In the standard (x, y) coordinate plane, the vertices of a right triangle are P(1, 2), Q(5… — ACT

In the standard $(x, y)$ coordinate plane, the vertices of a right triangle are $P(1, 2)$ , $Q(5, 5)$ , and $R(k, 9)$ . If the right angle of the triangle is at vertex $Q$ , what is the value of the constant $k$ ?

Answer: 2

Answer

The value of

k

2

Because the right angle of the triangle is at vertex

Q

, segment

PQ

must be perpendicular to segment

QR

. The slope of

PQ

\frac{5 - 2}{5 - 1} = \frac{3}{4}

. The slope of a perpendicular line is the negative reciprocal, so the slope of

QR

must be

-\frac{4}{3}

. Expressing the slope of

QR

using the coordinates of

Q(5, 5)

and

R(k, 9)

gives

\frac{9 - 5}{k - 5} = \frac{4}{k - 5}

. Setting this equal to

-\frac{4}{3}

and solving for

k

yields

k - 5 = -3

, which means

k = 2

Step-by-Step Solution

Use the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

to calculate the slope of the line segment

PQ

with endpoints

P(1, 2)

and

Q(5, 5)

m_{PQ} = \frac{5 - 2}{5 - 1} = \frac{3}{4}

This establishes the direction of the first leg of the right triangle.

Find the slope of segment

QR

. Because the right angle is at vertex

Q

, the segment

PQ

is perpendicular to segment

QR

m_{QR} = -\frac{4}{3}

Perpendicular lines have slopes that are negative reciprocals of each other (

m_1 \cdot m_2 = -1

Write the slope of segment

QR

in terms of

k

using coordinates

Q(5, 5)

and

R(k, 9)

m_{QR} = \frac{9 - 5}{k - 5} = \frac{4}{k - 5}

This sets up an equation to find the unknown coordinate value.

Equate the two expressions for the slope of

QR

and solve for

k

\frac{4}{k - 5} = -\frac{4}{3} \Rightarrow k - 5 = -3 \Rightarrow k = 2

Solving the rational equation yields the correct coordinate parameter.

Key Concept

Perpendicular lines in a coordinate plane have slopes that are negative reciprocals of each other.

Estimated Time:1m 30s

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