Length of a Line

Finding the Length of a Line segment is useful for finding the shortest 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
  • \(x_1\) and \(y_1\) represent the first set of coordinates
  • \(x_2\) and \(y_2\) represent 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:

\(d = \sqrt{(x₂ - x₁)^2 + (y₂ - y₁)^2}\)

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

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

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

\(d = \sqrt{29} \approx 5.4\)

Therefore, we can determine that the length of the line segment is approximately \(\boldsymbol{5.4}\).


Calculate the length of a line segment with endpoints \(\left(\cfrac{-3}{4}, \cfrac{-2}{5}\right)\) and \(\left(\cfrac{1}{4}, \cfrac{3}{5}\right)\).

We can first convert all the fractions in the coordinates to have the same denominator. In this instance, we can set the denominator of each fraction to 20:

\(\text{Coordinate}₁ = \left(\cfrac{-15}{20}, \cfrac{-8}{20}\right)\)

\(\text{Coordinate}₂ = \left(\cfrac{5}{20}, \cfrac{12}{20}\right)\)

We can now plug the coordinates into the formula to determine the line length:

\(d = \sqrt{(x₂ - x₁)^2 + (y₂ - y₁)^2}\)

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

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

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

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

\(d = \sqrt{2} \approx 1.4\)

Therefore, we can determine the length of the line segment is \(\boldsymbol{\approx 1.4}\).


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

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

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, \(\text{C}\), as coordinate \(1\) and endpoint \(\text{D}\) as coordinate \(2\):

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

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

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

\(r = \sqrt{13} \approx 3.61\)

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

\(d = (2)(3.61) \approx 7.22\)

Therefore, we can determine that the diameter of the circle is \(\boldsymbol{\approx 7.22}\).


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 \(2\):

\((2, 3) = \left(\cfrac{5 + x₂}{2}, \cfrac{5 + y₂}{2}\right)\)

We can cross-multiply between the endpoints and midpoint to determine the coordinate for endpoint \(\text{E}\).

First, we can cross-multiply to determine the \(x\)-coordinate:

\((2)(2) = (2)\left(\cfrac{5 + x₂}{2}\right)\)

\(4 = 5 + x₂\)

\(x₂ = 4 - 5\)

\(x₂ = -1\)

Next, we can cross-multiply to determine the \(y\)-coordinate:

\((2)(3) = (2)\left(\cfrac{5 + y₂}{2}\right)\)

\(6 = 5 + y₂\)

\(y₂ = 6 - 5\)

\(y₂ = 1\)

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


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

  1. Determine the exact length of each side of the triangle
  2. Classify this triangle

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

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

First, we can calculate the side length of side \(\text{XY}\):

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

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

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

\(d_{\text{XY}} = \sqrt{160} \approx 12.65\)

Next, we can calculate the side length of side \(YZ\):

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

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

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

\(d_{YZ} = \sqrt{136} \approx 11.66\)

Finally, we can calculate the side length of side \(\text{XZ}\):

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

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

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

\(d_{\text{XZ}} = \sqrt{104} \approx 10.2\)

Therefore, we can determine that \(\boldsymbol{d_{\text{\text{XY}}} \approx 12.65}\), \(\boldsymbol{d_{\text{YZ}} \approx 11.66}\) and \(\boldsymbol{d_{\text{XZ}} \approx 10.2}\).


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






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