Question

what is the midpoint of the shortest side of a triangle with vertices (1, 1), (1, 5), and (7, 1)?
a. (1, 2)
b. (1, 3)
c. (4, 1)
d. (4, 3)

Explanation:

Step1: Calculate the lengths of the sides

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

• Between \((1,1)\) and \((1,5)\): \(d_1=\sqrt{(1 - 1)^2+(5 - 1)^2}=\sqrt{0 + 16}=4\)
• Between \((1,1)\) and \((7,1)\): \(d_2=\sqrt{(7 - 1)^2+(1 - 1)^2}=\sqrt{36+0}=6\)
• Between \((1,5)\) and \((7,1)\): \(d_3=\sqrt{(7 - 1)^2+(1 - 5)^2}=\sqrt{36 + 16}=\sqrt{52}=2\sqrt{13}\approx7.21\)

Step2: Identify the shortest side

The shortest side is between \((1,1)\) and \((1,5)\)

Step3: Find the mid - point of the shortest side

Use the mid - point formula \((\frac{x_1+x_2}{2},\frac{y_1 + y_2}{2})\). For points \((1,1)\) and \((1,5)\), \(x=\frac{1 + 1}{2}=1\), \(y=\frac{1+5}{2}=3\)

Answer:

B. \((1,3)\)