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Difficulty: MediumDistance and Midpoint Formulas

In the standard (x,y)(x, y) coordinate plane, a triangle has vertices at L(−2,−1)L(-2, -1), M(2,10)M(2, 10), and N(6,4)N(6, 4). A line segment is drawn from vertex LL to the midpoint of side MNMN. What is the length of this line segment?

  1. A
    1414
  2. B
    272\sqrt{7}
  3. C
    65\sqrt{65}
  4. 1010Answer
  5. E
    40\sqrt{40}

Answer

The length of the line segment is 1010.
First, find the midpoint PP of side MNMN with vertices M(2,10)M(2, 10) and N(6,4)N(6, 4) using the midpoint formula: P=(2+62,10+42)=(4,7)P = \left(\frac{2 + 6}{2}, \frac{10 + 4}{2}\right) = (4, 7). Then, calculate the distance between vertex L(−2,−1)L(-2, -1) and the midpoint P(4,7)P(4, 7) using the distance formula: d=(4−(−2))2+(7−(−1))2=62+82=100=10d = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10.

Step-by-Step Solution

1
Find the coordinates of the midpoint of side MNMN.
The midpoint is P(4,7)P(4, 7).
Apply the midpoint formula P=(x1+x22,y1+y22)P = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) to the vertices M(2,10)M(2, 10) and N(6,4)N(6, 4), yielding P=(2+62,10+42)=(4,7)P = \left(\frac{2 + 6}{2}, \frac{10 + 4}{2}\right) = (4, 7).
2
Calculate the distance between vertex L(−2,−1)L(-2, -1) and the midpoint P(4,7)P(4, 7).
The distance is 1010.
Apply the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} to points L(−2,−1)L(-2, -1) and P(4,7)P(4, 7), yielding d=(4−(−2))2+(7−(−1))2=62+82=100=10d = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10.

Key Concept

Distance and Midpoint Formulas
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Topics

Distance and Midpoint Formulas

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