In the xy-plane, line L_1 is given by the equation ax + 3y = 12, where a is a constant. Li… — GRE General Test

In the $xy$ -plane, line $L_1$ is given by the equation $ax + 3y = 12$ , where $a$ is a constant. Line $L_2$ is perpendicular to line $L_1$ and passes through the point $(4, -1)$ . If the $x$ -intercept of line $L_2$ is double its $y$ -intercept, and both intercepts of line $L_2$ are non-zero, what is the value of $a$ ?

$-6$ Answer
B
$-\frac{3}{2}$
C
$\frac{3}{2}$
D
$3$
E
$6$

Answer

-6

Rearranging

L_1: ax + 3y = 12

into slope-intercept form gives

y = -\frac{a}{3}x + 4

, yielding a slope of

m_1 = -\frac{a}{3}

. The perpendicular line

L_2

has a slope of

m_2 = \frac{3}{a}

. Using the point

(4, -1)

, the equation for

L_2

y + 1 = \frac{3}{a}(x - 4)

, which simplifies to

y = \frac{3}{a}x - \frac{12+a}{a}

. Setting

x = 0

gives the

y

-intercept

y_0 = -\frac{12+a}{a}

, and setting

y = 0

gives the

x

-intercept

x_0 = \frac{12+a}{3}

. Equating

x_0 = 2 y_0

yields

\frac{12+a}{3} = -2\frac{12+a}{a}

. Canceling the non-zero term

(12+a)

gives

\frac{1}{3} = -\frac{2}{a}

, which leads to

a = -6

Step-by-Step Solution

Determine the slope of line

L_1

and line

L_2

Slope of

L_1

m_1 = -\frac{a}{3}

, so the slope of perpendicular line

L_2

m_2 = \frac{3}{a}

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

Write the equation of line

L_2

using point-slope form

y - (-1) = \frac{3}{a}(x - 4) \implies y = \frac{3}{a}x - \frac{12}{a} - 1 = \frac{3}{a}x - \left(\frac{12 + a}{a}\right)

Line

L_2

passes through

(4, -1)

with slope

m_2 = \frac{3}{a}

Find the

x

-intercept and

y

-intercept of line

L_2

The

y

-intercept is

y_0 = -\frac{12 + a}{a}

. Setting

y = 0

gives

0 = \frac{3}{a}x - \frac{12 + a}{a} \implies x_0 = \frac{12 + a}{3}

Intercepts occur where

x = 0

(for

y

-intercept) and

y = 0

(for

x

-intercept).

Set up and solve the given relation

x_0 = 2 y_0

\frac{12 + a}{3} = 2 \left( -\frac{12 + a}{a} \right)

. Since the intercepts are non-zero,

12 + a \neq 0

. Dividing both sides by

(12 + a)

gives

\frac{1}{3} = -\frac{2}{a} \implies a = -6

The problem states the

x

-intercept is double the

y

-intercept.

Key Concept

Perpendicular Slopes and Line Intercepts

Estimated Time:2m 0s

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