A rectangle has three of its vertices at points #(3,4), # #(5,–4),# and #(–7, –7).# What is the #y# coordinate of the unknown vertex?

The given points can be drawn in the #xy#-plane as follows:
graph{((x-3)^2+(y-4)^2-0.05)((x-5)^2+(y+4)^2-0.05)((x+7)^2+(y+7)^2-0.05)=0 [-14.95, 13.52, -8.37, 5.87]}

From here, it is easy to visualize roughly where the fourth vertex will be—somewhere in #"Quadrant II"#.

Since rectangles are composed of two sets of parallel sides, the #y# distance between two adjacent points will be the same as the #y# distance between the other two points.

What does this mean? Well, the point #(3,4)# is 8 up from its neighbour #(5,"-"4)# (since #4-"(-4)"=8#), so the fourth vertex will also be 8 up from #("-7","-7")#. And what is 8 up from -7? That's right: 1.

We can use this to find the #x#-coordinate as well; just use the #x#-coordinates instead of the #y#'s. The remaining vertex is at #("-9",1)#.

Hope this helps!

Use :

The vector #(x_1, y_1) to (x_2, y_2)# is # < x_2-x_1, y_2-y_1>#.

The vertices are labelled

#A( 3, 4 ), B( 5, -4 ), C(-7. -7) and D( x, y ).#.

Vector #AD = < x-3, y-4>#

= vector # BC=<-12, -3>#.

Solving,

(#x, y) = (-9, 1)#.