\frac12x+6$
Substitution at $x=2$ gives 5, and the slopes multiply to $-1$ .

Answer

$y=-\frac12x+6$ .

Key idea

An intersection can supply the point needed to construct a line with a required direction.
Problem 5: Mark complete
Problem 6 Three full lines

The figure shows full lines A, B, and C. Classify each pair as parallel, perpendicular, or neither, and for each pair say whether the test $m_1\cdot m_2=-1$ can be applied.

Lines A, B and C on a unit grid.

Text description of this figure

A square coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative five to six and the vertical y-axis from negative three to five, with gridlines, tick marks and number labels at every whole number, arrowheads on both axes, and the origin labeled 0. Three full lines are drawn, each with arrowheads at both ends and labeled with its letter only. Line A is horizontal, two units above the x-axis, through y equals two. Line B is vertical, three units left of the y-axis, through x equals negative three. Line C slants up to the right, rising one unit for each unit to the right, and passes through the grid points negative four, negative three; zero, one; and four, five. No points are marked, and no coordinates, equations or angle marks are shown.
1. Hint 1
  Decide first which of the three lines has a slope number and which does not.
2. Hint 2
  Read each nonvertical line's slope from two grid points on it; a vertical line has no slope number.
3. Hint 3
  Where both slopes exist, compare the slopes and their product; where a line is vertical, decide by sight.
Answer

A and B: perpendicular, and the test cannot be applied; A and C: neither, and the test can be applied; B and C: neither, and the test cannot be applied.

Full solution

A is the horizontal line $y=2$ , with slope 0.
B is the vertical line $x=-3$ , whose slope is undefined.
C passes through $(0,1)$ and $(1,2)$ , so it rises 1 for each 1 across and has slope 1.

A and B are a horizontal line and a vertical line, so they are perpendicular by sight.
The test cannot be applied to them, because B has no slope number.

A and C both have real slopes, so the test applies.
The slopes 0 and 1 are not equal, so the lines are not parallel, and their product is
$0\cdot1=0$
This is not $-1$ , so A and C are neither parallel nor perpendicular.

B is vertical and C is slanted, so the test cannot be applied to them.
A vertical line and a slanted line cross, but not at a right angle, so B and C are neither.

Answer

A and B: perpendicular, and the test cannot be applied; A and C: neither, and the test can be applied; B and C: neither, and the test cannot be applied.

Key idea

When both lines have slopes, compare the slopes and their product; when one line is vertical, classify the pair by sight.
Problem 6: Mark complete
Problem 7 Two unknown slopes

The lines $y=mx+4$ and $y=nx-2$ both contain $(2,1)$ . Find $m$ and $n$ , then classify the lines as parallel, perpendicular, or neither.
1. Hint 1
  The shared point fixes each missing slope separately.
2. Hint 2
  After substituting, compare the slopes and their product.
Answer

$m=-\frac32$ and $n=\frac32$ ; neither.

Full solution

The first line gives $1=2m+4$ , so
$m=-\frac32$
The second gives $1=2n-2$ , so
$n=\frac32$

The slopes differ and their product is
$-\frac32\cdot\frac32=-\frac94$
This is not $-1$ , so the lines are neither parallel nor perpendicular.
Substitution of both values returns height 1 at input 2.

Answer

$m=-\frac32$ and $n=\frac32$ ; neither.

Key idea

Two slopes that differ only in sign give perpendicular lines only when they are $1$ and $-1$ ; otherwise the lines are neither parallel nor perpendicular.
Problem 7: Mark complete
Problem 8 Repeated height gap

Two nonvertical lines, A and B, have heights that differ by 5 units at $x=-1$ , with A above B. Their heights again differ by 5 units at $x=4$ , with A above B. Must A and B be parallel? Justify your answer.
1. Hint 1
  Compare each line's rise over the same horizontal interval.
2. Hint 2
  Subtracting heights whose differences agree makes the two rises agree.
Answer

Yes, A and B must be parallel.

Full solution

Let B have heights $u$ and $v$ at the two inputs.
A then has heights $u+5$ and $v+5$ .
Both runs are 5, which is nonzero.
B has slope $(v-u)/5$ , and A has slope
$\frac{(v+5)-(u+5)}5$
This simplifies to the same value $(v-u)/5$ .

The lines have equal slopes and are distinct, since their heights differ at either input.
Therefore they are parallel.

Answer

Yes, A and B must be parallel.

Key idea

When one nonvertical line stays the same height above another at two distinct inputs, the two lines have equal slopes.
Problem 8: Mark complete
Problem 9 One point requirement

Two students each write an equation of a line perpendicular to $y=3x-2$ through $(4,1)$ . One claims their equations must describe distinct parallel lines. Is the claim correct? Explain.
1. Hint 1
  The perpendicular requirement supplies the same slope to both lines.
2. Hint 2
  Consider how many lines can have one specified point and one specified slope.
Answer

No; both equations describe the same line, $y-1=-\frac13(x-4)$ , or $y=-\frac13x+\frac73$ .

Full solution

Each line must have slope $-1/3$ , the negative reciprocal of 3.
With the required point, each equation is
$y-1=-\frac13(x-4)$
or an equivalent form such as
$y=-\frac13x+\frac73$
A point and a slope determine one line, so the two equations describe the same line.

Equal slopes alone do not establish distinctness.
Here the common point makes their intercepts equal as well.

Answer

No; both equations describe the same line, $y-1=-\frac13(x-4)$ , or $y=-\frac13x+\frac73$ .

Key idea

A shared slope and a shared point determine one line, rather than two distinct parallel lines.
Problem 9: Mark complete
Problem 10 Swapped coordinates

Line A passes through $(-2,1)$ and $(3,5)$ . Line B is drawn through the points formed by swapping the two coordinates of each of these points. A student claims A and B are perpendicular. Decide whether the claim is correct and explain.
1. Hint 1
  Write the swapped points explicitly before calculating their slope.
2. Hint 2
  Compare the product of the two slopes with the condition for a right angle.
Answer

No; the slopes are $\frac45$ and $\frac54$ .

Full solution

The swapped points are $(1,-2)$ and $(5,3)$ .
The original slope is
$m_A=\frac{5-1}{3-(-2)}$ $m_A=\frac45$
The new slope is
$m_B=\frac{3-(-2)}{5-1}$ $m_B=\frac54$

The product of the slopes is
$\frac45\cdot\frac54=1$
This is 1, not $-1$ , so A and B are not perpendicular.
Swapping the coordinates exchanged rise and run without supplying the sign change required for perpendicular slopes.

Answer

No; the slopes are $\frac45$ and $\frac54$ .

Key idea

For a line with nonzero slope, exchanging rise and run gives the reciprocal slope, while two slanted lines are perpendicular only when their slopes are negative reciprocals.
Problem 10: Mark complete

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