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Difficulty: HardProperties of Quadrilaterals

In the standard (x,y)(x, y) coordinate plane, a quadrilateral ABCDABCD is an isosceles trapezoid with parallel bases ABAB and CDCD. The coordinates of three of the vertices are A(−15,−20)A(-15, -20), B(15,20)B(15, 20), and C(−7,24)C(-7, 24). If ABCDABCD is NOT a parallelogram, what are the coordinates of the fourth vertex, DD?

  1. A
    (−37,−16)(-37, -16)
  2. (−25,0)(-25, 0)Answer
  3. C
    (−7,−24)(-7, -24)
  4. D
    (7,−24)(7, -24)
  5. E
    (23,64)(23, 64)

Answer

(−25,0)(-25, 0)
The correct answer is (−25,0)(-25, 0). First, we find the slope of the parallel bases ABAB and CDCD to be 43\frac{4}{3}, which gives the equation of the line containing CDCD as 4x−3y+100=04x - 3y + 100 = 0. Since the trapezoid is isosceles, the leg lengths are equal, meaning AD2=BC2=500AD^2 = BC^2 = 500. Substituting the line equation into the distance equation yields two potential coordinates for DD: (−25,0)(-25, 0) and (−37,−16)(-37, -16). Since the problem specifies that ABCDABCD is not a parallelogram, we eliminate (−37,−16)(-37, -16) (which makes AD∥BCAD \parallel BC) to conclude that DD must be (−25,0)(-25, 0).

Step-by-Step Solution

1
Calculate the slope of the base ABAB.
The slope of ABAB is 20−(−20)15−(−15)=4030=43\frac{20 - (-20)}{15 - (-15)} = \frac{40}{30} = \frac{4}{3}.
Since AB∥CDAB \parallel CD, the line containing base CDCD must also have a slope of 43\frac{4}{3}.
2
Set up the equation for the line containing CDCD.
Using the point-slope form with C(−7,24)C(-7, 24) and slope 43\frac{4}{3}, the equation of the line is y−24=43(x+7)  ⟹  4x−3y+100=0y - 24 = \frac{4}{3}(x + 7) \implies 4x - 3y + 100 = 0.
The fourth vertex D(x,y)D(x, y) must lie on this line, so its coordinates satisfy x=3y−1004x = \frac{3y - 100}{4}.
3
Set up the distance equation for the equal leg lengths.
The square of the leg length BC2=(−7−15)2+(24−20)2=(−22)2+42=484+16=500BC^2 = (-7 - 15)^2 + (24 - 20)^2 = (-22)^2 + 4^2 = 484 + 16 = 500. Since AD=BCAD = BC, the distance equation is (x+15)2+(y+20)2=500(x + 15)^2 + (y + 20)^2 = 500.
In an isosceles trapezoid, the non-parallel sides (legs) ADAD and BCBC must have equal lengths.
4
Solve the system of equations for the coordinates of DD.
Substitute x=3y−1004x = \frac{3y - 100}{4} into the distance equation: (3y−404)2+(y+20)2=500  ⟹  25y2+400y=0(\frac{3y - 40}{4})^2 + (y + 20)^2 = 500 \implies 25y^2 + 400y = 0. This yields y=0y = 0 or y=−16y = -16. The corresponding coordinates are D1(−25,0)D_1(-25, 0) and D2(−37,−16)D_2(-37, -16).
Substituting the linear relationship into the quadratic distance equation gives the two mathematically possible locations for DD.
5
Verify which solution satisfies the non-parallelogram constraint.
For D2(−37,−16)D_2(-37, -16), the slope of ADAD is −16−(−20)−37−(−15)=−211\frac{-16 - (-20)}{-37 - (-15)} = -\frac{2}{11}, which is equal to the slope of BCBC. This makes ABCDABCD a parallelogram. For D1(−25,0)D_1(-25, 0), the slope of ADAD is −2≠−211-2 \neq -\frac{2}{11}, which forms a trapezoid.
The problem states that ABCDABCD is not a parallelogram, so DD must be (−25,0)(-25, 0).

Key Concept

Identifying vertices of an isosceles trapezoid using coordinate geometry, slopes of parallel lines, distance formula, and distinguishing a trapezoid from a parallelogram.
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Topics

Properties of Quadrilaterals

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