Geometry · G.GPE.B.6 · G.GPE.B.5

The midpoint formula

Two points are given by their coordinates. What are the coordinates of the point halfway between them, and how do you run that backwards, or write down every point that is equally far from both?

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Parallel and perpendicular linesMath 8 Perpendicular bisector of a segmentMath 8

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The midpoint formula

Coordinate proofs with variables

When you reach for it

The midpoint of a segment is the point on it that is equally far from both ends. On a number line it sits at the average of the two ends. On a coordinate grid the two ends are pairs of numbers, and you want the midpoint as a pair of numbers too.

Reach for the midpoint formula whenever a question names two points by their coordinates and asks for the point halfway between them: the center of a circle from the two ends of a diameter, the place a median of a triangle lands on the opposite side, the crossing point of the diagonals of a parallelogram.

Reach for it backwards when you know the midpoint and one end and need the other end. That turns up as often as the forward question: the far end of a diameter, the fourth corner of a parallelogram, the mirror image of a point.

And reach for it as the first step whenever you need the equation of the perpendicular bisector of a segment, the line through the midpoint at right angles to the segment. That line is exactly the set of points equally far from the two ends, which is why it is the tool for finding the center of a circle through given points.

The procedure

Every step below is shown on one running example: A = (3, 2) and B = (9, 6).

The midpoint formula

Walk from A to B. Along the way you move 9 - 3 = 6 units right and 6 - 2 = 4 units up. The midpoint should be halfway along the walk, so make half of each move: 3 right and 2 up. That lands on

M = (3 + 3,\ 2 + 2) = (6, 4)

Why is this point really the midpoint? Figure 1 shows the two halves of \overline{AB}. Each is the long side of a right triangle 3 across and 2 up. The two triangles have equal legs and a right angle between them, so they are congruent (side, angle, side), and their long sides are equal: AM = MB. And M is on the segment, because both halves climb at the same slope, \displaystyle \frac{2}{3}, and meet at M, so they line up into one straight segment rather than bending there.

Figure 1Half the run and half the rise. Both halves of \overline{AB} are the long side of a triangle 3 across and 2 up, so they are equal.

Now do the same with letters. The ends are (x_1, y_1) and (x_2, y_2). Half the move across, added to the start, is

x_1 + \frac{x_2 - x_1}{2} = \frac{2x_1 + x_2 - x_1}{2} = \frac{x_1 + x_2}{2}

and the same working gives \displaystyle \frac{y_1 + y_2}{2} upwards. So the midpoint is the average of the x-coordinates paired with the average of the y-coordinates:

M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)

Which end you call the first does not matter, because x_1 + x_2 = x_2 + x_1 (the commutative property of addition). The congruent-triangle argument never used the signs of the moves, so the formula holds when the segment runs down or to the left, with negative or fractional coordinates. For the running example, \displaystyle \left(\frac{3 + 9}{2}, \frac{2 + 6}{2}\right) = (6, 4), as before.

Running it backwards

Suppose you know the midpoint M = (x_M, y_M) and one end A = (x_1, y_1), and want the other end B = (x_2, y_2). The formula says \displaystyle x_M = \frac{x_1 + x_2}{2}. That is an equation with one unknown: multiply both sides by 2 and subtract x_1.

x_2 = 2x_M - x_1 \qquad\qquad y_2 = 2y_M - y_1

There is a second way to see the same thing. M is halfway, so the move from M to B is the same as the move from A to M. Make that move once more, starting at M: x_M + (x_M - x_1) = 2x_M - x_1, the same answer. In the running example, from A = (3, 2) to M = (6, 4) is 3 right and 2 up, and once more from M gives (9, 6), which is B.

The equation of the perpendicular bisector

The perpendicular bisector of \overline{AB} passes through the midpoint and meets \overline{AB} at a right angle. Each of those facts gives one thing a line equation needs.

A point on it: the midpoint. Here M = (6, 4).

Its slope: the negative reciprocal of the slope of \overline{AB}. The slope of \overline{AB} is

\frac{6 - 2}{9 - 3} = \frac{4}{6} = \frac{2}{3}

and two lines that are not level or upright are perpendicular exactly when their slopes multiply to -1. So the bisector has slope \displaystyle -\frac{3}{2}, since \displaystyle \frac{2}{3} \cdot \left(-\frac{3}{2}\right) = -1.

The equation: point-slope form. A line through (x_M, y_M) with slope m is y - y_M = m(x - x_M), by point-slope form. So

y - 4 = -\frac{3}{2}(x - 6)

and multiplying out, \displaystyle y - 4 = -\frac{3}{2}x + 9, so

y = -\frac{3}{2}x + 13

Check it with distances. The perpendicular bisector theorem says every point on this line is equally far from A and B. Take x = 4: then y = -6 + 13 = 7, so P = (4, 7) is on the line. By the distance formula,

PA^2 = (4 - 3)^2 + (7 - 2)^2 = 1 + 25 = 26

PB^2 = (4 - 9)^2 + (7 - 6)^2 = 25 + 1 = 26

so \displaystyle PA = PB = \sqrt{26}, as Figure 2 shows.

Figure 2The perpendicular bisector \displaystyle y = -\frac{3}{2}x + 13. The point P on it is \displaystyle \sqrt{26} from A and from B.

Level and upright segments. The slope step needs a slope with a negative reciprocal. A level segment has slope 0, and 0 has no reciprocal; its perpendicular bisector is the upright line through M, whose equation is x = x_M. An upright segment has no slope at all; its perpendicular bisector is the level line y = y_M. Neither case needs any calculation beyond the midpoint.

A second route, straight from the distances. A point (x, y) is on the perpendicular bisector exactly when its squared distances to A and B are equal:

(x - 3)^2 + (y - 2)^2 = (x - 9)^2 + (y - 6)^2

Expand both sides. The x^2 and y^2 appear once on each side and cancel, which is why the answer is a straight line:

\begin{aligned} -6x + 9 - 4y + 4 &= -18x + 81 - 12y + 36 \\ 12x + 8y &= 104 \\ 3x + 2y &= 26 \end{aligned}

and solving for y gives \displaystyle y = -\frac{3}{2}x + 13 again. This route takes more algebra, but it never divides by a slope, so it also works for level and upright segments without a special case.

Midpoints of complex numbers

In the complex plane the number a + bi sits at the point (a, b): the real part reads across and the imaginary part reads up. The midpoint formula averages the parts separately, and adding complex numbers also works on the parts separately, so the midpoint of z and w is simply

\frac{z + w}{2}

For 3 + 5i and -7 + i, that is \displaystyle \frac{-4 + 6i}{2} = -2 + 3i, which is the midpoint (-2, 3) of the points (3, 5) and (-7, 1). Running it backwards works the same way: if m is the midpoint and z is one end, the other end is 2m - z.

Worked examples

Example 1

Finding midpoints

Find the midpoint of the segment joining each pair of points.

(a) (-3, 8) and (5, 2) \qquad (b) (-7, -1) and (4, -6) \qquad (c) (2.4, -1.5) and (-0.6, 3.1)

Solution

Average the x-coordinates, then average the y-coordinates.

(a) \displaystyle \left(\frac{-3 + 5}{2}, \frac{8 + 2}{2}\right) = \left(\frac{2}{2}, \frac{10}{2}\right), so the midpoint is (1, 5).

(b) \displaystyle \left(\frac{-7 + 4}{2}, \frac{-1 + (-6)}{2}\right), so the midpoint is \displaystyle \left(-\frac{3}{2}, -\frac{7}{2}\right).

The ends are grid points but the midpoint is not, and nothing has gone wrong. The midpoint of two points with integer coordinates is itself a grid point exactly when both sums, x_1 + x_2 and y_1 + y_2, are even. Here both sums are odd.

(c) \displaystyle \left(\frac{2.4 + (-0.6)}{2}, \frac{-1.5 + 3.1}{2}\right) = \left(\frac{1.8}{2}, \frac{1.6}{2}\right), so the midpoint is (0.9, 0.8).

Example 2

The other end of a segment

M = (2, -1) is the midpoint of \overline{AB}, and A = (-4, 3). Find B.

Solution

Call B = (x, y). The midpoint formula, one coordinate at a time, gives two equations with one unknown each:

\frac{-4 + x}{2} = 2 \qquad\qquad \frac{3 + y}{2} = -1

Multiply both sides of each by 2: -4 + x = 4 and 3 + y = -2. So x = 8 and y = -5, and B = (8, -5).

The move picture gives the same point. From A to M is 2 - (-4) = 6 right and -1 - 3 = -4, that is 4 down. B is the same move again from M: (2 + 6, -1 - 4) = (8, -5).

Figure 3From A to M is 6 right and 4 down. The same move again, from M, lands on B.

Check. \displaystyle \left(\frac{-4 + 8}{2}, \frac{3 + (-5)}{2}\right) = (2, -1), which is M.

Example 3

Fractions and unknowns in the coordinates

(a) \displaystyle M = \left(-2, \frac{5}{2}\right) is the midpoint of \overline{AB}, and \displaystyle A = \left(\frac{1}{2}, -3\right). Find B.

(b) The midpoint of the segment from (2p, 7) to (6, p + 1) is (5, q). Find p and q.

Solution

(a) Use x_2 = 2x_M - x_1 and y_2 = 2y_M - y_1. Doubling the midpoint clears its fraction:

\begin{aligned} x_2 &= 2(-2) - \frac{1}{2} = -\frac{9}{2} \\ y_2 &= 2 \cdot \frac{5}{2} - (-3) = 8 \end{aligned}

So \displaystyle B = \left(-\frac{9}{2}, 8\right). Check: \displaystyle \frac{1}{2} + \left(-\frac{9}{2}\right) = -4, and half of that is -2; \displaystyle \frac{-3 + 8}{2} = \frac{5}{2}.

(b) The x-coordinates involve only p, so start there:

\begin{aligned} \frac{2p + 6}{2} &= 5 \\ p + 3 &= 5 \\ p &= 2 \end{aligned}

Now the second point is (6, 2 + 1) = (6, 3), and the y-coordinate of the midpoint is \displaystyle q = \frac{7 + 3}{2} = 5. So p = 2 and q = 5.

Example 4

The perpendicular bisector of a slanted segment

Find the equation of the perpendicular bisector of the segment from A = (-6, 1) to B = (2, 5).

Solution

The midpoint. \displaystyle \left(\frac{-6 + 2}{2}, \frac{1 + 5}{2}\right) = (-2, 3).

The slope. The slope of \overline{AB} is \displaystyle \frac{5 - 1}{2 - (-6)} = \frac{4}{8} = \frac{1}{2}. The perpendicular slope is its negative reciprocal, -2, since \displaystyle \frac{1}{2} \cdot (-2) = -1.

The equation. Point-slope form through (-2, 3) with slope -2:

y - 3 = -2(x + 2)

so y - 3 = -2x - 4, and y = -2x - 1.

Check. Take a point on the line, say x = 0, which gives P = (0, -1). Then PA^2 = 6^2 + (-2)^2 = 40 and PB^2 = (-2)^2 + (-6)^2 = 40, so P is equally far from both ends, as every point of the perpendicular bisector must be.

Example 5

Upright segments and a slope of -1

Find the equation of the perpendicular bisector of each segment.

(a) From C = (-5, 7) to D = (-5, -1) \qquad (b) From E = (-3, 4) to F = (5, -4)

Solution

(a) Both ends have x-coordinate -5, so \overline{CD} is upright, and its slope would mean dividing by -5 - (-5) = 0: it has none. There is no reciprocal to take, and none is needed. A line at right angles to an upright segment is level, and this one passes through the midpoint \displaystyle \left(-5, \frac{7 + (-1)}{2}\right) = (-5, 3). A level line through a point with y-coordinate 3 is

y = 3

(b) The midpoint is \displaystyle \left(\frac{-3 + 5}{2}, \frac{4 + (-4)}{2}\right) = (1, 0). The slope of \overline{EF} is \displaystyle \frac{-4 - 4}{5 - (-3)} = \frac{-8}{8} = -1, and the negative reciprocal of -1 is 1, since (-1) \cdot 1 = -1. Through (1, 0) with slope 1:

y - 0 = 1 \cdot (x - 1), \qquad\text{so}\qquad y = x - 1

Check. (0, -1) is on the line. Its squared distance to E is 3^2 + 5^2 = 34 and to F is 5^2 + 3^2 = 34.

Example 6

From the bisector back to the missing end

The line \ell with equation y = 2x - 5 is the perpendicular bisector of \overline{AB}, and A = (-1, 3). Find B.

Solution

Neither the midpoint nor B is given, but two facts pin them down.

\overline{AB} is perpendicular to \ell. The slope of \ell is 2, so the slope of \overline{AB} is \displaystyle -\frac{1}{2}. The line through A with that slope is \displaystyle y - 3 = -\frac{1}{2}(x + 1), that is

y = -\frac{1}{2}x + \frac{5}{2}

The midpoint is on \ell. The midpoint lies on \overline{AB} and on \ell, so it is where the two lines cross. Set the two expressions for y equal and multiply both sides by 2:

\begin{aligned} 2x - 5 &= -\frac{1}{2}x + \frac{5}{2} \\ 4x - 10 &= -x + 5 \\ 5x &= 15 \end{aligned}

So x = 3, and y = 2(3) - 5 = 1. The midpoint is M = (3, 1).

Run the midpoint backwards. B = (2 \cdot 3 - (-1),\ 2 \cdot 1 - 3) = (7, -1).

Check. The midpoint of (-1, 3) and (7, -1) is (3, 1), and 2(3) - 5 = 1, so it is on \ell. The slope of \overline{AB} is \displaystyle \frac{-1 - 3}{7 - (-1)} = -\frac{1}{2}, and \displaystyle 2 \cdot \left(-\frac{1}{2}\right) = -1. Folding the plane along \ell carries A onto B, so B is the mirror image of A in \ell: this is how a reflection in a slanted line is computed.

Example 7

The fourth corner of a parallelogram

ABCD is a parallelogram with A = (1, 2), B = (6, 3) and C = (8, 7). Find D.

Solution

The corners are named in order around the shape, so A is opposite C and B is opposite D. The diagonals of a parallelogram bisect each other, so the diagonals \overline{AC} and \overline{BD} have the same midpoint M.

Find M from the diagonal you know. \displaystyle M = \left(\frac{1 + 8}{2}, \frac{2 + 7}{2}\right) = \left(\frac{9}{2}, \frac{9}{2}\right).

Run it backwards along the other diagonal. M is the midpoint of \overline{BD} and B = (6, 3), so

D = \left(2 \cdot \frac{9}{2} - 6,\ 2 \cdot \frac{9}{2} - 3\right) = (3, 6)

Figure 4The diagonals of a parallelogram bisect each other, so \overline{AC} and \overline{BD} share their midpoint M.

Check. From A to B is 5 right and 1 up, and from D to C is 8 - 3 = 5 right and 7 - 6 = 1 up. The same move means \overline{AB} and \overline{DC} are parallel and equal in length, which is what a parallelogram needs.

Boundaries

Two different points. If the two ends are the same point, the “midpoint” is that point and there is no segment, and so no direction for a perpendicular line to be square to. The perpendicular bisector does not exist, and the distance route agrees: with equal ends, the equation reduces to 0 = 0, which every point satisfies.

The negative reciprocal needs a slope that is not 0. For a level segment the perpendicular bisector is x = x_M; for an upright one it is y = y_M. The route through equal distances handles both without a special case.

Running backwards needs an end as well as the midpoint. A midpoint on its own fixes nothing else: every segment through M that M cuts in half has it as its midpoint, and there are infinitely many.

The formula finds the halfway point only. A point a third of the way along, or any other fixed fraction, uses the same idea of taking part of each move, and is the subject of partitioning a segment in a ratio.

Perpendicular slopes need square axes. The midpoint formula only halves moves, so it would still be right on a grid whose axes were drawn slanted or with different units across and up. The rule that perpendicular slopes multiply to -1, and the distance check, both need the usual grid: axes at right angles with the same unit on each.

In space, a plane, not a line. With three coordinates the midpoint is still the average of each coordinate, but the points equally far from two ends fill a whole plane at right angles to the segment.

Where this goes next

Averaging two points is the start of averaging more. The balance point of a triangle, its Centroid, is the average of its three corners, and it lies two thirds of the way along each median, which runs from a corner to a midpoint found with this formula. Taking part of each move rather than half of it gives partitioning a segment in a ratio.

The perpendicular bisector equation leads to the Circumcenter: write the perpendicular bisectors of two sides of a triangle and solve them together, and the crossing point is equally far from all three corners, the center of the circle through them. With the distance formula for the radius, that gives the equation of the circle.

And the formula with letters in place of numbers is the main tool of coordinate proofs. Put a parallelogram at (0, 0), (a, 0), (a + b, c) and (b, c), compute the midpoints of both diagonals, and both come out as \displaystyle \left(\frac{a + b}{2}, \frac{c}{2}\right): that one line proves the diagonals bisect each other for every parallelogram at once.

Exercises

Find the midpoint

(2, -7) and (8, 3)
(-9, 6) and (-1, -2)
(3, -4) and (-2, 5)
(1.5, 2.2) and (-3.5, 0.8)

Find the missing endpoint

M = (4, 1) is the midpoint of \overline{AB} and A = (1, -3). Find B.
M = (-3, 2) is the midpoint of \overline{AB} and A = (5, -6). Find B.
The origin is the midpoint of \overline{AB} and A = (-7, 4). Find B.
\displaystyle M = \left(\frac{1}{2}, -\frac{5}{2}\right) is the midpoint of \overline{AB} and A = (3, -1). Find B.
M = (2.5, -1.5) is the midpoint of \overline{AB} and B = (-4, 6). Find A.
The midpoint of the segment from (a, 3) to (5, b) is (-1, 7). Find a and b.

The perpendicular bisector

Find the equation of the perpendicular bisector of the segment from A to B.

A = (-3, -2), B = (5, 2)
A = (2, 7), B = (2, -3)
A = (-4, 6), B = (8, 6)
A = (1, -2), B = (4, 3)
A = (-5, 4), B = (1, -2)
Is P = (4, 6) on the perpendicular bisector of the segment from (0, 0) to (10, 2)? Answer it twice: once with the equation of the bisector, once with distances.

Using the midpoint

A diameter of a circle runs from (-3, 2) to (5, -4). Find the center and the radius of the circle.
PQRS is a parallelogram with P = (-2, -1), Q = (3, 0) and R = (5, 4). Find S.
The perpendicular bisector of \overline{AB} is the line y = x + 1, and A = (4, 1). Find B.
Find the midpoint of 6 - 2i and -4 + 8i in the complex plane.
In the complex plane, 2 - i is the midpoint of the segment from 5 + 3i to w. Find w.

Reasoning

Explain, using the move from one end to the other, why the x-coordinate of the midpoint is the average of the two x-coordinates.
Use the distance formula to show that \displaystyle M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) is equally far from (x_1, y_1) and (x_2, y_2), and that each distance is half the length of the segment.
The perpendicular bisector of a level segment cannot be written as y = mx + b. Explain why, and give its equation.
Show that setting the squared distances from (x, y) to (x_1, y_1) and to (x_2, y_2) equal always gives a straight line, and that the midpoint of the segment lies on it.

Answers

(5, -2). \displaystyle \frac{2 + 8}{2} = 5 and \displaystyle \frac{-7 + 3}{2} = -2.
(-5, 2). \displaystyle \frac{-9 + (-1)}{2} = -5 and \displaystyle \frac{6 + (-2)}{2} = 2.
\displaystyle \left(\frac{1}{2}, \frac{1}{2}\right). Both sums are 1, which is odd, so the midpoint is not a grid point.
(-1, 1.5). \displaystyle \frac{1.5 + (-3.5)}{2} = \frac{-2}{2} and \displaystyle \frac{2.2 + 0.8}{2} = \frac{3}{2}.
B = (7, 5). 2 \cdot 4 - 1 = 7 and 2 \cdot 1 - (-3) = 5.
B = (-11, 10). 2(-3) - 5 = -11 and 2 \cdot 2 - (-6) = 10.
B = (7, -4). Doubling (0, 0) gives (0, 0), so B is A with both signs changed.
B = (-2, -4). \displaystyle 2 \cdot \frac{1}{2} - 3 = -2 and \displaystyle 2 \cdot \left(-\frac{5}{2}\right) - (-1) = -4.
A = (9, -9). Run it backwards from B: 2(2.5) - (-4) = 9 and 2(-1.5) - 6 = -9.
a = -7, b = 11. \displaystyle \frac{a + 5}{2} = -1 gives a + 5 = -2; \displaystyle \frac{3 + b}{2} = 7 gives 3 + b = 14.
y = -2x + 2. The midpoint is (1, 0) and the slope of \overline{AB} is \displaystyle \frac{4}{8} = \frac{1}{2}, so the bisector has slope -2: y - 0 = -2(x - 1).
y = 2. The segment is upright, so the bisector is the level line through the midpoint (2, 2).
x = 2. The segment is level, so the bisector is the upright line through the midpoint (2, 6).
\displaystyle y = -\frac{3}{5}x + 2. The midpoint is \displaystyle \left(\frac{5}{2}, \frac{1}{2}\right) and the slope of \overline{AB} is \displaystyle \frac{5}{3}, so the bisector has slope \displaystyle -\frac{3}{5}: \displaystyle y - \frac{1}{2} = -\frac{3}{5}\left(x - \frac{5}{2}\right) = -\frac{3}{5}x + \frac{3}{2}.
y = x + 3. The midpoint is (-2, 1) and the slope of \overline{AB} is \displaystyle \frac{-6}{6} = -1, so the bisector has slope 1: y - 1 = x + 2.
Yes. The midpoint is (5, 1) and the segment has slope \displaystyle \frac{1}{5}, so the bisector is y - 1 = -5(x - 5), that is y = -5x + 26, and -5(4) + 26 = 6. By distances, P is \displaystyle \sqrt{4^2 + 6^2} = \sqrt{52} from (0, 0) and \displaystyle \sqrt{6^2 + 4^2} = \sqrt{52} from (10, 2).
Center (1, -1), radius 5. The center is the midpoint of the diameter. The diameter has length \displaystyle \sqrt{8^2 + 6^2} = 10, and the radius is half of it.
S = (0, 3). The diagonals \overline{PR} and \overline{QS} share the midpoint \displaystyle \left(\frac{3}{2}, \frac{3}{2}\right), so \displaystyle S = (2 \cdot \frac{3}{2} - 3,\ 2 \cdot \frac{3}{2} - 0).
B = (0, 5). \overline{AB} has slope -1, so it lies on y = -x + 5; that meets y = x + 1 at the midpoint (2, 3), and B = (2 \cdot 2 - 4,\ 2 \cdot 3 - 1).
1 + 3i. \displaystyle \frac{(6 - 2i) + (-4 + 8i)}{2} = \frac{2 + 6i}{2}.
w = -1 - 5i. w = 2(2 - i) - (5 + 3i) = 4 - 2i - 5 - 3i.
Going from (x_1, y_1) to (x_2, y_2) moves x_2 - x_1 across. The midpoint is halfway along, so it has made half that move: its x-coordinate is \displaystyle x_1 + \frac{x_2 - x_1}{2} = \frac{2x_1 + x_2 - x_1}{2} = \frac{x_1 + x_2}{2}, the average. The two halves of the segment are the long sides of congruent right triangles, which is what makes this point equally far from both ends.
From (x_1, y_1) to M the moves are \displaystyle \frac{x_2 - x_1}{2} across and \displaystyle \frac{y_2 - y_1}{2} up; from M to (x_2, y_2) they are the same two numbers. So both distances equal \displaystyle \sqrt{\left(\frac{x_2 - x_1}{2}\right)^2 + \left(\frac{y_2 - y_1}{2}\right)^2}. Taking the \displaystyle \frac{1}{4} out of the square root as \displaystyle \frac{1}{2}, that is \displaystyle \frac{1}{2}\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, half the length of the segment.
A level segment has slope 0, so a perpendicular line would need a slope m with 0 \cdot m = -1, and there is none. The bisector is upright, and an upright line is not the graph of any y = mx + b. Its equation is x = x_M, the x-coordinate of the midpoint.
Expanding (x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2, the x^2 and y^2 cancel, leaving 2(x_2 - x_1)x + 2(y_2 - y_1)y = x_2^2 - x_1^2 + y_2^2 - y_1^2. The two points differ, so at least one coefficient on the left is not 0, and this is a line. Substituting \displaystyle x = \frac{x_1 + x_2}{2} and \displaystyle y = \frac{y_1 + y_2}{2} makes the left side (x_2 - x_1)(x_1 + x_2) + (y_2 - y_1)(y_1 + y_2), which multiplies out to the right side, so the midpoint is on it.