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Section 9.2 Distance and Midpoint Formulas

Subsection 1. Use radicals

Because the distance formula is derived from the Pythagorean theorem, using it involves working with radicals.

Subsubsection Examples

Example 9.17.

In the right triangle shown, \(a=8\) and \(b=6\text{.}\)

  1. Is \(c=a+b\text{?}\) Why or why not?
  2. Find the length of \(c\text{.}\)
right triangle
Solution
  1. No, the length of \(c\) is shorter than the lengths of \(a\) and \(b\) combined, so \(c \lt a+b\text{.}\)

  2. We use the Pythagorean theorem:

    \begin{align*} c^2 \amp = a^2+b^2 \amp \amp \blert{\text{Substitute }a=8~ \text{and }b=6~ \text{and evaluate.}}\\ c^2 \amp = 8^2+6^2=64+36=100 \amp \amp \blert{\text{Take square roots.}}\\ c \amp = \sqrt{100} = 10 \end{align*}

    The hypotenuse is 10 cm long.

Example 9.18.

Is \(\sqrt{a^2+b^2} = a+b\text{?}\)

Solution

No. In the previous example, \(c=\sqrt{a^2+b^2}\text{,}\) and we saw that \(c \lt a+b\text{.}\) We cannot simplify \(\sqrt{a^2+b^2}\) by taking the square root of each term.

Example 9.19.

If \(r=\sqrt{14}\text{,}\) write and simplify expressions for:

  1. \(\displaystyle r \cdot r\)
  2. \(\displaystyle r+r\)
Solution
  1. \(r \cdot r = \left(\sqrt{14}\right)\left(\sqrt{14}\right) = 14\text{.}\) Or, \(r \cdot r = r^2 = \left(\sqrt{14}\right)^2 = 14\text{.}\)
  2. \(r+r = \sqrt{14} + \sqrt{14} = 2\sqrt{14}\text{.}\) Or, \(r+r = 2r = 2\sqrt{14}\text{.}\)

Subsubsection Exercises

Simplify if possible: \(~\sqrt{(x-4)^2+(y-2)^2}\)

Answer

cannot be simplified

Simplify if possible: \(~\sqrt{7^2+w^2}\)

Answer
\(\sqrt{49+w^2}\)

Simplify if possible: \(~\sqrt{7+w} \sqrt{7+w}\)

Answer
\(7+w\)

Subsection 2. Complete the square

To find the center and radius of a circle, we may need to complete the square.

Subsubsection Examples

Example 9.23.

Write an equivalent equation in which the left side is a perfect square: \(~x^2-6x=2\)

Solution

We want to find a constant \(p^2\) so that \(x^2\blert{-6x}+p^2\) is a perfect square, namely \((x+p)^2\text{.}\) Now, \(~(x+p)^2 = x^2\blert{+2px}+p^2~\text{,}\) so \(2p=-6\text{,}\) and \(p=-3\text{.}\) Thus, we add \(p^2=\alert{9}\) to both sides of the equation.

\begin{equation*} x^2-6x+\alert{9} = 2+\alert{9} \end{equation*}

Now we can write the left side as a perfect square:

\begin{equation*} (x-3)^2 = 11 \end{equation*}
Example 9.24.

Write an equivalent equation in which the left side is a perfect square: \(~y^2+7y=5\)

Solution

For this equation, \(2p=7\) so \(p = \dfrac{7}{2}\text{.}\) We add \(p^2=\left(\dfrac{7}{2}\right)^2=\alert{\dfrac{49}{4}}\) to both sides of the equation.

\begin{equation*} y^2+7y+\alert{\dfrac{49}{4}}=5+\alert{\dfrac{49}{4}} \end{equation*}

We write the left sides as a perfect square, \((x+p)^2\text{,}\) and simplify the right side.

\begin{equation*} \left(x+\dfrac{7}{2}\right)^2=\dfrac{69}{4}~~~~~~~~~~~~\blert{5+\dfrac{49}{4}=\dfrac{20}{4}+\dfrac{49}{4}=\dfrac{69}{4}} \end{equation*}

Subsubsection Exercises

Write an equivalent equation in which the left side is a perfect square: \(~x^2+12x=-6\)

Answer
\((x+6)^2=30\)

Write an equivalent equation in which the left side is a perfect square: \(~y^2-5y=3\)

Answer
\(\left(y-\dfrac{5}{2}\right)^2=\dfrac{37}{4}\)

Subsection 3. Use the equation for a circle

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

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

Subsubsection Examples

Example 9.27.

Does \((-1,5)\) lie on the circle \(~(x-1)^2+(y-3)^2=8~\text{?}\)

Solution

We substitute \(x=\alert{-1},~ y=\alert{5}\) into the equation for the circle.

\begin{equation*} (\alert{-1}-1)^2+(\alert{5}-3)^2 = (-2)^2+2^2 = 4+4 = 8 \end{equation*}

The point \((-1,5)\) satisfies the equation, so it does lie on the circle.

Example 9.28.

Find a point with \(x\)-coordinate \(-3\) that lies on the circle \((x+2)^2+(y-4)^2=5\text{.}\)

Solution

We substitute \(x=\alert{-3}\) into the equation, and solve for \(y\text{.}\)

\begin{align*} (\alert{-3}+2)^2+(y-4)^2 \amp = 5 \amp \amp \blert{\text{Simplify.}}\\ (-1)^2 +(y-4)^2 \amp = 5 \amp \amp \blert{\text{Solve by extracting roots.}}\\ (y-4)^2 \amp = 4 \amp \amp \blert{\text{Take square roots.}}\\ y-4 \amp = \pm 2 \end{align*}

We see that the solutions are \(y=6\) and \(y=2\text{.}\) Thus, the points \((-3,6)\) and \((-3,2)\) lie on the circle.

Subsubsection Exercises

Does \((2,6)\) lie on the circle \(~(x+3)^2+(y-5)^2=26~\text{?}\)

Answer
Yes

Find a point with \(y\)-coordinate \(-2\) that lies on the circle \((x-2)^2+(y+1)^2=9\text{.}\)

Answer

\((2+\sqrt{8}, -2)\) and \((2-\sqrt{8}, -2)\)