In the xy-plane, line k has the equation y = -2x + 10. Line m is parallel to line k and pa… — SAT

In the $xy$ -plane, line $k$ has the equation $y = -2x + 10$ . Line $m$ is parallel to line $k$ and passes through the point $(6, 8)$ . Line $l$ is perpendicular to line $k$ and intersects the $x$ -axis at the point $(a, 0)$ , where $a > 10$ . If the region bounded by lines $k$ , $m$ , $l$ , and the $y$ -axis has an area of $110$ , what is the value of $a$ ?

Answer: 25

Answer

To find the value of

a

, we determine the equations of the lines

m

and

l

based on their geometric relationships to line

k

. Line

m

is parallel to line

k

(

y = -2x + 10

), so its slope is

-2

. Using the point

(6, 8)

, its equation is

y = -2x + 20

. Line

l

is perpendicular to line

k

, so its slope is

\frac{1}{2}

. It intersects the

x

-axis at

(a, 0)

, giving the equation

y = \frac{1}{2}x - \frac{1}{2}a

. The bounded region formed by the parallel lines

k

and

m

, the perpendicular line

l

, and the

y

-axis is a trapezoid. Calculating the area of this trapezoid by dividing it into a parallelogram and a triangle yields the area formula

60 + 2a

. Setting this equal to the given area of

110

yields

60 + 2a = 110

, which solves to

a = 25

Step-by-Step Solution

Determine the equation of line

m

using the parallel slope and the given point.

Line

m

has the equation

y = -2x + 20

Parallel lines have equal slopes. Since line

k

has a slope of

-2

, line

m

also has a slope of

-2

. Substituting the point

(6, 8)

into the point-slope form gives

y - 8 = -2(x - 6)

Determine the equation of line

l

using the perpendicular slope and its

x

-intercept.

Line

l

has the equation

y = \frac{1}{2}x - \frac{1}{2}a

Perpendicular lines have negative reciprocal slopes. The negative reciprocal of

-2

\frac{1}{2}

. Using the point

(a, 0)

in the point-slope form gives

y - 0 = \frac{1}{2}(x - a)

Calculate the vertices of the bounded region by finding the intersection points of the boundary lines.

The vertices of the bounded region are

(0, 20)

(0, 10)

(4 + 0.2a, 2 - 0.4a)

, and

(8 + 0.2a, 4 - 0.4a)

The region is bounded by the parallel lines

k

and

m

, the perpendicular line

l

, and the

y

-axis (

x = 0

Find the area of the region as an algebraic expression in terms of

a

The area is equal to

60 + 2a

The region can be divided into a parallelogram with a vertical base of

10

and width

4 + 0.2a

, and a right triangle with a vertical base of

10

and width

4

. The sum of their areas is

10(4 + 0.2a) + \frac{1}{2}(10)(4) = 40 + 2a + 20 = 60 + 2a

Set the area expression equal to the given area of

110

and solve for

a

a = 25

Setting the area equal to

110

yields

60 + 2a = 110

, which simplifies to

2a = 50

, or

a = 25

Key Concept

Linear functions, parallel and perpendicular lines, finding line equations, and coordinate geometry area.

Alternative Method

The area can also be calculated using the geometric properties of a trapezoid. The height of the trapezoid is the perpendicular distance between the parallel lines

k

and

m

, which is

\frac{|20 - 10|}{\sqrt{(-2)^2 + 1^2}} = 2\sqrt{5}

. The bases of the trapezoid are the segments of lines

k

and

m

from the

y

-axis to their intersection points with line

l

. The length of the base on line

k

9\sqrt{5}

and the length of the base on line

m

13\sqrt{5}

(when

a = 25

). Using the formula for the area of a trapezoid,

\text{Area} = \frac{9\sqrt{5} + 13\sqrt{5}}{2} \times 2\sqrt{5} = 11\sqrt{5} \times 2\sqrt{5} = 110

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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