Question

consider the points a and b graphed on the coordinate plane. what are the coordinates of point c when its location is in quadrant iii and it is a vertex of square abcd? (-1,3) (4,-2) (-1,-2) (4,3)

Explanation:

Step1: Find side - length and slope

Let \(A=(0, 3)\) and \(B=(4, 3)\). The side - length of the square \(AB\) is \(d=\vert4 - 0\vert=4\), and the slope of \(AB\) is \(m_{AB}=\frac{3 - 3}{4 - 0}=0\).

Step2: Determine the perpendicular direction

Since \(ABCD\) is a square, the sides are perpendicular. A line perpendicular to a horizontal line (slope \(m = 0\)) is a vertical line.

Step3: Locate point C

Point \(A=(0,3)\). To get to point \(C\) in the third - quadrant, we move 4 units down from point \(A\) (because of the side - length of the square). The \(x\) - coordinate of \(C\) is the same as the \(x\) - coordinate of \(A\) (because of the perpendicular relationship), and the \(y\) - coordinate of \(C\) is \(3-5=-2\). So the coordinates of \(C\) are \((0 - 1,-2)=(-1,-2)\)

Answer:

\((-1,-2)\)