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Lines, Slopes, and Perpendicularity: 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 A different slope triangle

On a falling nonvertical line, one slope triangle has horizontal run 4 and signed rise $-3$ . A second slope triangle on that line has run 12. Find its signed rise and justify why the slope triangles give the same ratio.
1. Hint 1
  Compare the angles made by the line and the horizontal legs.
2. Hint 2
  AA similarity scales corresponding legs by the same positive factor. Compare the two runs.
Answer

Signed rise $-9$ ; AA similarity preserves the leg ratio, and both rises are negative.

Full solution

Both triangles have a right angle and the same acute angle with a horizontal line.
Thus they are similar by AA.
The horizontal leg length changes by a factor of 3, so the vertical leg length changes by that factor too.

Both left-to-right trips fall, so the larger signed rise is
$h=3(-3)$ $h=-9.$
Its rise-to-run ratio is $-9/12$ , the same as $-3/4$ .

Answer

Signed rise $-9$ ; AA similarity preserves the leg ratio, and both rises are negative.

Key idea

Similarity fixes the ratio while the line's direction fixes the sign.
Problem 2 Build a parallel through an external point

Line $AB$ contains
$\begin{aligned}A&=(-6,-2),\\B&=(2,4).\end{aligned}$
Find an equation of the line through $D=(0,7)$ parallel to $AB$ , and verify that the lines are distinct.
1. Hint 1
  Find the direction of $AB$ before choosing the new line's location.
2. Hint 2
  The new line has the same slope, but use its required point to find its intercept. Compare intercepts to check distinctness.
Answer

$3x-4y=-28$ ; its intercept 7 differs from the intercept $5/2$ of $AB$ .

Full solution

The slope of $AB$ is
$m=\frac{4-(-2)}{2-(-6)}$ $m=\frac34.$
Through $D$ , the parallel candidate has equation
$y=\frac34x+7.$
Clearing the fraction gives
$3x-4y=-28.$

Using $B$ to locate $AB$ gives
$y=\frac34x+\frac52.$
The intercepts differ, so the equal-slope lines are distinct.
The parallel criterion now applies.

Answer

$3x-4y=-28$ ; its intercept 7 differs from the intercept $5/2$ of $AB$ .

Key idea

A direction match establishes parallelism only after distinctness is checked.
Problem 3 A horizontal segment

Find the perpendicular bisector of the segment with endpoints
$\begin{aligned}E&=(-5,4),\\F&=(7,4).\end{aligned}$
Explain why a negative-reciprocal calculation is unnecessary.
1. Hint 1
  Locate the midpoint and identify the segment's orientation.
2. Hint 2
  A horizontal segment has a vertical perpendicular bisector. Use the midpoint's first coordinate.
Answer

$x=1$ ; the segment is horizontal, so its perpendicular bisector is vertical and has undefined slope.

Full solution

The midpoint is $M=(1,4)$ .
The endpoint heights agree, so $EF$ is horizontal.
Its perpendicular bisector is the vertical line through $M$ , with equation
$x=1.$

The vertical line has undefined slope.
Its right angle with $EF$ follows from the horizontal and vertical directions, so there is no slope product or negative reciprocal of zero to evaluate.

The derived vertical line crosses the segment at its midpoint.

Answer

$x=1$ ; the segment is horizontal, so its perpendicular bisector is vertical and has undefined slope.

Key idea

Orientation supplies the perpendicular direction when a slope number is unavailable.
Problem 4 A slanted perpendicular bisector

Find the perpendicular bisector of $EF$ for
$\begin{aligned}E&=(1,-6),\\F&=(7,2).\end{aligned}$
Give an equation and verify its midpoint and direction conditions.

Only the segment endpoints are given.

Text description of this figure

A slanted segment to bisect. Marked points E at (1, -6); F at (7, 2). Segments and any given line identify the configuration. The two coordinate axes use equal scales.
1. Hint 1
  A perpendicular bisector needs both a point and a direction.
2. Hint 2
  Use the midpoint of $EF$ and the negative reciprocal of its slope.
Answer

$3x+4y=4$ , or any equivalent equation.

Full solution

The midpoint is $M=(4,-2)$ .
The segment slope is
$m_{EF}=\frac{2-(-6)}{7-1}$ $m_{EF}=\frac43.$
The perpendicular slope is $-3/4$ .

Through $M$ , point-slope form gives
$y+2=-\frac34(x-4).$
Clearing the fraction and collecting terms gives
$3x+4y=4.$

At $M$ , the left side is $3(4)+4(-2)$ , which is 4.
The product of the segment and bisector slopes is $-1$ .
These checks establish the two defining conditions.

Answer

$3x+4y=4$ , or any equivalent equation.

Key idea

Midpoint and perpendicular slope are independent checks on a perpendicular bisector.
Problem 5 Two definitions from one vertex

Triangle $ABC$ has the vertices below. Find the median line from $A$ and the altitude line from $A$ . State whether they coincide.
$\begin{aligned}A&=(-1,7),\\B&=(-5,0),\\C&=(7,4).\end{aligned}$

The triangle supplies one vertex and one opposite side for both tasks.

Text description of this figure

Choose the defining condition for each line. Marked points A at (-1, 7); B at (-5, 0); C at (7, 4). Segments and any given line identify the configuration. The two coordinate axes use equal scales.
1. Hint 1
  The median uses the opposite midpoint; the altitude uses the opposite side's direction.
2. Hint 2
  Find the midpoint and slope of $BC$ separately, then write the two lines through $A$ .
Answer

Median line:
$5x+2y=9.$
Altitude line:
$y=-3x+4.$
They are distinct.

Full solution

The midpoint of $BC$ is $M=(1,2)$ .
The slope from $A$ to $M$ is $-5/2$ .
The median therefore satisfies
$y-7=-\frac52(x+1),$
giving
$5x+2y=9.$

The slope of $BC$ is $1/3$ , so the altitude slope is $-3$ .
Through $A$ its equation is
$y-7=-3(x+1),$
or
$y=-3x+4.$

The slopes $-5/2$ and $-3$ differ, so the lines are distinct.
At $M$ , the altitude's equation gives height 1 rather than 2, checking that it does not pass through the opposite midpoint.

Answer

Median line:
$5x+2y=9.$
Altitude line:
$y=-3x+4.$
They are distinct.

Key idea

A median and an altitude satisfy different geometric conditions even though both pass through the same vertex.
Problem 6 Three altitude checks

Triangle $ABC$ has
$\begin{aligned}A&=(1,2),\\B&=(7,2),\\C&=(3,6).\end{aligned}$
Find the intersection $H$ of the altitude lines from $B$ and $C$ . Verify that $H$ also lies on the altitude line from $A$ .
1. Hint 1
  One opposite side is horizontal, so one altitude is vertical.
2. Hint 2
  Find the perpendicular to $AC$ through $B$ , and intersect it with the vertical altitude from $C$ .
3. Hint 3
  For the final check, use the direction of $BC$ to write the altitude from $A$ .
Answer

$H=(3,4)$ . The altitude from $A$ is
$y=x+1.$
It contains $H$ .

Full solution

Because $AB$ is horizontal, the altitude from $C$ has equation
$x=3.$
Side $AC$ has slope 2, so the altitude from $B$ has slope $-1/2$ .
Its equation is
$y-2=-\frac12(x-7).$

At first coordinate 3, the second equation gives
$y=4.$
Thus $H=(3,4)$ .
It satisfies both altitude equations.

Side $BC$ has slope $-1$ , so the perpendicular through $A$ has slope 1.
Its equation is
$y=x+1.$
Substitution of $H$ gives $4=3+1$ , verifying the third altitude condition.

The vertical altitude and slanted altitude meet at H; the third altitude reaches the same point.

Answer

$H=(3,4)$ . The altitude from $A$ is
$y=x+1.$
It contains $H$ .

Key idea

An intersection calculation becomes a geometric verification when each line has been justified from its opposite side.
Problem 7 Locate a circumcenter with lines

Triangle $EFG$ has
$\begin{aligned}E&=(0,0),\\F&=(8,0),\\G&=(2,6).\end{aligned}$
Find the intersection $O$ of the perpendicular bisectors of $EF$ and $EG$ . Verify that it is equidistant from all three vertices using squared distances.

The diagram shows only the original triangle; the circumcenter is to be found.

Text description of this figure

Use two perpendicular bisectors. Marked points E at (0, 0); F at (8, 0); G at (2, 6). Segments and any given line identify the configuration. The two coordinate axes use equal scales.
1. Hint 1
  Each perpendicular bisector is found from its own midpoint and perpendicular direction.
2. Hint 2
  The bisector of $EF$ is vertical. Substitute its fixed first coordinate into the bisector equation for $EG$ .
3. Hint 3
  Compare squared distances from the intersection to the three vertices; equal nonnegative squared distances give equal distances.
Answer

$O=(4,2)$ ; all three squared distances are 20, so all three distances are $2\sqrt5$ .

Full solution

The midpoint of $EF$ is $(4,0)$ , and $EF$ is horizontal.
Its perpendicular bisector is
$x=4.$

The midpoint of $EG$ is $(1,3)$ and the segment slope is 3.
The second perpendicular bisector satisfies
$y-3=-\frac13(x-1),$
or
$x+3y=10.$
At first coordinate 4, the height is 2, giving $O=(4,2)$ .

The independent distance checks are
$\begin{aligned}OE^2&=4^2+2^2,\\OE^2&=20,\\OF^2&=(-4)^2+2^2,\\OF^2&=20,\\OG^2&=2^2+(-4)^2,\\OG^2&=20.\end{aligned}$
Each distance is the nonnegative square root of 20.
The point is therefore the triangle's circumcenter.

Answer

$O=(4,2)$ ; all three squared distances are 20, so all three distances are $2\sqrt5$ .

Key idea

Perpendicular-bisector intersections can be checked through endpoint distances without using a circle equation.
Problem 8 A meeting point on a path

A park map uses equal coordinate scales measured in meters. Gates are at
$\begin{aligned}A&=(-2,-4),\\B&=(4,2).\end{aligned}$
A meeting point must be equally far from the two gates and lie on the straight path below. Find its coordinates.
$y=-2x+5.$

The original gates and the given path are shown; the requested meeting point is not marked.

Text description of this figure

Locate an equally distant point on the path. Marked points A at (-2, -4); B at (4, 2). Solid and dashed lines distinguish related objects. The two coordinate axes use equal scales.
1. Hint 1
  Points equally far from two fixed points lie on their perpendicular bisector.
2. Hint 2
  Find that bisector, then solve its equation together with the path equation.
Answer

The meeting point is $(5,-5)$ .

Full solution

The midpoint of the gates is $(1,-1)$ , and $AB$ has slope 1.
The perpendicular bisector has slope $-1$ and equation
$y=-x.$

At its intersection with the path,
$\begin{aligned}-x&=-2x+5,\\x&=5,\\y&=-5.\end{aligned}$
The meeting point is $(5,-5)$ .

The squared distances to the gates are $7^2+(-1)^2$ and $1^2+(-7)^2$ , each 50.
Thus the distances agree, and substitution in the path equation gives height $-5$ as required.

Answer

The meeting point is $(5,-5)$ .

Key idea

An equidistance condition supplies a geometric line that can be intersected with a route constraint.
Problem 9 Recover a vertex from an altitude

Triangle $ABC$ has the vertices
$\begin{aligned}A&=(-2,1),\\B&=(4,7).\end{aligned}$
Vertex $C$ lies on $x=8$ . Its altitude to the full line of $AB$ lies on $y=-x+11$ .

Find $C$ , then find the median line from $C$ to $AB$ . Verify the altitude direction.
1. Hint 1
  The vertex lies on its altitude as well as the given vertical line.
2. Hint 2
  After finding the vertex, use the midpoint of $AB$ for the median. Keep the midpoint and perpendicular conditions separate.
Answer

$C=(8,3)$ . The median is
$x+7y=29.$
The slopes of $AB$ and the altitude are $1$ and $-1$ .

Full solution

At first coordinate $8$ , the altitude equation gives
$\begin{aligned}y&=-8+11,\\y&=3.\end{aligned}$
Therefore $C=(8,3)$ .

The midpoint of $AB$ is $M=(1,4)$ .
The median through $C$ and $M$ has slope $-1/7$ , so
$y-3=-\frac17(x-8).$
Clearing the fraction gives
$x+7y=29.$

The slope of $AB$ is 1, while the stated altitude slope is $-1$ .
Their product is $-1$ , and the recovered vertex lies on that altitude.
It is outside $AB$ , since the line of $AB$ is $y=x+3$ and at first coordinate 8 has height 11, not 3.

Answer

$C=(8,3)$ . The median is
$x+7y=29.$
The slopes of $AB$ and the altitude are $1$ and $-1$ .

Key idea

Inverse geometry problems require recovering the line's location before building another named line.
Problem 10 A median meets a perpendicular bisector

For a real number $t$ , triangle $ABC$ has the vertices below. Find the perpendicular bisector $p$ of $AB$ and the median line $n$ from $C$ . Determine all common points of $p$ and $n$ , including any exceptional value of $t$ .
$\begin{aligned}A&=(-4,1),\\B&=(4,1),\\C&=(t,7).\end{aligned}$
1. Hint 1
  The midpoint of $AB$ is fixed as $t$ varies. Compare its first coordinate with the first coordinate of $C$ .
2. Hint 2
  Separate the case where the median is vertical before dividing by a horizontal change.
3. Hint 3
  When the two line equations become identical, there is no unique intersection point.
Answer

$p$ is
$x=0$
. For $t\ne0$ , $n$ is
$y=\frac6t x+1$
and their only common point is $(0,1)$ . For
$t=0$
, both lines are
$x=0$
, with infinitely many common points.

Full solution

The base $AB$ is horizontal with midpoint $M=(0,1)$ .
Its perpendicular bisector is the fixed vertical line
$x=0.$
The vertex has height 7, so it is outside $AB$ for every real $t$ and the triangle never degenerates.

Case $t\ne0$ .
The median through $M$ and $C$ has nonzero run $t$ and rise 6, giving
$y=\frac6t x+1.$
Setting the first coordinate to zero forces height 1.
Thus the only common point is
$M=(0,1)$
.
Substitution of
$C=(t,7)$
also gives height 7, checking the median.

Case $t=0$ .
Both $M$ and $C$ have first coordinate zero, so the median is vertical with equation
$x=0.$
It coincides with the perpendicular bisector.
Every point on that vertical line is common to them.
The fraction $6/t$ is undefined here, so the nonvertical formula cannot be used in this case.

At t equal to zero, one vertical line represents both the median and perpendicular bisector.

Answer

$p$ is
$x=0$
. For $t\ne0$ , $n$ is
$y=\frac6t x+1$
and their only common point is $(0,1)$ . For
$t=0$
, both lines are
$x=0$
, with infinitely many common points.

Key idea

A geometric line can become vertical and coincide with another line at a parameter value that a slope formula excludes.

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