Question

rectangles have four right angles and opposite sides that are parallel.
a. is the figure shown a rectangle? explain.
b. if not, how could the points change so it would be a rectangle?
a. the slope of \\( \overline { a b } \\) is \\( \square \\) and the slope of \\( \overline { c d } \\) is 5, so \\( \overline { a b } \\) and \\( \overline { c d } \\) are not parallel. the slope of \\( \overline { b c } \\) is \\( - \frac { 1 } { 6 } \\) and the slope of \\( \overline { d a } \\) is \\( - \frac { 1 } { 7 } \\), so \\( \overline { b c } \\) and \\( \overline { d a } \\) are not parallel. these slopes also show that at least one pair of adjacent sides is not perpendicular, and so at least one of the angles is not a right angle. the figure shown is not a rectangle

Explanation:

Step1: Calculate the slope of \(AB\)

The formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \(A(-1,2)\) and \(B(-2,-3)\). Then \(m_{AB}=\frac{-3 - 2}{-2+1}=\frac{-5}{-1}=5\)

Step2: Analyze the slopes of opposite sides

For a rectangle, opposite sides must be parallel (equal slopes). We are given \(m_{CD} = 5\), \(m_{BC}=-\frac{1}{6}\), \(m_{DA}=-\frac{1}{7}\). Since \(m_{BC}
eq m_{DA}\), \(BC\) and \(DA\) are not parallel. Also, for adjacent sides to be perpendicular, the product of their slopes should be \(- 1\). \(5\times(-\frac{1}{6})
eq - 1\) and \(5\times(-\frac{1}{7})
eq - 1\), \(m_{BC}\times m_{CD}
eq - 1\) and \(m_{DA}\times m_{AB}
eq - 1\)

Step3: Answer part (b)

To make it a rectangle, we can adjust the points so that opposite sides have equal slopes and adjacent sides have slopes whose product is \(-1\). For example, if we change point \(C\) to \((5,-3)\) and point \(D\) to \((6,2)\).

• Slope of \(AB\): Let \(A(-1,2)\) and \(B(-2,-3)\), \(m_{AB}=\frac{-3 - 2}{-2 + 1}=5\)
• Slope of \(CD\): Let \(C(5,-3)\) and \(D(6,2)\), \(m_{CD}=\frac{2+3}{6 - 5}=5\)
• Slope of \(BC\): Let \(B(-2,-3)\) and \(C(5,-3)\), \(m_{BC}=\frac{-3+3}{5 + 2}=0\)
• Slope of \(DA\): Let \(D(6,2)\) and \(A(-1,2)\), \(m_{DA}=\frac{2 - 2}{-1-6}=0\)
• Adjacent sides: \(m_{AB}\times m_{BC}=5\times0 = 0\) (in a rectangle, adjacent sides can also be in a case where one has slope \(0\) and the other is undefined. Here, if we consider another - pair of adjacent sides. Let's assume a re - labeled rectangle. If we have \(AB\) with slope \(m_1\) and \(BC\) with slope \(m_2\), for a non - horizontal/vertical case, say \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\), \(D(x_4,y_4)\). Another way is to make \(A(-1,2)\), \(B(-2,-2)\), \(C(5,-2)\), \(D(6,2)\)
• Slope of \(AB\): \(m_{AB}=\frac{-2 - 2}{-2+1}=4\)
• Slope of \(CD\): \(m_{CD}=\frac{2 + 2}{6 - 5}=4\)
• Slope of \(BC\): \(m_{BC}=\frac{-2+2}{5 + 2}=0\)
• Slope of \(DA\): \(m_{DA}=\frac{2 - 2}{-1-6}=0\)

Answer:

a. The slope of \(\overline{AB}\) is \(5\) (calculated as \(\frac{-3 - 2}{-2+1}\)). The figure shown is not a rectangle because opposite sides \(BC\) and \(DA\) are not parallel (slopes \(m_{BC}=-\frac{1}{6}\) and \(m_{DA}=-\frac{1}{7}\)) and adjacent sides are not perpendicular (product of slopes of adjacent sides is not \(-1\))
b. One way: Change \(B\) to \((-2,-2)\), \(C\) to \((5,-2)\) and \(D\) to \((6,2)\) (so that \(AB\parallel CD\) with slope \(4\) (\(\frac{-2 - 2}{-2 + 1}=\frac{2+2}{6 - 5}\)) and \(BC\parallel DA\) with slope \(0\) (\(\frac{-2 + 2}{5+2}=\frac{2 - 2}{-1-6}\)) and adjacent sides are perpendicular (e.g., slope of \(AB = 4\) and slope of \(BC=0\), in a rectangle, when one side is horizontal (\(m = 0\)) and the other has a non - zero slope, we can also check other pairs. Another pair: if we consider a non - axis - parallel rectangle, move the points such that for \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\), \(D(x_4,y_4)\), \(m_{AB}=m_{CD}\), \(m_{BC}=m_{DA}\) and \(m_{AB}\times m_{BC}=-1\))