Coordinates, Distance, and Midpoint

Distance Formula

Distance Between Two Points

The distance between two points with the same xx- or yy-coordinate is the length of a horizontal or vertical line segment connecting those points. This can be found by counting grid squares or subtracting the coordinates that are not equal.
The distance between two points on a number line is the absolute value of the difference between the endpoints.
The distance between two points on a number line with coordinates xx and yy can be expressed as yx\lvert y-x\rvert or as xy\lvert x-y\rvert. The two expressions are equivalent because of the absolute value symbols.
On the coordinate plane, the same principle is applied when the xx-coordinates or the yy-coordinates of two points are equal.

The distance between pairs of points can be found by subtracting the coordinates or by counting squares on the grid.

The Distance between Points with an Identical Coordinate

A(5,11)A (5, 11) and B(5,4)B (5, -4) A(3,3)A (-3, 3) and B(5,3)B (5, 3)
The xx-values are equal. So, the distance between the points along the yy-axis is 411=15|-4-11|=15 units. The yy-values are equal. So, the distance between the points along the xx-axis is 5(3)=8|5-(-3)|=8 units.

There are two ways to determine the distance between two points with the same coordinate: subtracting the coordinates that are not the same or counting the number of units from the first point to the second point.

Using the Pythagorean Theorem to Find Distance

The distance between two points that do not have the same xx- or yy-coordinate can be found by drawing a right triangle with the line segment between the points as the hypotenuse and using the Pythagorean theorem.
For two points in a plane that do not have equal xx- or yy-coordinates, the distance between the points can be found by using the Pythagorean theorem: If aa and bb are the legs of a right triangle and cc is the hypotenuse, then:
a2+b2=c2a^2+b^2=c^2
First, sketch a right triangle. The line segment joining the two points is the hypotenuse, while horizontal and vertical line segments form the legs. Then apply the Pythagorean theorem to find the length of the hypotenuse.
Step-By-Step Example
Using the Pythagorean Theorem
Determine the distance between point AA (3,13)(-3, 13) and point BB (5,2)(5, -2) using the Pythagorean theorem.
Step 1
Plot point AA and point BB. Then draw a right triangle to locate point CC (3,2)(-3, -2) to draw a right triangle.
Step 2

Determine the length of the horizontal leg, or line segment aa, of the triangle.

Point BB (5,2)(5,-2) and point CC (3,2)(-3, -2) have the same yy-coordinates. So, calculate the distance between the xx-coordinates or count the number of squares between the points.
Line segmenta=x2x1=5(3)=5+3=8\begin{aligned}{\text{Line segment}}\;a &= |x_{2}-x_{1}|\\&=|5-(-3)|\\&=|5+3|\\&=|8|\end{aligned}
Step 3

Determine the length of the vertical leg, or line segment bb, of the triangle.

Point AA (3,13)(-3,13) and point CC (3,2)(-3, -2) have the same xx-coordinates. So, calculate the distance between the yy-coordinates or count the number of squares between the points.
Line segmentb=y2y1=13(2)=13+2=15\begin{aligned}{\text{Line segment}}\;b &= |y_{2}-y_{1}|\\&=|13-(-2)|\\&=|13+2|\\&=|15|\end{aligned}
Step 4
Use the Pythagorean theorem:
a2+b2=c2a^2+b^2=c^2
Line segment aa is 8 units. Line segment bb is 15 units.

The hypotenuse of the triangle, line segment cc, is the distance between the points.

Substitute the lengths of each leg in the theorem:
a2+b2=c282+152=c264+225=c2289=c2289=c17=c\begin{aligned}a^{2}+b^{2}&=c^{2}\\ 8^{2}+15^{2}&=c^{2} \\ 64+225&=c^{2} \\ 289&=c^{2}\\ \sqrt{289}&=c\\17&=c\end{aligned}
Solution
The distance between point AA (3,13)(-3, 13) and point BB (5,2)(5, -2) is 17 units.

Deriving and Applying the Distance Formula

Using a right triangle, a formula can be derived for the distance between any two points in the coordinate plane.
The distance formula is used to calculate the distance between the points AA (x2,y2)(x_2, y_2) and BB (x1,y1)(x_1, y_1) in the coordinate plane. The distance formula is derived from the Pythagorean theorem. For a right triangle ΔABC\Delta ABC with hypotenuse AB\overline{AB}, the lengths of the horizontal and vertical legs are:
x2x1y2y1\begin{aligned} |x_{2}-x_{1}|\\ |y_{2}-y_{1}|\end{aligned}
To find the distance between point AA and point BB, draw a vertical line segment from point AA and a horizontal line segment from point BB to form a right triangle. The point where the line segments meet is point CC. Then use the coordinates of point AA and point BB to determine the lengths of the legs of the right triangle. Lastly, apply the Pythagorean theorem to determine the length of the hypotenuse, or the distance between point AA and point BB.
Substitute the lengths of the legs into the Pythagorean theorem. Let dd represent the distance between the two points, or the hypotenuse, line segment cc.
d2=x2x12+y2y12d^2=|x_{2}-x_{1}|^{2}+|y_{2}-y_{1}|^{2}
Then take the square root of both sides of the equation to get the distance formula.
d=(x2x1)2+(y2y1)2d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}
Since the squared values are always positive, the absolute-value signs are no longer needed.
Step-By-Step Example
Using the Distance Formula
Calculate the distance between AA (3,3)(-3, 3) and BB (9,2)(9, -2) using the distance formula.
Step 1
Substitute the coordinates of the points in the distance formula.
d=(x2x1)2+(y2y1)2=[(9(3)]2+(23)2\begin{aligned}d&=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}} \\ &=\sqrt{[(9-(-3)]^{2}+(-2-3)^{2}} \end{aligned}
Step 2
Simplify the equation.
d=(9+3)2+(5)2=122+25=144+25=169=13\begin{aligned}d&=\sqrt{(9+3)^{2}+(-5)^{2}}\\ &=\sqrt{12^{2}+25}\\&=\sqrt{144+25}\\&=\sqrt{169}\\&=13\end{aligned}
Solution
The distance between point AA (3,3)(-3, 3) and point BB (9,2)(9, -2) is 13 units.