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Difficulty: Very hardGeometric Figures on the Coordinate Plane

In the standard (x,y)(x,y) coordinate plane, a right isosceles triangle PQRPQR has its right angle at vertex Q(2,−1)Q(2, -1) and another vertex at P(−1,3)P(-1, 3). If the third vertex R(x,y)R(x, y) is located in the first quadrant, what is the value of 3x−y3x - y?

  1. 16Answer
  2. B
    12
  3. C
    -2
  4. D
    8
  5. E
    21

Answer

16
The correct answer is 16. To find the coordinates of vertex R(x,y)R(x, y), we find that the vector QP⃗=(−3,4)\vec{QP} = (-3, 4) has a length of 5 and a slope of −4/3-4/3. Since △PQR\triangle PQR is a right isosceles triangle with the right angle at QQ, the segment QRQR must be perpendicular to QPQP (slope of 3/43/4) and equal in length (QR=5QR = 5). This yields two possible locations for RR: (6,2)(6, 2) and (−2,−4)(-2, -4). Because RR must lie in the first quadrant, its coordinates are (6,2)(6, 2). Evaluating 3x−y3x - y gives 3(6)−2=163(6) - 2 = 16.

Step-by-Step Solution

1
Calculate the vector QP⃗\vec{QP} and the length of segment QPQP.
QP⃗=P−Q=(−1−2,3−(−1))=(−3,4)\vec{QP} = P - Q = (-1 - 2, 3 - (-1)) = (-3, 4). The distance is QP=(−3)2+42=5QP = \sqrt{(-3)^2 + 4^2} = 5.
Since the triangle is right isosceles with the right angle at QQ, the leg QRQR must be perpendicular to QPQP and have the same length of 5.
2
Find the slope of QPQP and determine the slope of the perpendicular line containing QRQR.
Slope of QP=3−(−1)−1−2=−43QP = \frac{3 - (-1)}{-1 - 2} = -\frac{4}{3}. The perpendicular slope of QRQR is the negative reciprocal, which is 34\frac{3}{4}.
Perpendicular lines on the coordinate plane have slopes that are negative reciprocals of each other.
3
Determine the possible coordinates of RR by scaling the unit perpendicular vector.
The vector QR⃗\vec{QR} must be of the form (4k,3k)(4k, 3k) for some scalar kk. Since the length is 5, (4k)2+(3k)2=25  ⟹  25k2=25  ⟹  k=±1(4k)^2 + (3k)^2 = 25 \implies 25k^2 = 25 \implies k = \pm 1. This yields two possible coordinates for RR: R=(2+4,−1+3)=(6,2)R = (2+4, -1+3) = (6, 2) or R=(2−4,−1−3)=(−2,−4)R = (2-4, -1-3) = (-2, -4).
Using the slope components and the distance constraint ensures QRQR is perpendicular and equal in length to QPQP.
4
Apply the quadrant constraint and evaluate the final expression 3x−y3x - y.
Since RR is in the first quadrant, R=(6,2)R = (6, 2) with x=6,y=2x=6, y=2. Thus, 3x−y=3(6)−2=163x - y = 3(6) - 2 = 16.
The coordinates of a point in the first quadrant must both be positive.

Key Concept

Using slopes, vectors, and distance formulas to determine the vertices of geometric figures on the coordinate plane.
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Topics

Geometric Figures on the Coordinate Plane

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