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4 votes
On a grid Joe’s house is marked at ( -5, -3 ) and Andy’s house is marked at ( 3, -1 ). What is the distance, on the grid, between Joe’s house and Andy’s house?

2 Answers

4 votes
you will need to use the distance formula to solve this:
Square root (x2-x1)^2+(y2-y1)^2

So,
i will do it without square first.
(3--5)^2+(-1--3)^2

(8)^2+(2)^2 ==== 64+4=68
answer is square root 68


User Yatin
by
6.4k points
3 votes

Answer:

The distance is
√(68) =2√(17)

Explanation:

Locating the points on the plane we can form a right triangle between them and solve the problem using the Pythagorean theorem.

*see attached image*

The sides of the triangle formed measure 8 and 2, so by pythagoras, the unknow side
x wich is the distance between the houses is:


x^2=8^2+2^2\\x^2=64+4\\x^2=68\\x=√(68)=√(4*17)=2√(17)

So the distance, on the grid, between Joe’s house and Andy’s house is
2√(17)

On a grid Joe’s house is marked at ( -5, -3 ) and Andy’s house is marked at ( 3, -1 ). What-example-1
User Bengt
by
6.4k points