Find the area of the triangle below.

#### Guidance

When working on a coordinate plane, you can always find the distance between two points (or the length of a line segment) by creating a right triangle and using the Pythagorean Theorem.

For example, suppose you want to find the length of \begin{align*}\overline{AB}\end{align*}.

Draw a right triangle such that \begin{align*}\overline{AB}\end{align*} is the hypotenuse.

You can find the lengths of \begin{align*}\overline{BC}\end{align*} and \begin{align*}\overline{AC}\end{align*} by counting because they are vertical and horizontal line segments.

Use the Pythagorean Theorem to find the length of \begin{align*}\overline{AB}\end{align*}.

\begin{align*}(AB)^2 &=2^2+4^2 \\ AB &=\sqrt{20}=2\sqrt{5} \ un\end{align*}

Notice that \begin{align*}AB=\sqrt{BC^2+AC^2}\end{align*}. This generalizes to a formula known as the distance formula. The **distance formula** states that the distance between \begin{align*}(x_1,y_1)\end{align*} and \begin{align*}(x_2,y_2)\end{align*} is \begin{align*}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}. You will prove this generalized distance formula in Example A.

With the help of the distance formula, you can find the perimeter of polygons on a coordinate plane by finding the sum of the lengths of all the sides. You can also find the area of polygons by finding the lengths of key line segments like bases and heights.

**Example A**

Use the Pythagorean Theorem to show that the distance between the two points below is \begin{align*}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}.

**Solution:** Draw in a right triangle. The lengths of the legs are \begin{align*}(x_2-x_1)\end{align*} and \begin{align*}(y_2-y_1)\end{align*}. For now, let the length of the hypotenuse of the triangle (the distance you are trying to find) be “\begin{align*}k\end{align*}”.

By the Pythagorean Theorem, \begin{align*}(x_2-x_1)^2+(y_2-y_1)^2=k^2\end{align*}. Therefore, \begin{align*}k=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}. The distance between the two points is \begin{align*}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}.

**Example B**

Find the perimeter of the triangle below.

**Solution:** Use the distance formula to find the length of each side.

\begin{align*}AB &= \sqrt{2^2+3^2}=\sqrt{13}\approx3.61 \ un \\ BC &= \sqrt{1^2+2^2}=\sqrt{5}\approx2.24 \ un \\ AC &= \sqrt{2^2+4^2}=\sqrt{20}=2\sqrt{5}\approx4.47 \ un\end{align*}

The perimeter of the triangle is the distance around the triangle, which is the sum of the lengths of the three sides. The perimeter is \begin{align*}\sqrt{13}+\sqrt{5}+2\sqrt{5}=\sqrt{13}+3\sqrt{5}\approx10.31\end{align*} units.

**Example C**

Find the distance between \begin{align*}(2, 7)\end{align*} and \begin{align*}(-1, 5)\end{align*}.

**Solution:** You can use the distance formula without plotting the points. Let \begin{align*}(x_1,y_1)\end{align*}\begin{align*}=(2,7)\end{align*} and \begin{align*}(x_2,y_2)=(-1,5)\end{align*}.

\begin{align*}\text{Distance} = \sqrt{(-1-2)^2+(5-7)^2}=\sqrt{(-3)^2+(-2)^2}=\sqrt{9+4}=\sqrt{13} \ un\end{align*}

Note that it doesn't matter which point you choose for \begin{align*}(x_1,y_1)\end{align*}. You could have also calculated the distance the other way, with \begin{align*}(x_1,y_1)=(-1,5)\end{align*}:

\begin{align*}\text{Distance} = \sqrt{(-2-1)^2+(7-5)^2}=\sqrt{(-3)^2+(2)^2}=\sqrt{9+4}=\sqrt{13} \ un\end{align*}

**Concept Problem Revisited**

Consider the triangle with its three side lengths labeled (lengths found in Example B).

To find the area of the triangle, you can use the formula \begin{align*}A=\frac {bh}{2}\end{align*}. Any of the three sides can be the base, but you need a height that is perpendicular to the base you choose. Here, let \begin{align*}\overline{AC}\end{align*} be the base, so \begin{align*}b=4.47 \ un\end{align*}

Now, you need to find a line perpendicular to \begin{align*}\overline{AC}\end{align*} through point \begin{align*}B\end{align*}. \begin{align*}\overline{AC}\end{align*} has a slope of \begin{align*}\frac {1}{2}\end{align*}. A line perpendicular to \begin{align*}\overline{AC}\end{align*} will have a slope of -2. **You need to find a line with a slope of -2 that passes through the point \begin{align*}(1, 4)\end{align*}.**

**\begin{align*}4 &= -2(1)+b \\
6 &= b\end{align*}**

The equation is \begin{align*}y=-2x+6\end{align*}.

Next, you need to find where this line intersects \begin{align*}\overline{AC}\end{align*}. The equation of the line that contains \begin{align*}\overline{AC}\end{align*} is \begin{align*}y=\frac {1}{2}x+ \frac {3}{2}\end{align*}. Solve the system of equations to find the point of intersection.

\begin{align*}y=-2x+6\end{align*} and \begin{align*}y=\frac {1}{2}x+ \frac {3}{2}\end{align*}

\begin{align*}-2x+6 &= \frac {1}{2}x+ \frac {3}{2} \\ \frac {9}{2} &= \frac {5}{2}x \\ x &= 1.8 \\ y &= -2(1.8)+6 \\ y &= 2.4\end{align*}

The point of intersection is \begin{align*}(1.8, 2.4)\end{align*}.

Next, you need to find the length of the height, which is the distance from \begin{align*}(1.8, 2.4)\end{align*} to \begin{align*}(1, 4)\end{align*}.

\begin{align*}h=\sqrt{(1.8-1)^2+(2.4-4)^2}=1.79 \ un\end{align*}

Now that you know the lengths of the base and the height, you can find the area of the triangle. The area of the triangle is:

\begin{align*}A &=\frac {bh}{2} \\ A &=\frac {(4.47)(1.79)}{2} \\ A & \approx4 \ un^2\end{align*}

#### Vocabulary

The ** distance formula** is a formula that helps calculate the distance between two points or the length of a line segment in the coordinate plane. The distance between \begin{align*}(x_1,y_1)\end{align*} and \begin{align*}(x_2,y_2)\end{align*} is \begin{align*}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}.

#### Guided Practice

1. Find the perimeter of the rectangle below.

2. Find the area of the rectangle from #1.

3. Find the distance between \begin{align*}(-16, 4)\end{align*} and \begin{align*}(312, 211)\end{align*}. How does this calculation help to show why having the distance formula is helpful?

**Answers:**

1. Because the polygon is a rectangle, its opposite sides are the same length. Use the distance formula to find the length of two of the sides.

\begin{align*}AB &= \sqrt{3^2+1^2}=\sqrt{10} \\ BC &= \sqrt{6^2+2^2}=\sqrt{40}=2\sqrt{10}\end{align*}

The perimeter of the rectangle is:

\begin{align*}P=\sqrt{10}+2\sqrt{10}+\sqrt{10}+2\sqrt{10}=6\sqrt{10} \ un\end{align*}

2. To find the area of the rectangle, use the formula \begin{align*}A=bh\end{align*}. The base and the height lengths were found in #1.

\begin{align*}A &= \left ( \sqrt{10} \right ) \left ( 2\sqrt{10} \right ) \\ A &= 20 \ un^2\end{align*}

3. Let \begin{align*}(x_1,y_1)\end{align*}\begin{align*}=(-16,4)\end{align*} and \begin{align*}(x_2,y_2)=(312,211)\end{align*}.

\begin{align*}d=\sqrt{(312-(-16))^2+(211-4)^2}\approx387.01 \ un\end{align*}

Because these two points are so far apart, it would have been unrealistic to plot them and try to find the distance between them by drawing in a right triangle and using the Pythagorean Theorem. The distance formula makes these types of calculations much quicker.

#### Practice

Find the distance between each pair of points.

1. \begin{align*}(7, 11)\end{align*} and \begin{align*}(4, 23)\end{align*}.

2. \begin{align*}(19, 56)\end{align*} and \begin{align*}(-21, 45)\end{align*}.

3. \begin{align*}(-11,91)\end{align*} and \begin{align*}(89, 16)\end{align*}.

For #4-#5, use the rectangle below.

4. Find the perimeter of the rectangle.

5. Find the area of the rectangle.

For #6-#7, use the triangle below.

6. Find the perimeter of the triangle.

7. Find the area of the triangle.

For #8-#9, use the rectangle below.

8. Find the perimeter of the rectangle.

9. Find the area of the rectangle.

For #10-#11, use the triangle below.

10. Find the perimeter of the triangle.

11. Find the area of the triangle.

For #12-#14, find the perimeter of each polygon.

12.

13.

14.

15. What does the distance formula have to do with the Pythagorean Theorem?