In the standard (x, y) coordinate plane, the lines with equations y = \frac{1}{2}x + 1, y … — ACT

In the standard $(x, y)$ coordinate plane, the lines with equations $y = \frac{1}{2}x + 1$ , $y = -x + 7$ , and $y = k$ enclose a triangular region with an area of $6$ square units. What is a possible value of $k$ ?

A
2
B
3
C
4
5Answer
E
7

Answer

The correct value of k is 5.

The correct answer is 5. Finding the intersection of the two boundary lines yields the vertex

P(4, 3)

. Calculating the intersection points of the horizontal line

y = k

with the boundary lines gives the

x

-coordinates

2k - 2

and

7 - k

. The distance between these coordinates represents the base of the triangle,

3|k - 3|

, while the height is the vertical distance

|k - 3|

. Substituting these into the triangle area formula

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

results in

6 = \frac{3}{2}(k - 3)^2

. Solving this quadratic equation gives

(k - 3)^2 = 4

, which leads to

k = 5

k = 1

. Therefore, 5 is the correct possible value.

Step-by-Step Solution

Find the intersection point of the two non-horizontal lines by setting their equations equal to each other.

\frac{1}{2}x + 1 = -x + 7 \Rightarrow \frac{3}{2}x = 6 \Rightarrow x = 4

. Substituting

x = 4

back into either equation gives

y = 3

. The intersection point is

P(4, 3)

The intersection point serves as the third vertex of the triangle, and its

y

-coordinate helps determine the height of the triangle relative to the horizontal boundary line

y = k

Find the

x

-coordinates of the intersection points between the horizontal line

y = k

and the other two lines.

For

y = \frac{1}{2}x + 1

, setting

y = k

gives

k = \frac{1}{2}x + 1 \Rightarrow x = 2k - 2

. For

y = -x + 7

, setting

y = k

gives

k = -x + 7 \Rightarrow x = 7 - k

These two points define the base of the triangle along the line

y = k

Express the base length and height of the triangle in terms of

k

, then set up the area equation.

The base is the distance between the two

x

-coordinates:

|(2k - 2) - (7 - k)| = |3k - 9| = 3|k - 3|

. The height is the vertical distance from the line

y = k

to the point

P(4, 3)

|k - 3|

. The area is

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{3}{2}(k - 3)^2

This sets up a solvable algebraic equation using the given area of

6

square units.

Solve the area equation for

k

\frac{3}{2}(k - 3)^2 = 6 \Rightarrow (k - 3)^2 = 4 \Rightarrow k - 3 = \pm 2

. Thus,

k = 5

k = 1

. Since

5

is among the options, it is the correct choice.

This yields the possible values of

k

that satisfy the geometric conditions.

Key Concept

Linear Equations and Graphing

Alternative Method

Instead of solving the algebraic quadratic equation, you can test the given choices for

k

. For example, if you test the value

5

, the line is

y = 5

. The intersection of

y = 5

and

y = \frac{1}{2}x + 1

(8, 5)

, and the intersection of

y = 5

and

y = -x + 7

(2, 5)

. The base of the triangle is the horizontal distance from

x = 2

x = 8

, which is

6

. The height of the triangle is the vertical distance from

y = 3

(the intersection vertex) to

y = 5

, which is

2

. The area is

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

. This matches the problem statement, confirming that

5

is the correct answer.

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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