A tree is located 20yd to the east and 10yd to the north of a house.

Question

A tree is located 20yd to the east and 10yd to the north of a house. A second tree is located 10yd to the east and 50yd to the north of the house. The prize of a scavenger hunt is placed exactly midway between the trees

1) how far is the prize from the house?

2) what are the coordinates of the positions of the prize

Answer 1

Let's draw the scenario on a coordinate plane with the house at the origin. Then, one tree is at (20, 10) and the other at (10, 50). The midpoint of the segment is found by adding the two x-values then dividing the result by two and then repeating for the y-values.

x = (10 + 20)/2 = 15

y = (50 + 10)/2 = 30

So the midpoint of the segment is at the coordinate (15, 30)

To find the distance from the house, we use the Pythagorean Theorem. Distance = √(15² + 30²) = √(225 + 900) = √1125 = 15√5 yards ≈ 33.54 yards.

The coordinates of the prize again are (15, 30) or 15 yards east and 30 yards north