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Difficulty: MediumGeometric Figures on the Coordinate Plane

In the standard (x,y)(x, y) coordinate plane, a right triangle has vertices A(−2,−3)A(-2, -3), B(6,−3)B(6, -3), and C(6,y)C(6, y), where y>0y > 0. If the hypotenuse of the triangle has a length of 17 units, what is the value of yy?

  1. A
    6
  2. 12Answer
  3. C
    15
  4. D
    18
  5. E
    14

Answer

The correct value of yy is 1212.
The horizontal leg ABAB has a length of 6−(−2)=86 - (-2) = 8 units. By the Pythagorean theorem, the vertical leg BCBC has a length of 172−82=15\sqrt{17^2 - 8^2} = 15 units. Since the vertex BB is at (6,−3)(6, -3) and CC is at (6,y)(6, y) with y>0y > 0, the vertical distance is y−(−3)=15y - (-3) = 15, which yields y=12y = 12.

Step-by-Step Solution

1
Determine the length of the horizontal leg ABAB.
The length of ABAB is 88 units.
Since vertices A(−2,−3)A(-2, -3) and B(6,−3)B(6, -3) share the same yy-coordinate, the segment is horizontal. The length is the positive difference between their xx-coordinates: 6−(−2)=86 - (-2) = 8.
2
Use the Pythagorean theorem to calculate the length of the vertical leg BCBC.
The length of BCBC is 1515 units.
Since ABAB is horizontal and BCBC is vertical (vertices BB and CC share the same xx-coordinate of 66), the angle at BB is a right angle. The hypotenuse is AC=17AC = 17. By the Pythagorean theorem, AB2+BC2=AC2  ⟹  82+BC2=172  ⟹  64+BC2=289  ⟹  BC2=225  ⟹  BC=15AB^2 + BC^2 = AC^2 \implies 8^2 + BC^2 = 17^2 \implies 64 + BC^2 = 289 \implies BC^2 = 225 \implies BC = 15.
3
Set up an equation using the coordinates to find yy.
y=12y = 12
The length of the vertical segment BCBC is the difference in yy-coordinates: ∣y−(−3)∣=15|y - (-3)| = 15, which simplifies to y+3=15y + 3 = 15 since y>0y > 0. Solving for yy yields y=12y = 12.

Key Concept

Using coordinate differences and the Pythagorean theorem to determine unknown vertices of geometric figures in the coordinate plane.

Alternative Method

Alternatively, you can apply the distance formula directly between the vertices A(−2,−3)A(-2, -3) and C(6,y)C(6, y) with a distance of 17 units: (6−(−2))2+(y−(−3))2=17  ⟹  82+(y+3)2=289  ⟹  (y+3)2=225\sqrt{(6 - (-2))^2 + (y - (-3))^2} = 17 \implies 8^2 + (y + 3)^2 = 289 \implies (y + 3)^2 = 225, which yields y+3=15  ⟹  y=12y + 3 = 15 \implies y = 12 since y>0y > 0.
Estimated Time:1m 30s
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Topics

Geometric Figures on the Coordinate Plane

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