Question

Triangle HJK has vertices at H(2, 2) J(2, 4) and K(0, 2). What is the midpoint of the longest side of the triangle?

Answer 1

The coordinates of the vertices of triangle are given as H(2,2), (J(2, 4), K(0,2)

We would determine the longest side by applying the formula for finding the distance between two points which is expressed as

`\begin{gathered} \text{Distance = }\sqrt[]{x2-x1)^2+(y2-y1)^2} \\ \text{For HJ, x1 = 2, y1 = 2, x2 = 2, y2 = 4} \\ \text{Distance = }\sqrt[]{(2-2)^2+(4-2)^2}\text{ = }\sqrt[]{2^2}\text{ = 2} \\ \text{For JK, x1 = 2, y1 = 4, x2 = 0, y2 = 2} \\ \text{Distance = }\sqrt[]{(0-2)^2+(2-4)^2}=\sqrt[]{(4+4)}=\text{ }2.83 \\ \text{For HK, x1 = 2, y1 = 2, x2 = 0, y2 = 2} \\ \text{Distance = }\sqrt[]{0-2)^2+(2-2)^2}=\text{ }\sqrt[]{4}\text{ = 2} \end{gathered}`

Thus, the longest side is JK. The formula for finding midpoint is

Midpoint = (x1 + x2)/2, (y1 + y2)/2

Midpoint = (2 + 0)/2, (4 + 2)/2

Midpoint = 2/2, 6/2

Midpoint = 1, 3