Line segments

Topic summary

In coordinate geometry, we often need to find the length of a line segment between two points and the area between curves and the x-axis. While calculus is often used to find areas under curves, we will focus on using simpler methods like the triangle area formula for areas between lines and the x-axis.

1. Finding the Length of a Line Segment:

To find the length of a line segment between two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) on the coordinate plane, we use the distance formula. The formula is derived from the Pythagorean theorem and is given by:

\[ \text{Length} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]

Example:

Given the points \(A(2, 3)\) and \(B(6, 7)\), we can find the length of the line segment between them:

\[ \text{Length} = \sqrt{(6 – 2)^2 + (7 – 3)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66 \]

2. The Area Between Equations and the X-Axis (Using the Triangle Formula):

To find the area between a line and the x-axis, we can use the formula for the area of a triangle. If we have a line that intersects the x-axis at some point, the area under the line forms a triangle. The area of a triangle is given by:

\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]

Example 1:

Consider the line \(y = 2x\) between \(x = 0\) and \(x = 3\). To find the area between this line and the x-axis, we first determine the base and height of the triangle:

• The base of the triangle is the horizontal distance between \(x = 0\) and \(x = 3\), so the base is 3.
• The height is the value of \(y\) when \(x = 3\), which is \(y = 2(3) = 6\).

Now, apply the triangle area formula:

\[ \text{Area} = \frac{1}{2} \times 3 \times 6 = \frac{1}{2} \times 18 = 9 \]

So, the area under the line \(y = 2x\) between \(x = 0\) and \(x = 3\) is 9 square units.

Example 2:

Consider the line \(y = -x + 4\) between \(x = 0\) and \(x = 4\). We again apply the triangle formula:

• The base of the triangle is 4 (from \(x = 0\) to \(x = 4\)).
• The height is the value of \(y\) when \(x = 4\), which is \(y = -(4) + 4 = 0\).

So, we need to find the height when \(x = 0\), which is \(y = 4\). Now, applying the formula:

\[ \text{Area} = \frac{1}{2} \times 4 \times 4 = \frac{1}{2} \times 16 = 8 \]

Therefore, the area under the line \(y = -x + 4\) between \(x = 0\) and \(x = 4\) is 8 square units.

3. Summary:

• To find the length of a line segment between two points, use the distance formula: \[ \text{Length} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]
• To find the area between a line and the x-axis, use the triangle area formula: \[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]