Question

Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A(-8, -6), B(-3,-6), C(-3, -4), and D(-8, -4). Given these coordinates, what is the length of side AB of this rectangle?

Answer 1

5

Explanation:

Firstly, I'm not from the USA so I'm not sure if I've understood the question correctly (different phrasing of the question here in the UK), but I hope I've done this right and this helps! :)

Method 1:

I would draw the graph! It makes it easier to visualise. You don't have to do a neat proper graph, just scribble something down. If you then count the number of squares between point A and point B you would get 5.

Method 2:

This is probably a more MATHEMATICAL method, but I think Method 1 or Method 2 are both fine. Since we are only asked about point A and point B, we don't have to think about points C or D. Since this is a rectangle, we know that the line would be straight and would also likely be perpendicular / parallel to the x or y axis. If we compare the coordinates of points A and B, we can see that they both have -6 in them: this shows us that these points are along the same row. So all you have to do is find the difference between the tow other points, i.e. the -8 in A and the -3 in B. -3 minus -8 = 5. So that gives you the length of the side.

I hope this helped, sorry for the lengthy explanation and slightly dodgy vocabulary (?).

Bluey

Answer 2

Explanation:

You could use distance formula to find the length

A( -8, -6) ; B(-3,-6)

Distance=
`\sqrt{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)}`

`AB =\sqrt{(-3-[-8])^(2)+(-6-[-6])^(2)}\\\\=\sqrt{(-3+8)^(2)+(-6+6)^(2)}\\\\= \sqrt{(5)^(2)+0}\\\\= \sqrt{5^(2)}\\`

AB = 5 units