Length and area

​In a nutshell

The length between two points can be calculated using the distance formula. This formula allows you to solve length and area problems.

The distance formula

Pythagoras' theorem forms the basis of the distance formula. 

From Pythagoras' theorem, you can see that $$d^2 = (x_2-x_1)^2 + (y_2-y_1)^2$$. Taking the square root of $$d$$ gives the distance formula.

​$$\boxed{d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}$$​​

Procedure

Example 1

A line starts at the point $$(3,6)$$​ and ends at the point $$(6,10)$$​. What is the distance of the line?

Label the points

$$x_1 = 3\space, y_1 = 6\space, x_2=6\space, y_2 =10$$

Substitute into the distance formula. Solve for d.

​$$\begin{aligned}&d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\&d=\sqrt{(6-3)^2 + (10-6)^2}\\&d=\sqrt{(3)^2 + (4)^2}\\&d=\sqrt{9 + 16}\\&d=\sqrt{25}\\&\underline{d=5}\\\end{aligned}$$

Solving for area

A shape formed from several lines will have an area. With the distance formula, you can find that area. You must use the area formula for the relevant shape.

Example 2

A rectangle has the points $$A, (1,1)\space B,(0,3)\space C,(3,5)$$ and $$\space D,(4,3).$$ What is the area of the rectangle? Give your answer to two decimal places.

Find the distance of the lines using the distance formula.

$$\begin{aligned}&A\rightarrow B\\(&1,1)\space (0,3)\\x&_1 = 1\space, y_1 = 1\space, x_2=0\space, y_2 =3\\d&=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\d&=\sqrt{(0-1)^2 + (3-1)^2}\\d&=\sqrt{(-1)^2 + (2)^2}\\d&=\sqrt{1+4}\\d&=\sqrt5\\\end{aligned}$$

​$$\begin{aligned}&A\rightarrow D\\x&_1 = 1\space, y_1 = 1\space, x_2=4\space, y_2 =3\\d&=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\d&=\sqrt{(4-1)^2 + (3-1)^2}\\d&=\sqrt{(3)^2 + (2)^2}\\d&=\sqrt{9+4}\\d&=\sqrt{13}\\\end{aligned}$$​​

​

Find the area of the shape. Round to two decimal places.

$$\begin{aligned}Are&a\space= length\times height\\&A=\sqrt5\times\sqrt{13}\\&{A=\sqrt{65}}\\&\underline{A=8.06 \space(2\space d.p.)}\end{aligned}$$