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Distance and Midpoint: 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 Supply the perpendicular corner

The endpoints of $AB$ are $(-7,1)$ and $(-2,5)$ . Name the corner $H$ reached by moving horizontally from $A$ and then vertically to $B$ . Find both leg lengths and the exact length $AB$ .

Endpoints before the perpendicular corner is added.

Text description of this figure

Endpoints before the perpendicular corner is added. A is at (-7, 1); B is at (-2, 5). The axes use equal scales; the surrounding problem supplies the coordinates.
1. Hint 1
  A horizontal move keeps the starting height; the vertical move uses the ending column.
2. Hint 2
  Read the two gaps, then apply the right-triangle relation.
Answer

$H=(-2,1)$ ; horizontal leg $5$ units; vertical leg $4$ units; $AB=\sqrt{41}$ units.

Full solution

The corner uses B’s x-coordinate and A’s y-coordinate.
$H=(-2,1)$

The horizontal gap is $|-2-(-7)|=5$ units.
The vertical gap is $|5-1|=4$ units.

These legs are perpendicular, so
$AB^2=5^2+4^2$

Taking the nonnegative square root gives
$AB=\sqrt{41}$

The coordinate right triangle used in the solution.

Answer

$H=(-2,1)$ ; horizontal leg $5$ units; vertical leg $4$ units; $AB=\sqrt{41}$ units.

Key idea

A coordinate corner combines one endpoint’s height with the other endpoint’s horizontal position.
Problem 2 Two orientations

Case A: Find the length and midpoint of the segment joining $(-6,-4)$ and $(3,-4)$ .

Case B: Find the length and midpoint of the segment joining $(2,-7)$ and $(2,-1)$ .
1. Hint 1
  Identify which coordinate stays fixed in each case.
2. Hint 2
  The changing coordinate gives the distance; averaging coordinates gives the midpoint.
Answer

Case A: $9$ units, midpoint $(-3/2,-4)$ . Case B: $6$ units, midpoint $(2,-4)$ .

Full solution

Case A is horizontal.
Its length is
$|3-(-6)|=9$
Its midpoint keeps height $-4$ and averages the horizontal readings, giving $(-3/2,-4)$ .

Case B is vertical.
Its length is
$|-1-(-7)|=6$
Its midpoint keeps x-coordinate 2 and averages the heights, giving $(2,-4)$ .

Answer

Case A: $9$ units, midpoint $(-3/2,-4)$ . Case B: $6$ units, midpoint $(2,-4)$ .

Key idea

Horizontal and vertical cases use the same midpoint rule even though only one coordinate changes.
Problem 3 A midpoint between grid intersections

Find the midpoint of the segment joining $(-8,3)$ and $(-1,-6)$ . State whether it lies left or right of the y-axis and above or below the x-axis.
1. Hint 1
  Find the average of each coordinate pair.
2. Hint 2
  The signs of the two averages tell you the side of each axis.
Answer

Midpoint $(-9/2,-3/2)$ ; left of the y-axis and below the x-axis.

Full solution

The horizontal sum is $-9$ , so
$x_M=-9/2$

The vertical sum is $-3$ , so
$y_M=-3/2$

Both coordinates are negative.
The midpoint is left of the y-axis and below the x-axis.
Fractional coordinates describe valid positions between grid intersections.

Answer

Midpoint $(-9/2,-3/2)$ ; left of the y-axis and below the x-axis.

Key idea

The midpoint’s coordinates need not be integers, even when both endpoint coordinates are integers.
Problem 4 Recover and verify

Point $M$ at $(3/2,-1)$ is the midpoint of $AB$ . Endpoint $A$ is at $(-2,3)$ . Find $B$ . Then find the exact length $AB$ and check that each half has half that length.

One endpoint and the midpoint of a longer segment.

Text description of this figure

One endpoint and the midpoint of a longer segment. A is at (-2, 3); M is at (1.5, -1). The axes use equal scales; the surrounding problem supplies the coordinates.
1. Hint 1
  Recover the coordinate sums from twice the midpoint.
2. Hint 2
  Subtract the known endpoint to find the other endpoint.
3. Hint 3
  Compare the changes on each half with those on the full segment.
Answer

$B=(5,-5)$ ; $AB=\sqrt{113}$ units; each half has length $\sqrt{113}/2$ units.

Full solution

The midpoint gives the coordinate sums.
$\begin{aligned}x_B&=2(3/2)-(-2),\\x_B&=5,\\y_B&=2(-1)-3,\\y_B&=-5.\end{aligned}$

The full horizontal and vertical changes are 7 and $-8$ .
$AB^2=7^2+(-8)^2$
Thus $AB=\sqrt{113}$ units.

Each half has horizontal gap $7/2$ and vertical gap 4.
$AM^2=113/4$
The same is true of MB, so each half has length $\sqrt{113}/2$ units.

Averaging $(-2,3)$ and $(5,-5)$ also returns $(3/2,-1)$ , checking both coordinates.

Answer

$B=(5,-5)$ ; $AB=\sqrt{113}$ units; each half has length $\sqrt{113}/2$ units.

Key idea

Reconstruction can be checked with both coordinate averages and exact half-lengths.
Problem 5 A cable and its center marker

A straight cable joins positions $A$ at $(-3,2)$ and $B$ at $(5,7)$ on a plan. Each coordinate unit represents 2 meters. A marker is attached halfway along the cable. Find the exact cable length in meters and the marker’s coordinates on the plan.

The cable endpoints on the plan.

Text description of this figure

The cable endpoints on the plan. A is at (-3, 2); B is at (5, 7). The axes use equal scales; the surrounding problem supplies the coordinates.
1. Hint 1
  The plan uses a length scale, and the cable is a straight segment.
2. Hint 2
  Use the coordinate gaps for the plan length, then convert units.
3. Hint 3
  The marker is the segment’s midpoint.
Answer

Cable length $2\sqrt{89}$ meters; marker $(1,9/2)$ .

Full solution

The horizontal gap is 8 plan units and the vertical gap is 5.
$d^2=8^2+5^2$
Thus the plan length is $\sqrt{89}$ units.

Each plan unit is 2 meters, so the cable length is $2\sqrt{89}$ meters.

The midpoint averages the plan coordinates.
$\begin{aligned}x_M&=(-3+5)/2,\\x_M&=1,\\y_M&=(2+7)/2,\\y_M&=9/2.\end{aligned}$

Answer

Cable length $2\sqrt{89}$ meters; marker $(1,9/2)$ .

Key idea

Convert the computed length using the plan scale while keeping the midpoint in the plan’s coordinate system.
Problem 6 Two proposed midpoint locations

The segment endpoints are $A$ at $(-3,0)$ and $B$ at $(5,2)$ . Two marked points are $P$ at $(1,1)$ and $Q$ at $(0,5)$ . Decide separately whether $P$ and $Q$ are midpoints of $AB$ . Find the squared distances from each marked point to the two endpoints to explain why equal distances alone do not settle the question.

Two marked points compared with one segment.

Text description of this figure

Two marked points compared with one segment. A is at (-3, 0); B is at (5, 2); P is at (1, 1); Q is at (0, 5). The axes use equal scales; the surrounding problem supplies the coordinates.
1. Hint 1
  A midpoint must lie on the segment as well as have equal endpoint distances.
2. Hint 2
  Compute the endpoint coordinate averages, then check the four squared distances.
Answer

P is the midpoint; Q is not.
$\begin{aligned}PA^2&=17,\\PB^2&=17,\\QA^2&=34,\\QB^2&=34.\end{aligned}$

Full solution

The average of A and B is $(1,1)$ , exactly P.
Hence P is the midpoint.

Its endpoint distances have squares
$\begin{aligned}PA^2&=4^2+1^2,\\PA^2&=17,\\PB^2&=4^2+1^2,\\PB^2&=17.\end{aligned}$

For Q the squared distances are
$\begin{aligned}QA^2&=3^2+5^2,\\QA^2&=34,\\QB^2&=(-5)^2+3^2,\\QB^2&=34.\end{aligned}$

Q is equally far from the endpoints, but it differs from the coordinate midpoint.
It is not on segment AB: every point on that segment has height between 0 and 2, whereas Q has height 5.

Answer

P is the midpoint; Q is not.
$\begin{aligned}PA^2&=17,\\PB^2&=17,\\QA^2&=34,\\QB^2&=34.\end{aligned}$

Key idea

Equal endpoint distances must be combined with the segment condition when identifying a midpoint.
Problem 7 Check a proposed SSS order

Triangle $ABC$ has vertices $A$ at $(-5,-1)$ , $B$ at $(-1,1)$ , and $C$ at $(-4,4)$ . Triangle $DEF$ has vertices $D$ at $(5,-3)$ , $E$ at $(3,1)$ , and $F$ at $(0,-2)$ . Decide whether $\triangle ABC\cong\triangle DEF$ in that order. Justify all three side pairs with exact squared lengths.

A solid triangle and a dashed triangle with a proposed correspondence.

Text description of this figure

A solid triangle and a dashed triangle with a proposed correspondence. A is at (-5, -1); B is at (-1, 1); C is at (-4, 4); D is at (5, -3); E is at (3, 1); F is at (0, -2). The axes use equal scales; the surrounding problem supplies the coordinates.
1. Hint 1
  Match the side pairs using the stated vertex order.
2. Hint 2
  Calculate each pair of coordinate differences and compare their square sums.
Answer

Yes, by SSS. The corresponding squared lengths are $20$ for AB and DE, $26$ for AC and DF, and $18$ for BC and EF.

Full solution

The proposed correspondence is A to D, B to E, and C to F.

The first side pair has squares
$\begin{aligned}AB^2&=4^2+2^2,\\AB^2&=20,\\DE^2&=(-2)^2+4^2,\\DE^2&=20.\end{aligned}$

The second side pair has squares
$\begin{aligned}AC^2&=1^2+5^2,\\AC^2&=26,\\DF^2&=(-5)^2+1^2,\\DF^2&=26.\end{aligned}$

The last side pair has squares
$\begin{aligned}BC^2&=(-3)^2+3^2,\\BC^2&=18,\\EF^2&=(-3)^2+(-3)^2,\\EF^2&=18.\end{aligned}$

The positive corresponding lengths therefore match.
The triangles are nondegenerate: their three side lengths satisfy the strict triangle inequalities.
SSS proves the stated congruence.

Answer

Yes, by SSS. The corresponding squared lengths are $20$ for AB and DE, $26$ for AC and DF, and $18$ for BC and EF.

Key idea

A stated congruence order determines which three squared-distance comparisons are needed.
Problem 8 Every possible horizontal position

One endpoint is $A$ at $(-2,-3)$ . The other is $B$ at $(t,1)$ for a real number $t$ . The segment length is $\sqrt{52}$ units. Find every possible $B$ and the midpoint in each case.
1. Hint 1
  The vertical change is already fixed.
2. Hint 2
  Subtract its square from the whole length squared.
3. Hint 3
  A nonzero horizontal square allows both a left and a right position.
Answer

$B$ at $(4,1)$ with midpoint $(1,-1)$ , or $B$ at $(-8,1)$ with midpoint $(-5,-1)$ .

Full solution

The vertical change is 4.
The distance condition is
$\begin{aligned}(t+2)^2+16&=52,\\(t+2)^2&=36.\end{aligned}$

For the right position, solve
$\begin{aligned}t+2&=6,\\t&=4.\end{aligned}$
This places B at $(4,1)$ .
Averaging with A gives midpoint $(1,-1)$ .

For the left position, solve
$\begin{aligned}t+2&=-6,\\t&=-8.\end{aligned}$
This places B at $(-8,1)$ .
Averaging with A gives midpoint $(-5,-1)$ .

Both have gap sizes 6 and 4, giving squared length 52.
The equation has just these two real solutions, so the list is complete.

Answer

$B$ at $(4,1)$ with midpoint $(1,-1)$ , or $B$ at $(-8,1)$ with midpoint $(-5,-1)$ .

Key idea

Keep both signs of a coordinate difference while keeping the distance itself nonnegative.
Problem 9 Compare sides and locate a median endpoint

Triangle $ABC$ has vertices $A$ at $(0,0)$ , $B$ at $(6,2)$ , and $C$ at $(2,7)$ . Let $M$ be the midpoint of $BC$ . Find $M$ and the exact length $AM$ . Identify the longest side of $ABC$ , with an exact comparison.

The triangle before its side midpoint is located.

Text description of this figure

The triangle before its side midpoint is located. A is at (0, 0); B is at (6, 2); C is at (2, 7). The axes use equal scales; the surrounding problem supplies the coordinates.
1. Hint 1
  The midpoint of BC locates the endpoint of the segment from A.
2. Hint 2
  Compute the squared lengths before taking roots.
3. Hint 3
  Nonnegative lengths have the same ordering as their squares.
Answer

$M=(4,9/2)$ ; $AM=\sqrt{145}/2$ units; AC is the longest side.

Full solution

Average the coordinates of B and C.
$M=(4,9/2)$

Compute the segment from the origin A to M.
$\begin{aligned}AM^2&=4^2+(9/2)^2,\\AM^2&=145/4,\\AM&=\sqrt{145}/2.\end{aligned}$

The side-square comparisons are
$\begin{aligned}AB^2&=6^2+2^2,\\AB^2&=40,\\AC^2&=2^2+7^2,\\AC^2&=53,\\BC^2&=(-4)^2+5^2,\\BC^2&=41.\end{aligned}$

Since 53 is the largest square, AC is the longest side.
No rounded lengths are needed.

Answer

$M=(4,9/2)$ ; $AM=\sqrt{145}/2$ units; AC is the longest side.

Key idea

A midpoint and exact squared lengths can answer different questions about the same triangle.
Problem 10 Reconstruct a feasible triangle

Point $M$ at $(2,1)$ is the midpoint of $AB$ . Point $N$ at $(2,4)$ is the midpoint of $AC$ . Point $A$ lies on the x-axis, and $AB$ has length $2\sqrt{10}$ units.

Find every ordered triple $(A,B,C)$ satisfying these data. Check both midpoint conditions and the length in each case, and explain why each triple makes a nondegenerate triangle.
1. Hint 1
  Represent A using its one unknown horizontal coordinate.
2. Hint 2
  Each midpoint determines the other endpoint once A is known.
3. Hint 3
  The length condition gives a squared equation with two possible horizontal positions.
4. Hint 4
  Compare the common x-coordinate of B and C with A’s x-coordinate.
Answer

$(A,B,C)=((-1,0),(5,2),(5,8))$ or $((5,0),(-1,2),(-1,8))$ . Both are nondegenerate and satisfy every condition.

Full solution

Write $A=(a,0)$ .
The two midpoint conditions recover
$\begin{aligned}B&=(4-a,2),\\C&=(4-a,8).\end{aligned}$

The AB gaps are $4-2a$ horizontally and 2 vertically.
Thus
$\begin{aligned}(4-2a)^2+4&=40,\\(4-2a)^2&=36.\end{aligned}$

The positive horizontal change gives
$\begin{aligned}4-2a&=6,\\a&=-1.\end{aligned}$
This yields A at $(-1,0)$ , B at $(5,2)$ , and C at $(5,8)$ .

The negative horizontal change gives
$\begin{aligned}4-2a&=-6,\\a&=5.\end{aligned}$
This yields A at $(5,0)$ , B at $(-1,2)$ , and C at $(-1,8)$ .

In either case, averaging A with B gives $(2,1)$ and averaging A with C gives $(2,4)$ .
The AB gap sizes are 6 and 2, so
$AB^2=6^2+2^2$
This is 40, giving the required positive length $2\sqrt{10}$ .

B and C are distinct points on the same vertical line.
A has a different x-coordinate in each case, so it is off that line.
Each triangle is nondegenerate.
The squared equation has only the two solved cases, so no triples are missing.

The first solution is shown. The second reverses the horizontal placements around x equal to 2.

Answer

$(A,B,C)=((-1,0),(5,2),(5,8))$ or $((5,0),(-1,2),(-1,8))$ . Both are nondegenerate and satisfy every condition.

Key idea

Coupled midpoint conditions reconstruct the configuration; the distance and noncollinearity checks establish feasibility.

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