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Difficulty: MediumDistance and Midpoint Formulas

In the standard (x,y)(x, y) coordinate plane, a triangle has vertices at A(0,1)A(0, 1), B(2,3)B(2, 3), and C(8,11)C(8, 11). If point MM is the midpoint of side ABAB and point NN is the midpoint of side ACAC, what is the length of the line segment MNMN?

Answer: 5

Answer

The length of the line segment MNMN is 55.
The length of the line segment MNMN is 55. Calculating the coordinates of the midpoint of ABAB, we get M(1,2)M(1, 2). For ACAC, the midpoint is N(4,6)N(4, 6). Applying the distance formula between MM and NN yields (4−1)2+(6−2)2=9+16=5\sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9+16} = 5. Alternatively, by the Midsegment Theorem, the segment connecting the midpoints of two sides of a triangle is half the length of the third side. The length of the third side BCBC is (8−2)2+(11−3)2=36+64=10\sqrt{(8-2)^2 + (11-3)^2} = \sqrt{36 + 64} = 10, so the length of MNMN is 102=5\frac{10}{2} = 5.

Step-by-Step Solution

1
Find the coordinates of MM, the midpoint of side ABAB with endpoints A(0,1)A(0, 1) and B(2,3)B(2, 3).
M(1,2)M(1, 2)
The midpoint formula is M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).
2
Find the coordinates of NN, the midpoint of side ACAC with endpoints A(0,1)A(0, 1) and C(8,11)C(8, 11).
N(4,6)N(4, 6)
Applying the midpoint formula gives (0+82,1+112)=(4,6)\left(\frac{0 + 8}{2}, \frac{1 + 11}{2}\right) = (4, 6).
3
Calculate the distance between M(1,2)M(1, 2) and N(4,6)N(4, 6) using the distance formula.
55
The distance formula is d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. Here, d=(4−1)2+(6−2)2=32+42=25=5d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.

Key Concept

Midpoint and Distance Formulas
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Topics

Distance and Midpoint Formulas

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