In the xy-plane, line l_1 passes through the origin and has a positive slope. Line l_2 is … — SAT

In the $xy$ -plane, line $l_1$ passes through the origin and has a positive slope. Line $l_2$ is perpendicular to $l_1$ and intersects the $x$ -axis at $(d, 0)$ , where $d > 0$ . The two lines intersect at the point $(a, b)$ in the first quadrant. If the ratio of $a$ to $b$ is $3$ to $4$ , and the area of the triangle bounded by $l_1$ , $l_2$ , and the $x$ -axis is $A$ , which of the following expressions represents $d$ in terms of $A$ ?

A
$\frac{5\sqrt{6A}}{8}$
B
$\frac{25A}{6}$
$\frac{5\sqrt{6A}}{6}$ Answer
D
$\frac{10\sqrt{6A}}{9}$

Answer

The expression

\frac{5\sqrt{6A}}{6}

represents

d

in terms of

A

The correct expression is

\frac{5\sqrt{6A}}{6}

. The slope of line

l_1

passing through the origin and

(a,b)

is given by

m_1 = \frac{b}{a}

. Since the ratio of

a

b

3

4

, we have

\frac{a}{b} = \frac{3}{4}

, which implies the slope

m_1 = \frac{4}{3}

. Because line

l_2

is perpendicular to

l_1

, its slope is

m_2 = -\frac{3}{4}

. The equation of

l_2

passing through

(d,0)

y = -\frac{3}{4}(x - d)

. Solving the system of equations gives the intersection coordinates

a = \frac{9}{25}d

and

b = \frac{12}{25}d

. The area of the triangle bounded by the two lines and the

x

-axis is

A = \frac{1}{2} \times d \times b = \frac{6}{25}d^2

. Solving for

d

yields

d = \sqrt{\frac{25A}{6}} = \frac{5\sqrt{6A}}{6}

Step-by-Step Solution

Determine the slope and equation of line

l_1

Slope

m_1 = \frac{4}{3}

, and the equation of the line is

y = \frac{4}{3}x

Since

l_1

passes through

(0,0)

and the point

(a,b)

in the first quadrant, its slope is

m_1 = \frac{b}{a}

. Since the ratio of

a

b

3

4

, we have

\frac{a}{b} = \frac{3}{4} \implies \frac{b}{a} = \frac{4}{3}

Determine the equation of line

l_2

The equation of the line is

y = -\frac{3}{4}(x - d)

Line

l_2

is perpendicular to

l_1

, so its slope is the negative reciprocal of

m_1

, which is

m_2 = -\frac{3}{4}

. Using the point-slope form with the

x

-intercept

(d,0)

, the equation is

y - 0 = -\frac{3}{4}(x - d)

Find the coordinates of the intersection point

(a,b)

in terms of

d

a = \frac{9}{25}d

and

b = \frac{12}{25}d

Set the two equations equal to find the

x

-coordinate of the intersection:

\frac{4}{3}x = -\frac{3}{4}(x - d) \implies \frac{4}{3}x = -\frac{3}{4}x + \frac{3}{4}d \implies \frac{25}{12}x = \frac{3}{4}d \implies x = \frac{9}{25}d

. Substituting this back into the equation of

l_1

yields the

y

-coordinate:

y = \frac{4}{3}\left(\frac{9}{25}d\right) = \frac{12}{25}d

Express the area of the triangle in terms of

d

A = \frac{6}{25}d^2

The base of the triangle along the

x

-axis is

d

(from

x=0

x=d

). The height of the triangle is the

y

-coordinate of the intersection point,

b = \frac{12}{25}d

. The area of the triangle is given by

A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}d\left(\frac{12}{25}d\right) = \frac{6}{25}d^2

Solve for

d

in terms of

A

d = \frac{5\sqrt{6A}}{6}

Isolate

d^2

to get

d^2 = \frac{25A}{6}

. Taking the square root of both sides gives

d = \sqrt{\frac{25A}{6}} = \frac{5\sqrt{A}}{\sqrt{6}} = \frac{5\sqrt{6A}}{6}

Key Concept

Using the properties of perpendicular lines, setting up equations from coordinate parameters, and applying geometric formulas to relate parameters in linear systems.

Alternative Method

Using the geometric mean theorem (altitude rule) in a right triangle, the altitude

b

divides the hypotenuse

d

into segments

a

and

d-a

, such that

b^2 = a(d-a)

. Since

b = \frac{4}{3}a

, we can substitute this to find

a = \frac{9}{25}d

and

b = \frac{12}{25}d

directly without finding the line equations. Then,

A = \frac{1}{2} d b = \frac{6}{25}d^2

, which solves to

d = \frac{5\sqrt{6A}}{6}

Estimated Time:3m 0s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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