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Section 9.2 The Distance and Midpoint Formulas

Subsection 9.2.1 Distance in a Coordinate Plane

Figure (a) shows a line segment joining the two points \((-2, 7)\) and \((6, 3)\text{.}\) What is the distance between the two points?

line segment in coordinate plane
line segment as hypotenuse

The distance between two points is the length of the segment joining them.

If we make a right triangle as shown in Figure (b), we can use the Pythagorean theorem to find its length. First, notice that the coordinates at the right angle are \((-2, 3)\text{.}\) We can find the lengths of the two legs of the triangle, because they are horizontal and vertical segments.

\begin{align*} a\amp= \abs{3-7} = 4\\ b\amp=\abs{6-(-2)}=8 \end{align*}

The segment we want is the hypotenuse of the right triangle, so we apply the Pythagorean theorem.

\begin{align*} c^2 \amp= a^2+b^2\\ \amp= 4^2+8^2=80\amp\amp \blert{\text{Take square roots.}}\\ c\amp=\sqrt{80}\approx 8.9 \end{align*}

Subsection 9.2.2 The Distance Formula

We can also use the Pythagorean theorem to derive a formula for the distance between any two points, \(P_1\) and \(P_2\text{,}\) in terms of their coordinates. We first label a right triangle, as we did in the exmple above. Draw a horizontal line through \(P_1\) and a vertical line through \(P_2\text{.}\)

These lines meet at a point \(P_3\text{,}\) as shown in the figure below. The \(x\)-coordinate of \(P_3\) is the same as the \(x\)-coordinate of \(P_2\text{,}\) and the \(y\)-coordinate of \(P_3\) is the same as the \(y\)-coordinate of \(P_1\text{.}\) Thus, the coordinates of \(P_3\) are \((x_2, y_1)\text{.}\)

distance between two points in the plane

The distance between \(P_1\) and \(P_3\) is \(\abs{x_2-x_1}\) , and the distance between \(P_2\) and \(P_3\) is \(\abs{y_2-y_1} \text{.}\) These two numbers are the lengths of the legs of the right triangle. The length of the hypotenuse is the distance between \(P_1\) and \(P_2\text{,}\) which we'll call \(d\text{.}\) By the Pythagorean theorem,

\begin{gather*} d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \end{gather*}

Taking the (positive) square root of each side of this equation gives us the distance formula.

Distance Formula.

The distance \(d\) between points \(P_1(x_1,y_1)\) and \(P_2(x_2,y_2)\) is

\begin{gather*} \blert{d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}} \end{gather*}

True or false.

  1. The distance formula is a version of the Pythagorean theorem.

    • True

    • False

  2. \((u-v)^2 = (v-u)^2 = |u-v|^2=|v-u|^2\)

    • True

    • False

  3. We can find the length of a horizontal segment by subtracting the \(x\)-coordinates of its endpoints.

    • True

    • False

  4. The distance between two points is always a positive number (or zero).

    • True

    • False

Answer 1.

\(\text{True}\)

Answer 2.

\(\text{True}\)

Answer 3.

\(\text{True}\)

Answer 4.

\(\text{True}\)

Solution.
  1. True

  2. True

  3. True

  4. True

Find the distance between \((2,-1)\) and \((4,3)\text{.}\)

Solution.

Substitute \((2,-1)\) for \((x_1,y_1)\) and \((4,3)\) for \((x_2,y_2)\) in the distance formula to obtain

\begin{align*} d \amp = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \amp = \sqrt{(4-2)^2+\left[3-(-1)\right]^2}\\ \amp = \sqrt{4+16} =\sqrt{20} \approx 4.47 \end{align*}

It doesn't matter which point we call \(P_1\) and which is \(P_2\text{.}\) We obtain the same answer in the previous Example if we switch the two points and use \((4,3)\) for \(P_1\) and \((2,-1)\) for \(P_2\text{:}\)

\begin{align*} d \amp = \sqrt{(2-4)^2+[(-1)-3]^2}\\ \amp = \sqrt{4+16} =\sqrt{20} \end{align*}

Caution 9.2.3.

We cannot simplify \(\sqrt{4+16} \) as \(\sqrt{4}+\sqrt{16} \text{.}\) Remember that \(\sqrt{a^2+b^2}\ne a+b \text{.}\) You can easily see this by observing that

\begin{gather*} \sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{5} \end{gather*}

so it cannot be true that \(\sqrt{3^2+4^2} \) equals \(3+4\text{,}\) or 7. For the same reason, we cannot simplify the distance formula to \((x_2-x_1)+(y_2-y_1)\text{.}\)

  1. Find the distance between the points \((-5,3)\) and \((3,-9)\text{:}\)

  2. Plot the points, and illustrate how the Pythagorean theorem is used in calculating the distance.

Answer.

\(\sqrt{208}\)

Solution.
  1. \(\displaystyle \sqrt{208}\)

  2. A graph is below.

Graph for Practice 1:

distance formula

Subsection 9.2.3 Finding the Midpoint

The midpoint of a segment is the point halfway between its endpoints, so that the distance from the midpoint to either endpoint is the same. The \(x\)-coordinate of the midpoint is halfway between the \(x\)-coordinates of the endpoints, and likewise for the \(y\)-coordinate. For the points \((-2, 7)\) and \((6, 3)\) shown below, the \(x\)-coordinate of the midpoint is \(2\text{,}\) which is halfway between \(-2\) and \(6\text{.}\) The \(y\)-coordinate is halfway between \(7\) and \(3\text{,}\) or \(5\text{.}\) Thus, the midpoint is \((2, 5)\text{.}\)

line segment

Subsection 9.2.4 The Midpoint Formula

If we know the coordinates of two points, we can calculate the coordinates of the midpoint. Each coordinate of the midpoint is the average of the corresponding coordinates of the two points.

Midpoint Formula.

The midpoint of the line segment joining the points \(P_1(x_1,y_1)\) and \(P_2(x_2,y_2)\) is the point \(M(\overline{x},\overline{y})\text{,}\) where

\begin{gather*} \blert{\overline{x}=\dfrac{x_1+x_2}{2}}\qquad\text{ and }\qquad\blert{\overline{y}=\dfrac{y_1+y_2}{2}} \end{gather*}

Find the midpoint of the line segment joining the points \((-2,1)\) and \((4,3)\text{.}\)

Solution.

We substitute \((-2,1)\) for \((x_1,y_1)\) and \((4,3)\) for \((x_2,y_2)\) in the midpoint formula to obtain

\begin{align*} \overline{x}\amp=\dfrac{x_1+x_2}{2}=\dfrac{-2+4}{2}=1\\ \overline{y}\amp=\dfrac{y_1+y_2}{2}=\dfrac{1+3}{2}=2 \end{align*}

The midpoint of the segment is the point \((\overline{x},\overline{y})=(1,2)\text{.}\)

  1. Find the midpoint of the line joining the points \((-5,3)\) and \((3,-9)\text{:}\)

  2. Plot the points and draw a rectangle with the points as opposite vertices. Illustrate that the midpoint is the center of the rectangle.

Answer.

\(\left(-1,-3\right)\)

Solution.
  1. \(\displaystyle (-1,-3)\)

  2. A figure is below.

Graph for Practice 2:

rectangle

Subsection 9.2.5 Circles

A circle is the set of all points in a plane that lie at a given distance, called the radius, from a fixed point called the center. We can use the distance formula to find an equation for a circle. First consider the circle in Figure (a), whose center is the origin, \((0,0)\text{.}\)

circles

The distance from the origin to any point \(P(x,y)\) on the circle is \(r\text{.}\) Therefore,

\begin{gather*} \sqrt{(x-0)^2+(y-0)^2}=r \end{gather*}

or, squaring both sides,

\begin{gather*} (x-0)^2+(y-0)^2=r^2 \end{gather*}

Thus, the equation for a circle of radius \(r\) centered at the origin is

\begin{gather*} \blert{x^2+y^2=r^2} \end{gather*}

Now consider the circle in Figure (b), whose center is the point \((h,k)\text{.}\) Every point \(P(x,y)\) on the circle lies a distance \(r\) from \((h,k)\text{,}\) so the equation of the circle is given by the following formula.

Standard Form for a Circle.

The equation for a circle of radius \(r\) centered at the point \((h,k)\) is

\begin{gather*} \blert{(x-h)^2+(y-k)^2=r^2} \end{gather*}
  1. Does the center of a circle lie on the graph of the circle?

    • Yes

    • No

  2. What is the radius of the circle \(x^2 + y^2 = 81\text{?}\)

  3. What is the diameter of the circle in part (b)?

  4. Can we simplify the equation in part (b) to \(x + y = 9\text{?}\)

    • Yes

    • No

Answer 1.

\(\text{No}\)

Answer 2.

\(9\)

Answer 3.

\(18\)

Answer 4.

\(\text{No}\)

Solution.
  1. No

  2. 9

  3. 18

  4. No

The equation \((x-h)^2+(y-k)^2=r^2 \) is the standard form for a circle of radius \(r\) with center at \((h,k)\text{.}\) It is easy to graph a circle if its equation is given in standard form.

Graph the circles.

  1. \(\displaystyle (x-2)^2+(y+3)^2=16\)

  2. \(\displaystyle x^2+(y-4)^2=7\)

Solution.
  1. The graph of \((x-2)^2+(y+3)^2=16\) is a circle with radius 4 and center at \((2,-3)\text{.}\) To sketch the graph, we first locate the center of the circle. (The center is not part of the graph of the circle.)

    From the center, we move a distance of 4 units (the radius of the circle) in each of four directions: up, down, left, and right. This locates four points that lie on the circle: \((2,1)\text{,}\) \((2,-7)\text{,}\) \((-2,-3)\text{,}\) and \((6,-3)\text{.}\) We sketch the circle through these four points.

    circles
  2. The graph of \(x^2+(y-4)^2=7\) is a circle with radius \(\sqrt{7}\) and center at \((0,4)\text{.}\) From the center, we move approximately \(\sqrt{7}\text{,}\) or 2.6 units in each of the four coordinate directions to obtain the points \((0,6.6)\text{,}\) \((0,1.4)\text{,}\) \((-2.6,4)\text{,}\) and \((2.6,4)\text{.}\) We sketch the circle through these four points.

  1. State the center and radius of the circle

    \begin{equation*} (x+3)^2+(y+2)^2=16 \end{equation*}

    center: , radius:

  2. Graph the circle.

Answer 1.

\(\left(-3,-2\right)\)

Answer 2.

\(4\)

Solution.
  1. center \((-3,-2)\text{,}\) radius 4

  2. A graph is below.

\((x+3)^2+(y+2)^2=16\)

circle

Subsection 9.2.6 General Form for Circles

The equations of circles often appear in a general quadratic form, rather than the standard form described above. For example, we can expand the squares of binomials in part (a) of the previous Example,

\begin{gather*} (x-2)^2+(y+3)^2=16 \end{gather*}

to obtain

\begin{gather*} x^2-4x+4+y^2+6y+9=16 \end{gather*}

or

\begin{gather*} x^2+y^2-4x+6y-3=0 \end{gather*}

This is a quadratic equation in two variables. Such an equation describes a circle if the coefficients of the quadratic, or squared, terms are equal.

Conversely, an equation of the form \(x^2+y^2+ax+by+c=0\) can be converted to standard form by completing the square in both variables. Once this is done, the center and radius of the circle can be determined directly from the equation.

Write the equation of the circle

\begin{gather*} x^2+y^2+8x-2y+6=0 \end{gather*}

in standard form, and graph the equation.

Solution.

We prepare to complete the square in both variables by writing the equation as

\begin{gather*} (x^2+8x+\fillinmath{XXXXXX})+(y^2-2y+\fillinmath{XXXXXX})=-6 \end{gather*}

We complete the square in \(x\) by adding 16 to each side of the equation, and complete the square in \(y\) by adding 1 to each side, to get

\begin{gather*} (x^2+8x+\alert{16})+(y^2-2y+\alert{1})=-6+\alert{16}+\alert{1} \end{gather*}

from which we obtain the standard form,

\begin{gather*} (x+4)^2+(y-1)^2=11 \end{gather*}

Thus, the circle has its center at \((-4,1)\text{,}\) and its radius is \(\sqrt{11}\text{,}\) or approximately 3.3. The graph is shown at right.

circle

Fill in the blanks.

  1. The standard form for a circle is useful because we can see the

    • area and circumference

    • center and radius

    • diameter and arclength

    of the circle in the equation.

  2. A quadratic equation in two variables describes a circle if the coefficients of the quadratic terms are

    • squared

    • opposites

    • equal

    • zero

    .

  3. We convert the equation for a circle into standard form by

    • completing the square

    • cross-multiplying

    • factoring

    in both variables.

Answer 1.

\(\text{center and radius}\)

Answer 2.

\(\text{equal}\)

Answer 3.

\(\text{completing the square}\)

Solution.
  1. center and radius

  2. equal

  3. completing the square

Write the equation of the circle \(x^2+y^2-14x+4y+25=0\) in standard form:

\begin{equation*} \begin{gathered} (x-h)^2+(y-k)^2=r^2 \end{gathered} \end{equation*}

\(h=\), \(k=\), \(r=\),

Answer 1.

\(7\)

Answer 2.

\(-2\)

Answer 3.

\(\sqrt{28}\)

Solution.

\((x-7)^2+(y+2)^2=28\)

We can write an equation for any circle if we can find its center and radius.

Find an equation for the circle whose diameter has endpoints \((7,5)\) and \((1,-1)\text{.}\)

Solution.

The center of the circle is the midpoint of its diameter. We use the midpoint formula to find the center:

\begin{gather*} h=\overline{x}=\dfrac{7+1}{2}=4\\ k=\overline{y}=\dfrac{5-1}{2}=2 \end{gather*}

Thus, the center is the point \((h,k)=(4,2)\text{.}\)

circle with diameter (7,5) to (1,-1)

The radius is the distance from the center to either of the endpoints of the diameter, say the point \((7,5)\text{.}\) We use the distance formula with the points \((7,5)\) and \((4,2)\) to find the radius.

\begin{align*} r \amp = \sqrt{(7-4)^2+(5-2)^2}\\ \amp = \sqrt{3^2+3^2}=\sqrt{18} \end{align*}

Finally, we substitute 4 for\(h\) and 2 for \(k\) (the coordinates of the center) and \(\sqrt{18}\) for \(r\) (the radius) into the standard form

\begin{gather*} (x-h)^2+(y-k)^2=r^2 \end{gather*}

to obtain

\begin{gather*} (x-4)^2+(y-2)^2=18 \end{gather*}

Find an equation for the circle whose diameter has endpoint \((-7,-5)\) and \((-1,-1) \text{.}\)

Answer.

\(\left(x-\left(-4\right)\right)^{2}+\left(y-\left(-3\right)\right)^{2} = 13\)

Solution.

\((x+4)^2 + (y+3)^2 = 13\)

Exercises 9.2.7 Problem Set 9.2

Warm Up

1.

Choose values for \(x\) and \(y\) to decide whether the statements are true.

  1. \(\displaystyle \sqrt{x^2+y^2}=x+y\)

  2. \(\displaystyle (x+y)^2 = x^2+y^2\)

Answer.
  1. False

  2. False

2.

Leanne is sailing 3 miles west and 5 miles south of the harbor. She heads directly towards an island that is 8 miles west and 7 miles north of the harbor.

  1. How far is Leanne from the island?

  2. How far will Leanne be from the harbor when she is halfway to the island?

Exercise Group.

For Problems 3 and 4, complete the table of values, then sketch the graph.

3.

\(x^2+y^2=16\)

\(x\) \(-4\) \(-3\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(y\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\)
Answer.
\(x\) \(-4\) \(-3\) \(-2\) \(-1\) \(0\) \(1\) \(2\) \(3\) \(4\)
\(y\) \(0\) \(\pm\sqrt{7}\) \(\pm\sqrt{12}\) \(\pm\sqrt{15}\) \(\pm 4\) \(\pm\sqrt{15}\) \(\pm\sqrt{12}\) \(\pm\sqrt{7}\) \(0\)
circle of radius 4 centered at origin
4.

\((x-2)^2+(y+2)^2=9\)

\(x\) \(-1\) \(0\) \(1\) \(2\) \(3\) \(4\) \(5\)
\(y\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\) \(\qquad\)
Exercise Group.

For Problems 5 and 6, solve by completing the square.

5.

\(x^2-5x-2=0\)

Answer.

\(\dfrac{5\pm \sqrt{33}}{2} \)

6.

\(x^2+6x=11 \)

Skills Practice

Exercise Group.

For Problems 7–10, find the distance between each of the given pairs of points, and find the midpoint of the segment joining them.

7.

\((2,-3) \text{,}\) \((-2,-1) \)

Answer.

distance: \(\sqrt{20} \text{;}\) midpoint: \(\left(0,-2 \right) \)

8.

\((1,1) \text{,}\) \((4,5) \)

9.

\((-2,-5) \text{,}\) \((-2,3) \)

Answer.

distance: \(8\text{;}\) midpoint: \(\left(-2,-1 \right) \)

10.

\((5,-4) \text{,}\) \((-1,1) \)

Exercise Group.

For Problems 11–14, state the center and radius of the circle.

11.

\(2x^2+2y^2=50\)

Answer.

center: \((0,0) \text{;}\) radius: \(5 \)

12.

\(x^2+y^2=16 \)

13.

\((x+3)^2+y^2=10 \)

Answer.

center: \((-3,0) \text{;}\) radius: \(\sqrt{10} \)

14.

\((x-4)^2+(y+2)^2=9 \)

Applications

15.

Find the perimeter of the triangle with vertices \((-1,5)\text{,}\) \((8,-7)\text{,}\) \((4,1)\text{.}\) (Hint: The length of a side is the distance between its endpoints.)

Answer.

\(15+\sqrt{80}+\sqrt{41} \approx 30.3\)

16.

Show that the points \((-2,1)\text{,}\) \((0,-1) \text{,}\) and \((\sqrt{3}-1, \sqrt{3}) \) are the vertices of an equilateral triangle. (Hint: Use the distance formula to show that the lengths of the three sides are equal.)

17.

The points \((1,6)\text{,}\) \((5,2)\text{,}\) \((-2,3)\text{,}\) and \((2,-1)\) are the vertices of a quadrilateral. Show that its diagonals are of equal length.

Answer.

\(\sqrt{50} \approx 7.1\)

18.

Two opposite vertices of a square are \(A(-9,-5)\) and \(C(3,3)\text{.}\)

  1. Find the length of a diagonal of the square.

  2. Find the length of the side of the square.

19.

Find the equation of the perpendicular bisector of the line segment joining \(A(2,1)\) and \(B(1,3)\text{.}\)

Answer.

\(y=\dfrac{1}{2}x + \dfrac{5}{4}\)

20.

Brie took off from Oldfield Airport in a small plane and is now 253 miles east and 124 miles north of the airport. She knows that Preston Airport is 187 miles east and 201 miles south of Oldfield Airport. Which airport is closer to Brie’s present location?

  1. Use the diagram and the Pythagorean theorem to find the distance from Brie's position to Oldfield Airport.

    Brie to Oldfield
    Brie to Preston
  2. Use the diagram and the distance formula to find the distance from Brie's position to Preston Airport.

  3. Answer the question in the problem.

21.

Suppose that Brie flies from her present position at (253, 124) towards Preston Airport at \((187,-201)\text{.}\)

  1. What are her coordinates when she is halfway to Preston?

    Brie halfway to Preston
  2. Show that the distance from Brie's present position to the halfway point is the same as the distance from the halfway point to Preston Airport.

Answer.
  1. \(\displaystyle (220,-38.5)\)

  2. Both distances are 165.8 mi.

Exercise Group.

For Problems 22-27, graph the equation.

22.

\(x^2+y^2=25\)

24.

\((x-4)^2+(y+2)^2=9\)

26.

\((x-1)^2+(y-3)^2=16\)

Exercise Group.

For Problems 28–31, write the equation in standard form. State the center and radius of the circle

28.

\(x^2+y^2+2x-4y-6=0\)

29.

\(x^2+y^2+8x=4\)

Answer.

\((x+4)^2+y^2=20 \text{;}\) center: \((-4,0) \text{;}\) radius: \(\sqrt{20} \)

30.

\(x^2+y^2-6x+2y-4=0\)

31.

\(x^2+y^2-10y=2\)

Answer.

\(x^2+(y-5)^2=27 \text{;}\) center: \((0,5) \text{;}\) radius: \(\sqrt{27} \)

Exercise Group.

For Problems 32–35, write an equation for the circle with the given properties.

32.

Center at \((-2,5) \text{,}\) radius \(2\sqrt{3} \text{.}\)

33.

Center at \(\left(\dfrac{3}{2},-4\right) \text{,}\) one point on the circle is \((4,-3) \text{.}\)

Answer.

\(\left(x-\dfrac{3}{2}\right)^2+(y+4)^2=\dfrac{29}{4}\)

34.

Endpoints of a diameter at \((1,5)\) and \((3,-1).\)

35.

Center at \((-3,-1)\text{,}\) the \(x\)-axis is tangent to the circle.

Answer.

\(\left(x+3\right)^2+(y+1)^2=1\)

36.

Find an equation for the circle that passes through the points \((2,3),~(3,2)\text{,}\) and \((-4,-5)\text{.}\) (Hint: Find values for \(a, ~b\text{,}\) and \(c\) so that the three points lie on the graph of \(x^2+y^2+ax+by+c=0\text{.}\))

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