Question

1. higher order thinking a rectangle on a coordinate plane has one vertex at (-5, -6) and a perimeter of 30 units. what could be the coordinates of the other 3 vertices?

Explanation:

Step1: Use the perimeter formula of a rectangle

The perimeter formula of a rectangle is \(P = 2(l + w)\), where \(P = 30\). So \(2(l + w)=30\), then \(l + w=\frac{30}{2}=15\). Let's assume \(l = 10\) and \(w = 5\) (there are multiple pairs of \(l\) and \(w\) that satisfy \(l + w=15\)).

Step2: Find the coordinates of the other vertices

If one vertex is \((- 5,-6)\), and we assume the length is parallel to the \(x\) - axis and the width is parallel to the \(y\) - axis.
If we move along the \(x\) - axis (length direction):
The \(x\) - coordinate of the adjacent vertex (along the length) is \(-5 + 10=5\), and the \(y\) - coordinate remains \(-6\), so one vertex is \((5,-6)\).
If we move along the \(y\) - axis (width direction) from \((-5,-6)\):
The \(y\) - coordinate of the adjacent vertex (along the width) is \(-6 + 5=-1\), and the \(x\) - coordinate remains \(-5\), so one vertex is \((-5,-1)\).
The fourth vertex can be found by moving from \((5,-6)\) along the \(y\) - axis (width direction) or from \((-5,-1)\) along the \(x\) - axis (length direction). The fourth vertex is \((5,-1)\)

Answer:

The other three vertices could be \((5,-6)\), \((-5,-1)\) and \((5,-1)\)