Length and midpoint of a line segment

The distance between two points as Pythagoras on the horizontal and vertical gaps (left in exact surd form), and the midpoint as the average of the coordinates — plus working backwards from a known midpoint.

Pure · Coordinate geometry

Length and midpoint of a line segment

Why it works

Drop a right-angled triangle between two points

A(x_1, y_1)

and

B(x_2, y_2)

: the horizontal leg is the change in

x

, the vertical leg is the change in

y

, and the segment

AB

is the hypotenuse. Pythagoras then gives the length

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Because both gaps are squared, their signs don't matter — you get the same length whichever point you call first. Leave the answer as an exact surd unless asked to round;

\sqrt{45}

should be simplified to

3\sqrt{5}

, not left or decimalised.

The midpoint is the point exactly halfway along, so each of its coordinates is the average of the two endpoints':

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

That's an average — you add and halve, you don't subtract. (Subtracting and halving gives you half the gap, which is a useful quantity but not the midpoint.)

These two ideas run straight into circles: a radius is the length from centre to a point, and the centre of a circle is the midpoint of any diameter.

A common reverse question gives you the midpoint

M

and one end

A

, and asks for the other end

B

. Since

M

is the average,

B

is "as far past

M

A

is before it":

B = (2M_x - x_A, \ 2M_y - y_A)

— double the midpoint, subtract the known end.