1. Lessons
  2. Gr10
  3. Math
  4. Midpoint

Midpoint

The Midpoint is the halfway point between \(2\) endpoints of a line segment. Certain line segments, such as medians and right bisectors are found using the midpoint. It can be found using the following formula:

\(\text{midpoint} = \left(\cfrac{x₁ + x₂}{2}, \cfrac{y₁ + y₂}{2}\right)\)

  • \(x₁\) represents the x-coordinate of the first point
  • \(x₂\) represents the x-coordinate of the second point
  • \(y₁\) represents the y-coordinate of the first point
  • \(y₂\) represents the y-coordinate of the second point

Example

Find the midpoint of a line segment with endpoints \(\text{C}(-4, 3)\) and \(\text{D}(2, -5)\).

We can plug the coordinates into the formula to determine the midpoint. In this instance, \(\text{C}\) will represent the first coordinate and \(\text{D}\) will represent the second coordinate:

\(\text{midpoint} = \left(\cfrac{x₁ + x₂}{2}, \cfrac{y₁ + y₂}{2}\right)\)

\(\text{midpoint}_{\text{CD}} = \left(\cfrac{-4 + 2}{2}, \cfrac{3 + (-5)}{2}\right)\)

\(\text{midpoint}_{\text{CD}} = \left(\cfrac{-2}{2}, \cfrac{-2}{2}\right)\)

\(\text{midpoint}_{\text{CD}} = (-1, -1)\)

Therefore, we can determine that the midpoint between points \(\text{C}\) and \(\text{D}\) is \(\boldsymbol{(-1, -1)}\).


Find the midpoint of a line segment with endpoints \(\text{A}\left(\cfrac{-2}{5}, \cfrac{-3}{4}\right)\) and \(\text{B}\left(\cfrac{4}{5}, \cfrac{3}{4}\right)\).

First, we need to ensure that all fractions have the same denominator so that the formula works properly. In this case, all fractions will have a denominator of \(20\):

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

\(\text{Coordinate}_{\text{B}} = \left(\cfrac{16}{20}, \cfrac{15}{20}\right)\)

Next, we can plug the coordinates into the formula to find the midpoint. In this instance, \(A\) will represent the first coordinate and \(B\) will represent the second coordinate:

\(\text{midpoint} = \left(\cfrac{x₁ + x₂}{2}, \cfrac{y₁ + y₂}{2}\right)\)

\(\text{midpoint}_{\text{AB}} = \left(\cfrac{\cfrac{-8}{20} + \cfrac{16}{20}}{2}, \cfrac{\cfrac{-15}{20} + \cfrac{15}{20}}{2}\right)\)

\(\text{midpoint}_{\text{AB}} = \left(\cfrac{\cfrac{8}{20}}{2}, \cfrac{\cfrac{0}{20}}{2}\right)\)

\(\text{midpoint}_{\text{AB}} = \left(\cfrac{4}{20}, 0\right)\)

\(\text{midpoint}_{\text{AB}} = \left(\cfrac{1}{5}, 0\right)\)

Therefore, we can determine that the midpoint between points \(\text{A}\) and \(\text{B}\) is \(\boldsymbol{\left(\cfrac{1}{5}, 0\right)}\).

The endpoints of the diameter of a circle are \(\text{A}(-5, -3)\) and \(\text{B}(3, 7)\). Find the coordinates of the circle's origin.

We can use the midpoint formula to determine the coordinates of the circle's origin. In this instance, \(\text{A}\) represents coordinate \(1\) and \(\text{B}\) represents coordinate \(2\):

\(\text{midpoint} = \left(\cfrac{x₁ + x₂}{2}, \cfrac{y₁ + y₂}{2}\right)\)

\(\text{origin} = \left(\cfrac{-5 + 3}{2}, \cfrac{-3 + 7}{2}\right)\)

\(\text{origin} = \left(\cfrac{-2}{2}, \cfrac{4}{2}\right)\)

\(\text{origin} = (-1, 2)\)

Therefore, we can determine that the coordinates for the circle's origin are \(\boldsymbol{(-1, 2)}\).

For \(\triangle \text{ABC}\) with vertices \(\text{A}(-8, 0)\), \(\text{B}(0, 0)\) and \(\text{C}(0, -8)\):

  1. Draw a triangle
  2. Construct the midpoints \(\text{AB}\), \(\text{BC}\), and \(\text{AC}\) and label them \(\text{D}\), \(\text{E}\), and \(\text{F}\) respectively. Join the midpoints to form \(\triangle \text{DEF}\)
  3. Show that line segment \(\text{DE}\) is parallel to the line segment \(\text{AC}\)

i. Based on the set of coordinates given in the question, we can draw the triangle:

Graph of a triangle ABC based on the coordinates given. It forms a right triangle.

ii. We can construct the midpoints using the midpoint formula:

\(\text{midpoint} = \left(\cfrac{x₁ + x₂}{2}, \cfrac{y₁ + y₂}{2}\right)\)

First, we can calculate the midpoint of side \(AB\):

\(\text{midpoint}_{\text{AB}} = \left(\cfrac{-8 + 0}{2}, \cfrac{0 + 0}{2}\right)\)

\(\text{midpoint}_{\text{AB}} = (\cfrac{-8}{2}, \cfrac{0}{2})\)

\(\text{midpoint}_{\text{AB}} = (-4, 0)\)

Next, we can calculate the midpoint of side \(BC\):

\(\text{midpoint}_{\text{BC}} = \left(\cfrac{0 + 0}{2}, \cfrac{0 - 8}{2}\right)\)

\(\text{midpoint}_{\text{BC}} = \left(\cfrac{0}{2}, \cfrac{-8}{2}\right)\)

\(\text{midpoint}_{\text{BC}} = (0, -4)\)

Finally, we can calculate the midpoint of side \(\text{AC}\):

\(\text{midpoint}_{\text{AC}} = \left(\cfrac{-8 + 0}{2}, \cfrac{0 - 8}{2}\right)\)

\(\text{midpoint}_{\text{AC}} = \left(\cfrac{-8}{2}, \cfrac{-8}{2}\right)\)

\(\text{midpoint}_{\text{AC}} = (-4, -4)\)

Using these coordinates, we can now graph \(\triangle DEF\):

A smaller triangle DEF encased within the original triangle is created by connecting its midpoints.

iii. To show that lines \(\text{AC}\) and \(DE\) are parallel to one another, we can compare their slopes:

\(m = \cfrac{y_2 - y_1}{x_2 - x_1}\)

First, we can determine the slope of line \(\text{\text{AC}}\):

\(m_{\text{AC}} = \cfrac{-8 - 0}{0 - (-8)}\)

\(m_{\text{AC}} = \cfrac{-8}{8} = -1\)

Next, we can determine the slope of line \(\text{DE}\):

\(\text{DE} = \cfrac{-4 - 0}{0 - (-4)}\)

\(\text{DE} = \cfrac{-4}{4} = -1\)

Finally, we can compare the slopes of the 2 lines to determine if they're parallel to each other:

\(\text{AC} = \text{DE}\)

\(-1 = -1\)

As lines \(\text{AC}\) and \(\text{DE}\) have the same slope, we can confirm that they are parallel to each other.



Review these lessons:
Try these questions: