Examkin

Question

Difficulty: HardProperties of Quadrilaterals

In the standard (x,y)(x, y) coordinate plane, quadrilateral ABCDABCD is a kite that is not a rhombus, with AB=BCAB = BC and AD=CDAD = CD. The vertices AA and CC are located at (2,5)(2, 5) and (8,13)(8, 13), respectively. If the vertex BB is located at (9,6)(9, 6), which of the following could be the coordinates of vertex DD?

  1. A
    (1,12)(1, 12)
  2. B
    (−1,17)(-1, 17)
  3. (−3,15)(-3, 15)Answer
  4. D
    (−3,3)(-3, 3)
  5. E
    (−5,10)(-5, 10)

Answer

The coordinates (−3,15)(-3, 15)
The correct answer is the coordinates (−3,15)(-3, 15). The diagonals of a kite are perpendicular, and the diagonal BDBD perpendicularly bisects the diagonal ACAC. The midpoint of ACAC is M(5,9)M(5, 9) and the slope of ACAC is 43\frac{4}{3}. Therefore, the line containing diagonal BDBD must pass through M(5,9)M(5, 9) with a perpendicular slope of −34-\frac{3}{4}. The equation of this line is 3x+4y=513x + 4y = 51. Testing the coordinates (−3,15)(-3, 15) shows that the point lies on this line. Since the distance from (−3,15)(-3, 15) to M(5,9)M(5, 9) is 1010 units while the distance from B(9,6)B(9, 6) to M(5,9)M(5, 9) is 55 units, the diagonals do not bisect each other, confirming the kite is not a rhombus.

Step-by-Step Solution

1
Find the midpoint MM of diagonal ACAC.
M=(2+82,5+132)=(5,9)M = \left(\frac{2 + 8}{2}, \frac{5 + 13}{2}\right) = (5, 9)
In a kite where AB=BCAB = BC and AD=CDAD = CD, the diagonal BDBD perpendicularly bisects the diagonal ACAC. Therefore, the line containing BDBD must pass through the midpoint of ACAC.
2
Calculate the slope of diagonal ACAC.
mAC=13−58−2=86=43m_{AC} = \frac{13 - 5}{8 - 2} = \frac{8}{6} = \frac{4}{3}
The slope is needed to find the perpendicular slope of the line containing diagonal BDBD.
3
Determine the slope of the line containing diagonal BDBD.
mBD=−1mAC=−34m_{BD} = -\frac{1}{m_{AC}} = -\frac{3}{4}
Since the diagonals of a kite are perpendicular, the slope of the line containing diagonal BDBD is the negative reciprocal of the slope of diagonal ACAC.
4
Write the equation of the line containing diagonal BDBD and test the coordinates of vertex DD.
y−9=−34(x−5)⇒3x+4y=51y - 9 = -\frac{3}{4}(x - 5) \Rightarrow 3x + 4y = 51
Using the point-slope formula with the midpoint M(5,9)M(5, 9) and slope −34-\frac{3}{4} gives the line equation. Testing the correct option (−3,15)(-3, 15) gives 3(−3)+4(15)=−9+60=513(-3) + 4(15) = -9 + 60 = 51, which lies on this line. Furthermore, the distance from (−3,15)(-3, 15) to M(5,9)M(5, 9) is (−3−5)2+(15−9)2=10\sqrt{(-3-5)^2 + (15-9)^2} = 10, which is different from the distance of 55 between (9,6)(9, 6) and M(5,9)M(5, 9), ensuring the kite is not a rhombus.

Key Concept

Diagonals of a kite are perpendicular, and the diagonal connecting the vertices between the unequal adjacent sides perpendicularly bisects the other diagonal.
Rate this question

Topics

Properties of Quadrilaterals

More questions from this topic