Distance and Midpoint

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Learning goals

Derive straight-line distance from perpendicular coordinate gaps and the Pythagorean theorem
Locate a midpoint and explain why both coordinates are averaged
Recover missing endpoint coordinates and check all geometric conditions
Justify numerical triangle congruence by matching three exact side lengths

From coordinates to a right triangle

An ordered pair gives horizontal position first, then vertical position. Use the same unit length on both perpendicular axes. For the points $(-4,2)$ and $(2,-6)$ , the horizontal change is $6$ and the vertical change is $-8$ . The negative sign means down; the vertical length is $8$ .

A horizontal step and a vertical step form a right angle at H. The direct segment is AB.

The two perpendicular legs let you measure the direct segment with the Pythagorean theorem:

\begin{aligned} AB^2&=6^2+8^2,\\ AB^2&=100,\\ AB&=10. \end{aligned}

Adding $6$ and $8$ would measure a bent route through $H$ , not the straight segment.

Check your understanding

From $(-2,3)$ to $(5,-1)$ , what are the horizontal and vertical leg lengths of the coordinate right triangle?

3

and

2

7

and

4

7

and

-4

5

and

1

Distance from perpendicular coordinate gaps#

Take a fresh pair $U=(-3,1)$ and $V=(9,6)$ . The corner $K=(9,1)$ gives perpendicular legs of lengths $12$ and $5$ . The Pythagorean theorem gives

\begin{aligned} UV^2&=12^2+5^2,\\ UV^2&=169. \end{aligned}

Thus the direct distance is $13$ .

Take endpoints $A$ at $(x_A,y_A)$ and $B$ at $(x_B,y_B)$ . The corner is $H$ at $(x_B,y_A)$ . When both coordinates change, the legs are perpendicular and have lengths $|x_B-x_A|$ and $|y_B-y_A|$ . The theorem gives

AB^2=(x_B-x_A)^2+(y_B-y_A)^2.

Absolute values disappear on squaring because a number and its opposite have the same square. Distance is nonnegative, so

AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}.

The actual coordinate values did not matter; the perpendicular gaps did. If the segment is horizontal or vertical, one gap is zero and this expression reduces to the other gap’s absolute value. No nondegenerate right triangle is needed in those cases. If the endpoints coincide, both gaps are zero and the distance is zero. Reversing the endpoints changes both signs but leaves both squares unchanged.

This is the distance formula. It measures a length, so its output is a number in the coordinate unit, not an ordered pair.

Worked example 1 Keep a slanted length exact

Problem. The endpoints of segment $AB$ are $(-5,-3)$ and $(1,6)$ . Find its exact length in simplest radical form.

The endpoints are given. Draw the horizontal and vertical legs when solving.

Solution. The horizontal gap is $6$ units and the vertical gap is $9$ units. Thus

\begin{aligned} AB^2&=6^2+9^2,\\ AB^2&=117,\\ AB&=\sqrt{117},\\ AB&=3\sqrt{13}. \end{aligned}

The square factor $9$ comes out of the radical as $3$ . The radical is an exact length; no decimal is needed.

Check your understanding

Find the exact distance between $(-4,-1)$ and $(0,5)$ .

10

52

2\sqrt{13}

\sqrt{10}

Why a midpoint averages both coordinates

A midpoint is the point on a segment that divides it into two equal lengths. Halfway along a straight segment means making half its horizontal change and half its vertical change together.

For instance, from $(-6,-2)$ to $(4,6)$ the changes are $10$ right and $8$ up. Starting at the first endpoint, half of each change gives $(-1,2)$ . A dilation centered at the first endpoint with scale factor $1/2$ puts this point on the segment and makes its distance from that endpoint half the whole length. The other half is the remaining segment.

The half-size right triangle ends at M on AB. Its horizontal and vertical changes are half the original changes.

Midpoint coordinates from a half-size dilation#

For a fresh pair $U=(-5,9)$ and $V=(11,-3)$ , the signed coordinate changes are $16$ and $-12$ . A half-size dilation from $U$ halves both changes:

\begin{aligned} x_M&=-5+8,\\ x_M&=3,\\ y_M&=9-6,\\ y_M&=3. \end{aligned}

It puts $M=(3,3)$ on $UV$ at half its length, rather than merely somewhere equally far from the endpoints.

For general distinct endpoints $A$ at $(x_A,y_A)$ and $B$ at $(x_B,y_B)$ , the same dilation halves both coordinate changes. Name the midpoint $M$ and its coordinates $(x_M,y_M)$ . Then

\begin{aligned} x_M&=x_A+\frac{x_B-x_A}{2},\\ x_M&=\frac{x_A+x_B}{2}. \end{aligned}

The vertical calculation has the same form:

\begin{aligned} y_M&=y_A+\frac{y_B-y_A}{2},\\ y_M&=\frac{y_A+y_B}{2}. \end{aligned}

No step depended on the example’s values or on the signs of the changes. A dilation works in any direction, including horizontal and vertical segments. The point stays on the segment and cuts its length in half.

The midpoint formula is therefore

M=\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right).

Average matching coordinates; do not average an x-coordinate with a y-coordinate. Fractional coordinates are allowed. If the endpoints coincide, the same average returns that point, although there are no two positive half-segments.

Worked example 2 Move a vertical endpoint

Problem. Segment $AB$ has endpoints $A$ at $(-3,-5)$ and $B$ at $(-3,7)$ . Find its length and midpoint. Then move $B$ to $C$ at $(-3,3)$ and find the length and midpoint of $AC$ .

The original endpoint is B; the moved endpoint is C. Both segments remain vertical. No midpoint is marked before it is found.

Solution. For the original segment, the x-coordinate stays $-3$ . Call its midpoint $M$ . Its length and midpoint are

\begin{aligned} AB&=|7-(-5)|,\\ AB&=12,\\ M&=(-3,1). \end{aligned}

The midpoint’s height is the average of $-5$ and $7$ . Each half is $6$ units.

For the moved endpoint, the segment is still vertical. Call the new midpoint $N$ :

\begin{aligned} AC&=|3-(-5)|,\\ AC&=8,\\ N&=(-3,-1). \end{aligned}

Here each half is $4$ units. Moving just the upper endpoint down $4$ units moved the midpoint down $2$ units.

Check your understanding

What is the midpoint of the segment joining $(5,-6)$ and $(-2,3)$ ?

(3,-3)

(7/2,-9/2)

(-3/2,3/2)

(3/2,-3/2)

Recover an endpoint

A midpoint tells you how far to continue beyond it. Repeat the horizontal and vertical changes from the known endpoint to the midpoint, continuing in the same direction.

Worked example 3 Use the midpoint in reverse

Problem. Point $M$ at $(2,-3)$ is the midpoint of $AB$ . One endpoint is $A$ at $(-4,5)$ . Find $B$ and check the midpoint.

M is the midpoint of the full segment AB. The unknown endpoint is beyond M from A.

Solution. The two coordinate averages must equal the given midpoint coordinates. Solve them separately:

\begin{aligned} \frac{-4+x_B}{2}&=2,\\ -4+x_B&=4,\\ x_B&=8. \end{aligned}

\begin{aligned} \frac{5+y_B}{2}&=-3,\\ 5+y_B&=-6,\\ y_B&=-11. \end{aligned}

So $B$ is at $(8,-11)$ . Averaging the endpoint coordinates gives $(2,-3)$ again. The changes from $M$ to $B$ are the same as from $A$ to $M$ , which also checks the straight continuation.

In general, multiply each midpoint coordinate by $2$ , then subtract the known endpoint coordinate. Check both coordinates afterward.

Check your understanding

The midpoint of $AB$ is $(-1,4)$ . Endpoint $A$ is at $(3,-2)$ . Where is $B$ ?

(-5,10)

(1,2)

(-2,8)

(5,-10)

Use exact lengths to check congruence

Coordinates can establish all three side pairs in the SSS congruence test. Equal squared lengths are enough: nonnegative lengths with equal squares are equal. Rounded decimals that merely look equal do not establish congruence.

Worked example 4 A numerical SSS check

Problem. Triangle $ABC$ has vertices $A$ at $(-1,-2)$ , $B$ at $(3,-1)$ , and $C$ at $(0,2)$ . Triangle $DEF$ has vertices $D$ at $(8,0)$ , $E$ at $(7,4)$ , and $F$ at $(4,1)$ . Decide whether $\triangle ABC\cong\triangle DEF$ in that order. Justify all three side matches.

Solid ABC and dashed DEF have the proposed order A to D, B to E, C to F. No equality or right-angle marks are assumed.

Solution. The proposed correspondence pairs $AB$ with $DE$ , $AC$ with $DF$ , and $BC$ with $EF$ . Compute their squares:

\begin{aligned} AB^2&=4^2+1^2,\\ AB^2&=17,\\ DE^2&=(-1)^2+4^2,\\ DE^2&=17. \end{aligned}

\begin{aligned} AC^2&=1^2+4^2,\\ AC^2&=17,\\ DF^2&=(-4)^2+1^2,\\ DF^2&=17. \end{aligned}

\begin{aligned} BC^2&=(-3)^2+3^2,\\ BC^2&=18,\\ EF^2&=(-3)^2+(-3)^2,\\ EF^2&=18. \end{aligned}

Each named pair has the same exact length. Both triples form nondegenerate triangles. Their two equal sides have length $\sqrt{17}$ , and the third has length $\sqrt{18}$ . The two equal sides add to more than the third because $68>18$ when those positive sums are squared. The other two inequalities each compare $\sqrt{17}+\sqrt{18}$ with $\sqrt{17}$ , so they also hold. SSS proves the stated congruence. Two matching sides alone would not have completed this check.

Check your understanding

For two nondegenerate triangles with a stated vertex correspondence, which coordinate calculation completes an SSS congruence check?

A Two corresponding side lengths match B The triangle perimeters match C All three corresponding squared side lengths match exactly D One pair of midpoints has equal coordinates

A distance condition can allow two endpoints

A known length need not locate one endpoint uniquely. A point can lie on either side of a fixed vertical line, so keep both signs when solving for a coordinate.

Worked example 5 Find every allowed endpoint

Problem. One endpoint is $A$ at $(1,-4)$ . The other is $B$ at $(t,2)$ , where $t$ is real, and $AB$ is $10$ units. Find every possible $B$ and the midpoint for each segment.

B must lie at height 2. Its horizontal position is still unknown; the dashed horizontal guide is not a segment length.

Solution. The vertical gap is $6$ , so the distance condition gives

\begin{aligned} (t-1)^2+6^2&=100,\\ (t-1)^2&=64. \end{aligned}

For the left-hand possibility,

\begin{aligned} t-1&=-8,\\ t&=-7. \end{aligned}

This gives $B$ at $(-7,2)$ and midpoint $(-3,-1)$ .

For the right-hand possibility,

\begin{aligned} t-1&=8,\\ t&=9. \end{aligned}

This gives $B$ at $(9,2)$ and midpoint $(5,-1)$ . Each has horizontal gap $8$ and vertical gap $6$ , so each length checks as $10$ units. These are all possibilities because the squared equation has only these two real solutions.

If the prescribed length were less than the fixed vertical gap, there would be no endpoint. If it equaled that gap, the horizontal gap would be zero and there would be one endpoint. This checks feasibility before you trust a sketch.

Distance and Midpoint

Learning goals

From coordinates to a right triangle

Why a midpoint averages both coordinates

Recover an endpoint

Use exact lengths to check congruence

A distance condition can allow two endpoints

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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