In the xy-plane, the graph of a linear function f has a negative slope and a positive y-in… — SAT

In the $xy$ -plane, the graph of a linear function $f$ has a negative slope and a positive $y$ -intercept. The graph of $f$ intersects the $x$ -axis at point $A$ and the $y$ -axis at point $B$ . A second line, $L$ , is perpendicular to the graph of $f$ and passes through the origin. If line $L$ intersects the graph of $f$ at point $P$ such that the ratio of the area of triangle $OAP$ to the area of triangle $OBP$ is $4:9$ , what is the slope of the graph of $f$ ?

$-\frac{3}{2}$ Answer
B
$-\frac{2}{3}$
C
$-\frac{9}{4}$
D
$-\frac{4}{9}$

Answer

The slope of the graph of

f

-\frac{3}{2}

The correct answer is

-\frac{3}{2}

. By setting up the equations of the lines

f(x) = mx + b

and

y = -\frac{1}{m}x

, we can express the coordinates of the intercepts

A

and

B

, as well as the intersection point

P

, in terms of

m

and

b

. Calculating the areas of triangle

OAP

and triangle

OBP

gives

\text{Area}(OAP) = \frac{1}{2} \left(-\frac{b}{m}\right) y_0

and

\text{Area}(OBP) = \frac{1}{2} b x_0

. Since

P

lies on the perpendicular line,

y_0 = -\frac{1}{m}x_0

. Substituting this into the area ratio yields

\frac{\text{Area}(OAP)}{\text{Area}(OBP)} = \frac{1}{m^2}

. Equating this to the given ratio

\frac{4}{9}

results in

m^2 = \frac{9}{4}

. Given that the slope is negative,

m

must be

-\frac{3}{2}

Step-by-Step Solution

Define the linear function

f(x)

and the line

L

perpendicular to it.

Let the function be

f(x) = mx + b

, where

m < 0

and

b > 0

. The line

L

perpendicular to the graph of

f

that passes through the origin has the equation

y = -\frac{1}{m}x

This establishes the equations of both lines in terms of the slope

m

and

y

-intercept

b

Determine the coordinates of the intercepts

A

and

B

, and the intersection point

P(x_0, y_0)

The

x

-intercept is

A\left(-\frac{b}{m}, 0\right)

and the

y

-intercept is

B(0, b)

. The intersection point

P(x_0, y_0)

is the solution to

mx + b = -\frac{1}{m}x

, which gives

x_0 = -\frac{mb}{m^2+1}

and

y_0 = \frac{b}{m^2+1}

Finding these points allows us to express the dimensions of triangles

OAP

and

OBP

Calculate the areas of triangles

OAP

and

OBP

and find their ratio in terms of

m

The base of triangle

OAP

along the

x

-axis is

OA = -\frac{b}{m}

and its height is

y_0

. Thus,

\text{Area}(OAP) = \frac{1}{2} \left(-\frac{b}{m}\right) y_0

. The base of triangle

OBP

along the

y

-axis is

OB = b

and its height is

x_0

. Thus,

\text{Area}(OBP) = \frac{1}{2} b x_0

. The ratio of their areas is

\frac{\text{Area}(OAP)}{\text{Area}(OBP)} = \frac{-\frac{b}{m} y_0}{b x_0} = -\frac{y_0}{m x_0}

. Since

P

lies on

L

, we have

\frac{y_0}{x_0} = -\frac{1}{m}

. Substituting this gives the ratio

\frac{\text{Area}(OAP)}{\text{Area}(OBP)} = -\frac{1}{m}\left(-\frac{1}{m}\right) = \frac{1}{m^2}

This simplifies the geometric relationship to a direct relation between the area ratio and the slope of the function.

Solve for the slope

m

using the given ratio of

4:9

Setting

\frac{1}{m^2} = \frac{4}{9}

yields

m^2 = \frac{9}{4}

. Since the problem states the slope is negative, we take the negative square root to get

m = -\frac{3}{2}

This identifies the correct slope value matching the given constraints.

Key Concept

Using coordinate geometry and system of linear equations to determine slopes and intersections, and relating those to geometric areas on the coordinate plane.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic