In the xy-plane, line \ell_1 is defined by the equation 3x - 4y = 12. Line \ell_2 is perpe… — GRE General Test

In the $xy$ -plane, line $\ell_1$ is defined by the equation $3x - 4y = 12$ . Line $\ell_2$ is perpendicular to line $\ell_1$ and intersects the positive $y$ -axis at the point $(0, d)$ . If the area of the triangular region bounded by line $\ell_1$ , line $\ell_2$ , and the $y$ -axis is $24$ , what is the value of $d$ ?

A
$5$
B
$6$
$7$ Cevap
D
$9$
E
$13$

Cevap

The value of

d

7

The line

\ell_1

has equation

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

, placing its

y

-intercept at

(0, -3)

. Line

\ell_2

is perpendicular, so its slope is

-\frac{4}{3}

, giving the equation

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

. The vertical base of the triangle along the

y

-axis spans from

(0, -3)

(0, d)

, with a length of

d + 3

. The intersection of the two lines occurs at an

x

-coordinate of

\frac{12(d+3)}{25}

, which serves as the height of the triangle. Setting the area

\frac{1}{2} \times (d+3) \times \frac{12(d+3)}{25} = 24

simplifies to

(d+3)^2 = 100

. Because

d > 0

d + 3 = 10

, giving

d = 7

Adım Adım Çözüm

Find the slope and

y

-intercept of line

\ell_1

Converting

3x - 4y = 12

into slope-intercept form yields

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

. The slope of

\ell_1

m_1 = \frac{3}{4}

and its

y

-intercept is

(0, -3)

Knowing the slope and

y

-intercept of

\ell_1

is essential to determine the equation of line

\ell_2

and the vertices of the triangular region along the

y

-axis.

Determine the equation of line

\ell_2

Since

\ell_2

is perpendicular to

\ell_1

, its slope is the negative reciprocal of

\frac{3}{4}

, which is

m_2 = -\frac{4}{3}

. Given that

\ell_2

intersects the

y

-axis at

(0, d)

, its equation is

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

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

Find the

x

-coordinate of the intersection point of

\ell_1

and

\ell_2

Set the two equations equal:

\frac{3}{4}x - 3 = -\frac{4}{3}x + d \implies \left(\frac{3}{4} + \frac{4}{3}\right)x = d + 3 \implies \frac{25}{12}x = d + 3 \implies x = \frac{12(d+3)}{25}

The

x

-coordinate of the intersection point represents the horizontal altitude (height) of the triangle with respect to the vertical base along the

y

-axis.

Express the area of the triangular region in terms of

d

and solve for

d

The vertical base along the

y

-axis stretches from

(0, -3)

(0, d)

, having length

d - (-3) = d + 3

. The height is

h = \frac{12(d+3)}{25}

. Using the area formula

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

, we set

\frac{1}{2}(d+3)\left(\frac{12(d+3)}{25}\right) = 24 \implies \frac{6(d+3)^2}{25} = 24 \implies (d+3)^2 = 100

. Since

d > 0

d + 3 = 10

, so

d = 7

The problem states that the area of the triangle is

24

Anahtar Kavram

Perpendicular line slopes and geometric area calculations using coordinate geometry

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