In the xy-coordinate plane, point P has coordinates (0, 0) and point Q has coordinates (6,… — GRE General Test

In the $xy$ -coordinate plane, point $P$ has coordinates $(0, 0)$ and point $Q$ has coordinates $(6, 8)$ . Point $R$ is positioned such that triangle $PQR$ is a right triangle with hypotenuse $PQ$ . Which of the following could be the coordinates of point $R$ ? Select all such points.

$(8, 4)$ Answer
$(-2, 4)$ Answer
$(3, -1)$ Answer
D
$(4, 8)$
E
$(6, 6)$

Answer

The valid coordinates for point

R

are

(8, 4)

(-2, 4)

, and

(3, -1)

According to the Pythagorean theorem and Thales's theorem, any point forming a right angle with segment

PQ

as hypotenuse must satisfy

PR^2 + RQ^2 = PQ^2 = 100

, placing it on a circle centered at

(3, 4)

with radius

5

. The points

(8, 4)

(-2, 4)

, and

(3, -1)

each lie on this circle because their squared distances to

P

and

Q

sum to

100

Step-by-Step Solution

Calculate the square of hypotenuse

PQ

using the distance formula.

PQ^2 = (6 - 0)^2 + (8 - 0)^2 = 36 + 64 = 100

Since

PQ

is given as the hypotenuse of right triangle

PQR

, the Pythagorean theorem requires

PR^2 + RQ^2 = PQ^2 = 100

Express the condition

PR^2 + RQ^2 = 100

in terms of coordinates

(x, y)

of point

R

(x - 0)^2 + (y - 0)^2 + (x - 6)^2 + (y - 8)^2 = 100 \implies (x - 3)^2 + (y - 4)^2 = 25

By Thales's Theorem, any point

R

that forms a right triangle with hypotenuse

PQ

lies on a circle whose diameter is

PQ

, centered at the midpoint

(3, 4)

with radius

R_{circle} = 5

Test each candidate coordinate pair to verify if it satisfies

(x - 3)^2 + (y - 4)^2 = 25

Points

(8,4)

(-2,4)

, and

(3,-1)

satisfy

(5)^2 + 0^2 = 25

(-5)^2 + 0^2 = 25

, and

0^2 + (-5)^2 = 25

, respectively. Points

(4,8)

and

(6,6)

yield

(4-3)^2 + (8-4)^2 = 17 \neq 25

and

(6-3)^2 + (6-4)^2 = 13 \neq 25

Only coordinates located on the circle of diameter

PQ

form a right angle

\angle PRQ = 90^\circ

Key Concept

Pythagorean Theorem and Right Triangles in Coordinate Geometry

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