Question

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?

Answer 1

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

Answer 2

Answer:

The distance is $\sqrt{68} =2\sqrt{17}$

Step-by-step 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:

$x^2=8^2+2^2\\x^2=64+4\\x^2=68\\x=\sqrt{68}=\sqrt{4*17}=2\sqrt{17}$

So the distance, on the grid, between Joe’s house and Andy’s house is $2\sqrt{17}$