Question

three vertices of isosceles trapezoid qrst are located at q(5,-3), r(1,-3), and s(2,-1), with rq representing one of the bases. use the distance formula to find the location of point t.
a. (4,-1)
b. (4,-3)
c. (6,-1)
d. (3,-1)
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Explanation:

Step1: Recall the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For an isosceles trapezoid \(QRST\) with \(RQ\) as a base (\(Q(5,-3)\), \(R(1,-3)\)), the non - parallel sides \(QT\) and \(RS\) are equal. \(S(2,-1)\).

Step2: Calculate the length of \(RS\)

Let \(R(1,-3)\) and \(S(2,-1)\). Then \(RS=\sqrt{(2 - 1)^2+(-1+ 3)^2}=\sqrt{1 + 4}=\sqrt{5}\).

Step3: Calculate the distance for each option

• Option A: Let \(T(4,-1)\) and \(Q(5,-3)\). Then \(QT=\sqrt{(4 - 5)^2+(-1 + 3)^2}=\sqrt{1+4}=\sqrt{5}\).
• Option B: Let \(T(4,-3)\) and \(Q(5,-3)\). Then \(QT=\sqrt{(4 - 5)^2+(-3+3)^2}=1\).
• Option C: Let \(T(6,-1)\) and \(Q(5,-3)\). Then \(QT=\sqrt{(6 - 5)^2+(-1 + 3)^2}=\sqrt{1 + 4}=\sqrt{5}\), but also check the slope condition (since \(RQ\) is horizontal \(y=-3\), for it to be a trapezoid, the other base should be parallel. The slope of \(RS\) is \(m_{RS}=\frac{-1+3}{2 - 1}=2\), the slope of \(QT\) for \(T(6,-1)\): \(m_{QT}=\frac{-1 + 3}{6 - 5}=2\), and the slope of \(RQ\) is \(m_{RQ}=0\). But for a trapezoid, only one pair of sides should be parallel.
• Option D: Let \(T(3,-1)\) and \(Q(5,-3)\). Then \(QT=\sqrt{(3 - 5)^2+(-1 + 3)^2}=\sqrt{4 + 4}=2\sqrt{2}\).

Since \(QT = RS=\sqrt{5}\) for option A and it satisfies the trapezoid condition (only one pair of parallel sides, \(RQ\) is parallel to the line passing through the mid - points of the non - parallel sides in an isosceles trapezoid and the non - parallel sides \(QT\) and \(RS\) are equal in length and the slope of \(RQ\) ( \(y=-3\), slope \(m = 0\)) and the non - parallel nature of \(QT\) and \(RS\) (slope of \(RS = 2\), slope of \(QT\) for \(T(4,-1)\) is \(\frac{-1+3}{4 - 5}=-2\))

Answer:

A. (4,-1)