In the standard (x, y) coordinate plane, the triangular region R is bounded by the lines y… — ACT

In the standard $(x, y)$ coordinate plane, the triangular region $R$ is bounded by the lines $y = 2x$ , $y = -x + 9$ , and the $x$ -axis. A vertical line $x = k$ (where $0 < k < 9$ ) divides region $R$ into two sub-regions. If the area of the sub-region to the right of the line $x = k$ is exactly $8$ , what is the value of $k$ ?

Answer: 5

Answer

The boundary lines intersect to form a triangle with vertices at

(0, 0)

(9, 0)

, and

(3, 6)

. Since the area to the right of

x = k

8

, and the total area of the triangle is

27

k

must be greater than

3

. The region to the right of

x = k

is a right triangle with a base of

9 - k

and a height of

9 - k

. Setting its area

\frac{1}{2}(9 - k)^2

equal to

8

yields

(9 - k)^2 = 16

, which gives

9 - k = 4

(since

k < 9

), and thus

k = 5

Step-by-Step Solution

Find the vertices of the triangular region

R

by finding the intersection points of the boundary lines

y = 2x

y = -x + 9

, and

y = 0

The vertices of the triangle are

A(0, 0)

B(9, 0)

, and

C(3, 6)

This establishes the boundaries and shape of the triangular region.

Determine which side of the peak

x = 3

the vertical line

x = k

must lie. Calculate the total area and the area of the left portion.

The total area of the triangle is

27

. The area to the left of the peak

x = 3

9

. Since the area of the region to the right of

x = k

8

, which is less than

18

k

must be greater than or equal to

3

This determines the geometric shape of the sub-region to the right of

x = k

as a right triangle.

Set up the area formula for the right-hand triangle with vertices

(k, 0)

(9, 0)

, and

(k, -k + 9)

, and set it equal to

8

The area is

\frac{1}{2}(9 - k)^2 = 8

, which simplifies to

(9 - k)^2 = 16

This relates the given area to the unknown coordinate

k

Solve the equation

(9 - k)^2 = 16

for

k

, keeping in mind that

k < 9

Taking the square root gives

9 - k = 4 \implies k = 5

This yields the final value of

k

Key Concept

Finding the area of a region defined by linear boundary equations and dividing it with a vertical line.

Alternative Method

Using similar triangles: The right-hand triangle formed by the line

x = 3

, the line

y = -x + 9

, and the

x

-axis has vertices at

(3,0)

(9,0)

, and

(3,6)

, with an area of

\frac{1}{2} \times 6 \times 6 = 18

. The smaller triangle to the right of

x = k

has an area of

8

and is similar to the larger triangle. The ratio of their areas is

\frac{8}{18} = \frac{4}{9}

, which means the ratio of their linear dimensions is

\sqrt{\frac{4}{9}} = \frac{2}{3}

. The base of the larger triangle is

9 - 3 = 6

, so the base of the smaller triangle must be

6 \times \frac{2}{3} = 4

. This gives

9 - k = 4 \implies k = 5

Estimated Time:2m 30s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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