In the xy-plane, line \ell_1 passes through the points (2, k) and (k, 14), where k is a co… — GRE General Test

In the $xy$ -plane, line $\ell_1$ passes through the points $(2, k)$ and $(k, 14)$ , where $k$ is a constant. Line $\ell_2$ passes through the point $(k, 14)$ and has a $y$ -intercept at $(0, 22)$ . If line $\ell_1$ is perpendicular to line $\ell_2$ , what is the sum of all possible values of $k$ ?

A
-22
B
-14
-6Answer
D
6
E
8

Answer

The sum of all possible values of

k

is -6.

The correct answer is -6 because the slope of line

\ell_1

m_1 = \frac{14-k}{k-2}

and the slope of line

\ell_2

m_2 = -\frac{8}{k}

. Since the lines are perpendicular, their slopes multiply to

-1

, yielding

\frac{8(14-k)}{k(k-2)} = 1

. Solving the resulting quadratic equation

k^2 + 6k - 112 = 0

yields

k = 8

and

k = -14

. The sum of these values is

8 + (-14) = -6

Step-by-Step Solution

Calculate the slope of line

\ell_1

in terms of

k

m_1 = \frac{14 - k}{k - 2}

for

k \neq 2

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

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

Calculate the slope of line

\ell_2

using

(k, 14)

and the

y

-intercept

(0, 22)

m_2 = \frac{14 - 22}{k - 0} = -\frac{8}{k}

for

k \neq 0

The

y

-intercept

(0, 22)

provides a second point on line

\ell_2

to find its slope.

Set up the perpendicularity condition

m_1 \cdot m_2 = -1

\left(\frac{14 - k}{k - 2}\right)\left(-\frac{8}{k}\right) = -1 \implies \frac{8(14 - k)}{k(k - 2)} = 1$.

Two non-vertical lines are perpendicular if and only if the product of their slopes equals

-1

Expand and rearrange the equation into standard quadratic form.

112 - 8k = k^2 - 2k \implies k^2 + 6k - 112 = 0.

Multiplying both sides by

k(k - 2)

clears the denominator to form a quadratic equation.

Factor the quadratic equation to find all possible values of

k

(k + 14)(k - 8) = 0 \implies k = -14 \text{ or } k = 8.

The quadratic expression factors neatly, giving two valid non-zero values for

k

Calculate the sum of all possible values of

k

(-14) + 8 = -6.

Summing the two solutions gives the required final numerical value.

Key Concept

Perpendicular Slopes and Quadratic Line Equations

Estimated Time:2m 30s

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