Finding the **Length of a Line** segment is useful for finding the distance between 2 separate coordinates/endpoints on a graph. This is very similar to finding the hypoteneuse of a triangle. It can be expressed using the formula:

\(d = \sqrt{(x₂ - x₁)^2 + (y₂ - y₁)^2}\)
- \(d\) represents the distance
- \(x₁\) and \(y₁\) represent the first set of coordinates
- \(x₂\) and \(y₂\) reprsent the second set of coordinates

**Example**
Calculate the length of a line sgement with endpoints \((4,3)\) and \((2,-2)\).

All we need to do to find the line length is plug the coordinates into the formula:

\(d = \sqrt{(2 - 4)^2 + (-2 - 3)^2}\)

\(d = \sqrt{(-2)^2 + (-5)^2}\)

\(d = \sqrt{4 + 25}\)

\(d = \sqrt{29} ≅ 5.4\)
Therefore, we can determine that the length of the line segment is roughly \(5.4\).

Calculate the length of a line segment with endpoints \((-3/4, -2/5)\) and \((1/4, 3/5)\)

Show Answer
We can first convert all the fractions in the coordinates to have the same denominator:

\(\text{Coordinate}₁ = (-15/20, -8/20)\)

\(\text{Coordinate}₂ = (5/20, 12/20)\)
We can now plug the coordinates into the formula to find the line length:

\(d = \sqrt{(\cfrac{5}{20} - \cfrac{-15}{20})^2 + (\cfrac{12}{20} - \cfrac{-8}{20})^2}\)

\(d = \sqrt{(\cfrac{20}{20})^2 + (\cfrac{20}{20})^2}\)

\(d = \sqrt{(1)^2 + (1)^2}\)

\(d = \sqrt{1 + 1}\)

\(d = \sqrt{2} ≅ 1.4\)

Therefore, we can determine the length of the line segment is roughly \(1.4\).

The endpoint of the radius of a circle with centre \(\text{C}(2,3)\) is \(\text{D}(5,5)\).

i. Determine the length of the diameter of the circle.

ii. Determine the coordinates of the endpoint \(\text{E}\) of the diameter \(\text{DE}\) of the circle.

Show Answer
i. As we have both the origin and an endpoint, we can calculate the length of the radius. In this instance we will classify the origin as coordinate \(1\) and endpoint \(\text{D}\) as coordinate \(2\):

\(r = \sqrt{(5 - 2)^2 + (5 - 3)^2}\)

\(r = \sqrt{(3)^2 + (2)^2}\)

\(r = \sqrt{9 + 4}\)

\(r = \sqrt{13} ≅ 3.6\)

All we need to do to determine the diameter is multiply the radius by \(2\):

\(d = (2)(3.6) ≅ 7.2\)
Therefore, we can determine that the diameter of the circle is roughly \(7.2\).

ii. We can use the midpoint formula to identify the other endpoint of the circle. In this instance, we will classify endpoint \(\text{D}\) as coordinate 1 and endpoint \(\text{E}\) as endpoint 2:

\((2, 3) = (\cfrac{5 + x₂}{2}, \cfrac{5 + y₂}{2})\)
We can cross-multiply between the endpoints and midpoint to determine the coordinate for endpoint \(\text{E}\):

\((2)(2) = (2)(\cfrac{5 + x₂}{2})\)

\(4 = 5 + x₂\)

\(x₂ = 4 - 5\)

\(x₂ = -1\)

\((2)(3) = (2)(\cfrac{5 + y₂}{2})\)

\(6 = 5 + y₂\)

\(y₂ = 6 - 5\)

\(y₂ = 1\)
Therefore, we can determine that the coordinate for endpoint \(\text{E}\) is \((-1, 1)\).

The vertices of △\(\text{XYZ}\) are \(\text{X}(-6, 8)\), \(\text{Y}(-2, -4)\) and \(\text{Z}(4,6)\).

i. Determine the exact length of each side of the triangle.

ii. Classify this triangle.

Show Answer
i. We can use the Length of a Line formula to calculate each of the side lengths:

d_{XY} \(= \sqrt{(-2 - (-6))^2 + (-4 - 8)^2}\)

d_{XY} \(= \sqrt{(4)^2 + (-12)^2}\)

d_{XY} \(= \sqrt{16 + 144}\)

d_{XY} \(= \sqrt{160} ≅ 12.65\)

d_{YZ} \(= \sqrt{(4 - (-2))^2 + (6 - (-4))^2}\)

d_{YZ} \(= \sqrt{(6)^2 + (10)^2}\)

d_{YZ} \(= \sqrt{36 + 100}\)

d_{YZ} \(= \sqrt{136} ≅ 11.66\)

d_{XZ} \(= \sqrt{(-6 - 4)^2 + (8 - 6)^2}\)

d_{XZ} \(= \sqrt{(-10)^2 + (2)^2}\)

d_{XZ} \(= \sqrt{100 + 4}\)

d_{XZ} \(= \sqrt{104} ≅ 10.2\)

Therefore, we can determine that d_{XY} \(≅ 12.65\), d_{YZ} \(≅ 11.66\) and d_{XZ} \(≅ 10.2\)

ii. As all 3 side lengths are unique from one another, we can identify that this is a Scalene Triangle.