Coordinate Geometry Practice Questions — ACT — Page 11

Question 201Question →

Consider two lines in a coordinate plane: Line $A$ , represented by the equation $3x + ky = 8$ for some constant $k$ , and Line $B$ , which contains the points $(2, -1)$ and $(5, 8)$ . If Line $A$ is perpendicular to Line $B$ , what is the value of $k$ ?

9

-9

1

-1

3

Show answer & explanation

Answer: 9

Answer

The value of

k

is

9

.

The correct answer is the value

9

. First, we find the slope of Line

B

using the coordinate points

(2, -1)

and

(5, 8)

, which gives a slope of

3

. Since Line

A

is perpendicular to Line

B

, its slope must be the negative reciprocal of

3

, which is

-\frac{1}{3}

. By rewriting the equation of Line

A

,

3x + ky = 8

, in slope-intercept form, we identify its slope as

-\frac{3}{k}

. Setting these two slope values equal gives

-\frac{3}{k} = -\frac{1}{3}

, which solves to

k = 9

.

Step-by-Step Solution

1

Calculate the slope of Line

B

using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

with the points

(2, -1)

and

(5, 8)

.

The slope of Line

B

is

m_B = \frac{8 - (-1)}{5 - 2} = \frac{9}{3} = 3

.

Finding the slope of the given line is necessary to determine the perpendicular slope.

2

Determine the perpendicular slope of Line

A

using the relationship

m_A = -\frac{1}{m_B}

.

The slope of Line

A

must be

m_A = -\frac{1}{3}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Express the slope of Line

A

in terms of

k

by rewriting

3x + ky = 8

in slope-intercept form (

y = mx + b

).

Subtracting

3x

from both sides gives

ky = -3x + 8

, and dividing by

k

yields

y = -\frac{3}{k}x + \frac{8}{k}

. The slope is

-\frac{3}{k}

.

This allows us to set up an equation to solve for

k

.

4

Equate the slope of Line

A

to the perpendicular slope found in Step 2 and solve for

k

.

-\frac{3}{k} = -\frac{1}{3} \Rightarrow \frac{3}{k} = \frac{1}{3} \Rightarrow k = 9

.

Solving this equation gives the value of

k

that makes the lines perpendicular.

Key Concept

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

Estimated Time:1m 30s

Question 202Question →

In the standard $(x, y)$ coordinate plane, a line segment has endpoints at $M(1, -3)$ and $N(5, 5)$ . If a second line is perpendicular to segment $MN$ at its midpoint, what is the $y$ -intercept of this second line?

Show answer & explanation

Answer: 2.5

Answer

The y-intercept of the second line is 2.5.

The slope of segment

MN

is

2

, meaning the perpendicular line must have a slope of

-0.5

. Since this perpendicular line bisects the segment, it must pass through the midpoint

(3, 1)

. Substituting these values into the slope-intercept form

y = mx + b

yields

1 = -0.5(3) + b

, which simplifies to

b = 2.5

.

Step-by-Step Solution

1

Calculate the coordinates of the midpoint of segment

MN

.

The midpoint is

(3, 1)

.

The perpendicular bisector must pass through the midpoint of the bisected segment.

2

Calculate the slope of segment

MN

.

The slope of

MN

is

2

.

The slope is needed to find the perpendicular slope.

3

Calculate the slope of the perpendicular line.

The perpendicular slope is

-0.5

.

Perpendicular lines have slopes that are negative reciprocals of each other.

4

Find the

y

-intercept (

b

) using the slope

-0.5

and the point

(3, 1)

.

The

y

-intercept is

2.5

.

Substituting the coordinates of the midpoint and the perpendicular slope into the equation

y = mx + b

gives

1 = -0.5(3) + b

, which simplifies to

b = 2.5

.

Key Concept

Finding the equation and y-intercept of a perpendicular bisector using midpoint and negative reciprocal slope.

Question 203Question →

In the standard $(x, y)$ coordinate plane, the midpoint of a line segment with endpoints $A(r, -2)$ and $B(10, s)$ is $M(6, 3)$ . What is the distance between the point $A$ and the origin $(0, 0)$ ?

2\sqrt{2}

3\sqrt{5}

2\sqrt{17}

4

2\sqrt{41}

Show answer & explanation

Answer: $2\sqrt{2}$

Answer

The distance between the point

A

and the origin is

2\sqrt{2}

.

To find the coordinates of point

A(r, -2)

, we set up the midpoint formula with the coordinates of

B(10, s)

and the midpoint

M(6, 3)

. For the

x

-coordinate,

\frac{r + 10}{2} = 6 \implies r + 10 = 12 \implies r = 2

. For the

y

-coordinate,

\frac{-2 + s}{2} = 3 \implies -2 + s = 6 \implies s = 8

. Thus, point

A

is

(2, -2)

. The distance from

A(2, -2)

to the origin

(0, 0)

is

\sqrt{(2 - 0)^2 + (-2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}

.

Step-by-Step Solution

1

Use the midpoint formula to set up equations for the coordinates of the midpoint

M(6, 3)

given the endpoints

A(r, -2)

and

B(10, s)

.

The equations are

\frac{r + 10}{2} = 6

and

\frac{-2 + s}{2} = 3

.

The midpoint coordinates are the averages of the corresponding coordinates of the endpoints.

2

Solve these equations for

r

and

s

.

r + 10 = 12 \implies r = 2

, and

-2 + s = 6 \implies s = 8

.

This determines the coordinates of point

A

as

(2, -2)

and point

B

as

(10, 8)

.

3

Calculate the distance between point

A(2, -2)

and the origin

(0, 0)

using the distance formula.

d = \sqrt{(2 - 0)^2 + (-2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}

.

The distance formula is used to find the length of the segment connecting the point to the origin.

Key Concept

Finding the endpoint of a line segment using the midpoint formula, and then finding the distance between a point and the origin using the distance formula.

Alternative Method

Instead of solving for the

y

-coordinate

s

of point

B

, we only need the

x

-coordinate of

A

, which is

r

, to find the distance between

A(r, -2)

and the origin. Since

A

's

y

-coordinate is already given as

-2

, we only set up the equation for the

x

-coordinate of the midpoint:

\frac{r+10}{2} = 6

, which gives

r = 2

. Thus,

A

is

(2, -2)

, and the distance is

\sqrt{2^2 + (-2)^2} = 2\sqrt{2}

. This saves time by not calculating

s

.

Estimated Time:1m 30s

Question 204Question →

On a coordinate grid, a line segment connects the points $A(1, 3)$ and $B(9, 15)$ . The midpoint of this segment is $M$ , and the midpoint of the segment connecting $A$ and $M$ is $C$ . What is the $y$ -coordinate of point $C$ ?

Show answer & explanation

Answer: 6

Answer

The

y

-coordinate of point

C

is 6.

Applying the midpoint formula to the endpoints

A(1, 3)

and

B(9, 15)

yields

M(5, 9)

. Applying the midpoint formula again to

A(1, 3)

and

M(5, 9)

yields

C(3, 6)

. The

y

-coordinate of point

C

is therefore 6.

Step-by-Step Solution

1

Calculate the coordinates of the midpoint

M

of segment

AB

.

M = (5, 9)

The midpoint formula is defined as

\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

.

2

Calculate the coordinates of the midpoint

C

of segment

AM

.

C = (3, 6)

Apply the midpoint formula using the coordinates of

A(1, 3)

and the newly found midpoint

M(5, 9)

.

3

Identify the

y

-coordinate of point

C

.

6

For the point

C(3, 6)

, the

x

-coordinate is 3 and the

y

-coordinate is 6.

Key Concept

Midpoint Formula

Estimated Time:1m 15s

Question 205Question →

In the standard $(x, y)$ coordinate plane, a vertex $V$ of a square is translated $3$ units to the left and $4$ units up, and then reflected across the $x$ -axis. If the coordinates of the final image of $V$ are $(2, -5)$ , what are the coordinates of the original vertex $V$ ?

(-1, 9)

(5, 9)

(1, -9)

(5, 1)

(5, -9)

Show answer & explanation

Answer: $(5, 1)$

Answer

The coordinates of the original vertex V are (5, 1)

The correct answer is the coordinates

(5, 1)

. Starting with the original vertex

V(x, y)

, a horizontal translation of

3

units left results in

x - 3

, and a vertical translation of

4

units up results in

y + 4

, yielding the intermediate point

(x - 3, y + 4)

. Reflecting this point across the

x

-axis negates the entire y-coordinate expression, giving

(x - 3, -y - 4)

. Equating this to the final coordinates

(2, -5)

results in

x - 3 = 2

, which solves to

x = 5

, and

-y - 4 = -5

, which simplifies to

-y = -1

, or

y = 1

.

Step-by-Step Solution

1

Represent the transformations algebraically starting from the original vertex

V(x, y)

.

Translating

3

units left and

4

units up maps

V(x, y)

to an intermediate point

V'(x - 3, y + 4)

.

Horizontal translation left subtracts from the x-coordinate, and vertical translation up adds to the y-coordinate.

2

Apply the second transformation, reflecting the intermediate point

V'

across the

x

-axis.

Reflecting

V'(x - 3, y + 4)

across the

x

-axis negates the y-coordinate, resulting in the final image

V''(x - 3, -(y + 4)) = (x - 3, -y - 4)

.

A reflection across the x-axis transforms any point

(a, b)

to

(a, -b)

.

3

Set the algebraic coordinates of the final image equal to the given coordinates

(2, -5)

and solve for

x

and

y

.

x - 3 = 2 \implies x = 5

and

-y - 4 = -5 \implies -y = -1 \implies y = 1

.

This determines the coordinates of the original pre-image vertex

V(5, 1)

.

Key Concept

Performing composite transformations in reverse order to find the pre-image coordinates.

Question 206Question →

A parallelogram with vertices $P(-3, -2)$ , $Q(5, -2)$ , $R(8, 2)$ , and $S(0, 2)$ is plotted on a standard coordinate grid. What is the perimeter of parallelogram $PQRS$ ?

16 + 2\sqrt{7}

26

30

32

34

Show answer & explanation

Answer: 26

Answer

The correct answer is 26, which is the perimeter of the parallelogram.

The horizontal base segments have a length of

5 - (-3) = 8

. The slanted side has a length of

\sqrt{(0 - (-3))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = 5

. Summing the four sides gives a perimeter of

2(8) + 2(5) = 26

.

Step-by-Step Solution

1

Calculate the length of the horizontal base PQ using the x-coordinates of P(-3, -2) and Q(5, -2).

PQ = 5 - (-3) = 8

Horizontal segments on the coordinate plane have a length equal to the difference in their x-coordinates.

2

Apply the distance formula to find the length of the slanted side PS with vertices P(-3, -2) and S(0, 2).

PS = \sqrt{(0 - (-3))^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5

The distance formula determines the straight-line distance between two coordinates.

3

Calculate the total perimeter of parallelogram PQRS by summing the lengths of all four sides.

\text{Perimeter} = 2 \times \text{base} + 2 \times \text{slanted side} = 2(8) + 2(5) = 16 + 10 = 26

The perimeter of a parallelogram is the sum of all its outer boundary lengths.

Key Concept

Calculating the perimeter of geometric figures on the coordinate plane by determining horizontal distances and applying the distance formula for skewed lines.

Alternative Method

Instead of using the distance formula, you can sketch a right triangle for the slanted side PS. The horizontal leg has length 3 and the vertical leg has length 4. Recognizing this as a standard 3-4-5 right triangle, the hypotenuse (the slanted side) must be 5.

Estimated Time:1m 30s

Question 207Question →

On a map of a city, Oak Street is represented by a straight line with the equation $3x - 4y = 12$ . A new road, Pine Street, is planned to be perpendicular to Oak Street and will pass through a park located at the coordinates $(2, -1)$ . Which of the following is an equation that represents Pine Street?

3x + 4y = 2

4x - 3y = 11

4x + 3y = 5

4x + 3y = 7

4x + 3y = -9

Show answer & explanation

Answer: $4x + 3y = 5$

Answer

4x + 3y = 5

The slope of Oak Street is

\frac{3}{4}

. The perpendicular line representing Pine Street must have a slope that is the negative reciprocal of

\frac{3}{4}

, which is

-\frac{4}{3}

. Using the point-slope form with the point

(2, -1)

, we get

y - (-1) = -\frac{4}{3}(x - 2)

. Simplifying this yields

y + 1 = -\frac{4}{3}x + \frac{8}{3}

, which becomes

y = -\frac{4}{3}x + \frac{5}{3}

. Multiplying the entire equation by 3 and moving the

x

term to the left side gives the standard form equation

4x + 3y = 5

.

Step-by-Step Solution

1

Find the slope of Oak Street by converting its equation to slope-intercept form (

y = mx + b

).

The equation

3x - 4y = 12

becomes

-4y = -3x + 12

, which simplifies to

y = \frac{3}{4}x - 3

. The slope of Oak Street is

\frac{3}{4}

.

To find the slope of a perpendicular line, we must first determine the slope of the original line.

2

Determine the perpendicular slope of Pine Street by taking the negative reciprocal of Oak Street's slope.

The negative reciprocal of

\frac{3}{4}

is

-\frac{4}{3}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Use the point-slope formula

y - y_1 = m(x - x_1)

with the point

(2, -1)

and the perpendicular slope

-\frac{4}{3}

to find the equation of Pine Street.

Substituting the values gives

y - (-1) = -\frac{4}{3}(x - 2)

, which simplifies to

y + 1 = -\frac{4}{3}x + \frac{8}{3}

.

The point-slope formula allows us to write the equation of a line given its slope and a point it passes through.

4

Convert the equation to standard form (

Ax + By = C

).

Subtract 1 from both sides:

y = -\frac{4}{3}x + \frac{8}{3} - \frac{3}{3} \Rightarrow y = -\frac{4}{3}x + \frac{5}{3}

. Multiply the entire equation by 3:

3y = -4x + 5

. Add

4x

to both sides:

4x + 3y = 5

.

Converting to standard form matches the format of the options provided in the question.

Key Concept

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other (

m_1 \cdot m_2 = -1

). Once the perpendicular slope is found, the point-slope formula can be used to write the equation of the line passing through a given point.

Question 208Question →

A segment in the coordinate plane has one endpoint at $A(1, 9)$ and its midpoint at $M(5, 6)$ . Let $B$ represent the other endpoint of the segment. What is the distance between point $B$ and the point $C(3, -5)$ ?

14

2\sqrt{10}

10

5\sqrt{5}

\sqrt{5}

Show answer & explanation

Answer: $10$

Answer

The distance between point

B

and point

C

is

10

.

First, we find the coordinates of point

B(x, y)

using the midpoint formula. Since

M(5, 6)

is the midpoint between

A(1, 9)

and

B(x, y)

, we set up the equations:

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

which simplifies to

10 = 1+x

and

x = 9

; and

6 = \frac{9+y}{2}

which simplifies to

12 = 9+y

and

y = 3

. Thus, point

B

is

(9, 3)

. Next, we apply the distance formula to find the distance between

B(9, 3)

and

C(3, -5)

:

d = \sqrt{(9-3)^2 + (3-(-5))^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10

. This is the correct distance.

Step-by-Step Solution

1

Find the coordinates of endpoint

B(x_B, y_B)

using the midpoint formula.

x_B = 9

and

y_B = 3

, so point

B

is

(9, 3)

.

Since

M(5, 6)

is the midpoint of segment

AB

, we can solve for

B

's coordinates using

5 = \frac{1 + x_B}{2}

and

6 = \frac{9 + y_B}{2}

.

2

Calculate the distance between

B(9, 3)

and

C(3, -5)

using the distance formula.

d = 10

Applying the distance formula:

d = \sqrt{(9-3)^2 + (3-(-5))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10

.

Key Concept

Using the midpoint formula to find a missing endpoint, and then applying the distance formula between two coordinate points.

Question 209Question →

In the standard $(x, y)$ coordinate plane, square $PQRS$ has adjacent vertices at $P(1, 2)$ and $Q(4, 6)$ . The line containing the side $QR$ has a y-intercept of $b$ . What is the value of $b$ ?

Show answer & explanation

Answer: 9

Answer

The correct answer is 9.

The slope of segment

PQ

is

\frac{6 - 2}{4 - 1} = \frac{4}{3}

. Since adjacent sides of a square are perpendicular, the line containing

QR

is perpendicular to

PQ

and passes through

Q(4, 6)

. The slope of this perpendicular line is the negative reciprocal of

\frac{4}{3}

, which is

-\frac{3}{4}

. Using the slope-intercept form

y = mx + b

with point

Q(4, 6)

, we substitute the values:

6 = -\frac{3}{4}(4) + b \Rightarrow 6 = -3 + b \Rightarrow b = 9

.

Step-by-Step Solution

1

Calculate the slope of side

PQ

using the coordinates of

P(1, 2)

and

Q(4, 6)

.

The slope of

PQ

is

m_{PQ} = \frac{6 - 2}{4 - 1} = \frac{4}{3}

.

To determine the direction of side

PQ

so we can find the perpendicular slope for

QR

.

2

Find the slope of the line containing side

QR

.

The slope of

QR

is

m_{QR} = -\frac{3}{4}

.

Because adjacent sides of a square are perpendicular, the slope of

QR

is the negative reciprocal of the slope of

PQ

.

3

Find the equation of the line containing

QR

using the slope-intercept form and the coordinates of vertex

Q(4, 6)

.

Substituting the slope

m = -\frac{3}{4}

and point

(4, 6)

into

y = mx + b

gives

6 = -\frac{3}{4}(4) + b

, which simplifies to

6 = -3 + b

, so

b = 9

.

The line containing side

QR

must pass through vertex

Q

, which allows us to determine the y-intercept

b

.

Key Concept

The slopes of perpendicular lines are negative reciprocals of each other:

m_1 \cdot m_2 = -1

.

Question 210Question →

In the standard $(x, y)$ coordinate plane, a trapezoid has vertices at $A(-5, -4)$ , $B(7, -4)$ , $C(7, 1)$ , and $D(-5, 6)$ . What is the perimeter of this trapezoid?

Show answer & explanation

Answer: 40

Answer

The perimeter of the trapezoid is 40.

The perimeter of the trapezoid is found by summing the lengths of its four sides. Two sides are vertical (AD of length 10 and BC of length 5), one side is horizontal (AB of length 12), and the final side is skewed (CD of length 13, found using the distance formula). Summing these gives 10 + 5 + 12 + 13 = 40.

Step-by-Step Solution

1

Calculate the lengths of the vertical sides AD and BC

AD = 10, BC = 5

Since vertices A(-5, -4) and D(-5, 6) share the same x-coordinate of -5, the side AD is vertical. Its length is the difference in their y-coordinates: 6 - (-4) = 10. Similarly, vertices B(7, -4) and C(7, 1) share the same x-coordinate of 7, so the side BC is vertical. Its length is the difference in their y-coordinates: 1 - (-4) = 5.

2

Calculate the length of the horizontal side AB

AB = 12

Since vertices A(-5, -4) and B(7, -4) share the same y-coordinate of -4, the side AB is horizontal. Its length is the difference in their x-coordinates: 7 - (-5) = 12.

3

Calculate the length of the skewed side CD using the distance formula

CD = 13

Using the coordinates of C(7, 1) and D(-5, 6), the distance is calculated as: CD = \sqrt{(-5 - 7)^2 + (6 - 1)^2} = \sqrt{(-12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13.

4

Sum the lengths of all sides to find the perimeter

Perimeter = 40

The perimeter of a polygon is the sum of the lengths of all its sides: Perimeter = AB + BC + CD + DA = 12 + 5 + 13 + 10 = 40.

Key Concept

Calculating side lengths of a geometric figure on a coordinate plane to find its perimeter

Question 211Question →

On a coordinate map of an airport, Runway A is represented by the line $4x - 3y = 12$ . Runway B is designed to be perpendicular to Runway A, and its path crosses the $y$ -axis at $(0, 7)$ . If Runway B passes through a guidance beacon located at the coordinate $(p, 1)$ , what is the value of $p$ ?

-8

-\frac{9}{2}

-\frac{32}{3}

8

\frac{9}{2}

Show answer & explanation

Answer: $8$

Answer

8

The correct answer is

8

. First, find the slope of Runway A by rewriting

4x - 3y = 12

as

y = \frac{4}{3}x - 4

, which gives a slope of

\frac{4}{3}

. The slope of Runway B must be the negative reciprocal,

-\frac{3}{4}

. Using the given

y

-intercept of

7

, the equation for Runway B is

y = -\frac{3}{4}x + 7

. Substituting the point

(p, 1)

gives

1 = -\frac{3}{4}p + 7

. Subtracting

7

from both sides yields

-6 = -\frac{3}{4}p

. Multiplying both sides by

-\frac{4}{3}

gives

p = 8

.

Step-by-Step Solution

1

Convert the equation of Runway A,

4x - 3y = 12

, into slope-intercept form (

y = mx + b

) to find its slope.

The equation becomes

y = \frac{4}{3}x - 4

, which shows the slope of Runway A is

m_1 = \frac{4}{3}

.

Finding the slope of the first line is necessary to determine the slope of any perpendicular line.

2

Calculate the perpendicular slope (

m_2

) for Runway B by taking the negative reciprocal of the slope of Runway A (

m_1

).

The perpendicular slope is

m_2 = -\frac{1}{m_1} = -\frac{3}{4}

.

Perpendicular lines in a coordinate plane have slopes that are negative reciprocals of each other.

3

Write the equation of Runway B using its slope

m_2 = -\frac{3}{4}

and its

y

-intercept of

7

.

The equation of Runway B is

y = -\frac{3}{4}x + 7

.

The slope-intercept form of a line is

y = mx + b

, where

m

is the slope and

b

is the

y

-intercept.

4

Substitute the point

(p, 1)

into the equation of Runway B and solve for

p

.

1 = -\frac{3}{4}p + 7 \implies -6 = -\frac{3}{4}p \implies 24 = 3p \implies p = 8

.

Since the guidance beacon lies on Runway B, its coordinates must satisfy the equation of the line.

Key Concept

Perpendicular lines have slopes that are negative reciprocals of each other, meaning their product is

-1

. Once the perpendicular slope and y-intercept are known, the line's equation can be written and solved for a missing coordinate.

Estimated Time:1m 0s

Question 212Question →

In the standard $(x, y)$ coordinate plane, the line $L_1$ is parallel to the line $y = 3x - 7$ . The line $L_2$ is perpendicular to $L_1$ and passes through the point $(6, 5)$ . If the $y$ -intercept of $L_2$ is $(0, c)$ , what is the value of $c$ ?

Show answer & explanation

Answer: 7

Answer

The value of

c

is

7

.

Since the line

L_1

is parallel to

y = 3x - 7

, its slope is

3

. The line

L_2

is perpendicular to

L_1

, so its slope is the negative reciprocal of

3

, which is

-\frac{1}{3}

. Using the point-slope formula with the point

(6, 5)

yields the equation of

L_2

:

y - 5 = -\frac{1}{3}(x - 6)

, which simplifies to

y = -\frac{1}{3}x + 7

. The

y

-intercept of this line is

(0, 7)

, which gives

c = 7

.

Step-by-Step Solution

1

Determine the slope of line

L_1

.

Slope of

L_1

is

3

.

Parallel lines have equal slopes, and the given reference line

y = 3x - 7

has a slope of

3

.

2

Determine the slope of line

L_2

.

Slope of

L_2

is

-\frac{1}{3}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Find the equation of line

L_2

.

The equation is

y = -\frac{1}{3}x + 7

.

Use the point-slope form

y - y_1 = m(x - x_1)

with point

(6, 5)

and slope

-\frac{1}{3}

.

4

Identify the

y

-intercept constant

c

.

c = 7

.

The equation of

L_2

is in slope-intercept form

y = mx + b

, meaning the

y

-intercept is

(0, 7)

.

Key Concept

The relationship between the slopes of parallel lines (which are equal) and perpendicular lines (which are negative reciprocals).

Question 213Question →

On a coordinate map of a harbor, a lighthouse is located at $L(-4, 9)$ and a dock is located at $D(8, -7)$ . A buoy is positioned at the midpoint of the straight-line segment connecting the lighthouse and the dock. If a boat is anchored at $B(-1, 5)$ , what is the distance, in coordinate units, between the boat and the buoy?

Show answer & explanation

Answer: 5

Answer

The distance between the boat and the buoy is 5 units.

To find the distance between the boat and the buoy, first determine the location of the buoy. Since the buoy is at the midpoint of the segment connecting the lighthouse at

L(-4, 9)

and the dock at

D(8, -7)

, we use the midpoint formula:

M = \left(\frac{-4 + 8}{2}, \frac{9 + (-7)}{2}\right) = (2, 1)

. Next, find the distance between the boat at

B(-1, 5)

and the buoy at

M(2, 1)

using the distance formula:

d = \sqrt{(2 - (-1))^2 + (1 - 5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Find the coordinates of the buoy, which is the midpoint

M(x_m, y_m)

of the segment connecting the lighthouse

L(-4, 9)

and the dock

D(8, -7)

.

M(2, 1)

The midpoint formula is given by

x_m = \frac{x_1 + x_2}{2}

and

y_m = \frac{y_1 + y_2}{2}

. Substituting the coordinates of

L

and

D

, we get

x_m = \frac{-4 + 8}{2} = 2

and

y_m = \frac{9 + (-7)}{2} = 1

.

2

Calculate the distance

d

between the boat at

B(-1, 5)

and the buoy at

M(2, 1)

.

5

units

The distance formula is given by

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

. Substituting the coordinates of

B

and

M

yields

d = \sqrt{(2 - (-1))^2 + (1 - 5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

.

Key Concept

Applying the midpoint and distance formulas sequentially to solve a coordinate geometry word problem.

Question 214Question →

A water tank is being drained at a constant rate. After $3$ hours of draining, the tank contains $120$ gallons of water. After $5$ hours of draining, the tank contains $80$ gallons of water. If the volume of water in the tank, $y$ (in gallons), is modeled as a linear function of the time spent draining, $x$ (in hours), what is the slope of the line representing this function in the standard $(x, y)$ coordinate plane?

20

-\frac{1}{20}

-20

-11

53

Show answer & explanation

Answer: $-20$

Answer

-20

The correct answer is

-20

. The slope

m

of a line passing through two points

(x_1, y_1)

and

(x_2, y_2)

is given by the formula

m = \frac{y_2 - y_1}{x_2 - x_1}

. Identifying the points from the problem as

(3, 120)

and

(5, 80)

, we calculate the slope as

m = \frac{80 - 120}{5 - 3} = \frac{-40}{2} = -20

. This represents a constant rate of change of

-20

gallons per hour.

Step-by-Step Solution

1

Identify the coordinates of the two data points from the problem statement.

The two points on the line are

(3, 120)

and

(5, 80)

, where

x

represents the time in hours and

y

represents the volume of water in gallons.

To calculate the slope of a line, we first need to identify the coordinate points

(x_1, y_1)

and

(x_2, y_2)

that lie on the line.

2

Substitute the coordinates into the slope formula,

m = \frac{y_2 - y_1}{x_2 - x_1}

.

m = \frac{80 - 120}{5 - 3}

The slope represents the constant rate of change, which is the change in the dependent variable (

y

) divided by the change in the independent variable (

x

).

3

Simplify the expression to find the final value of the slope.

m = \frac{-40}{2} = -20

Performing the subtraction and division gives the slope of the line, showing that the volume of water decreases by

20

gallons per hour.

Key Concept

The slope of a line, representing a constant rate of change, is calculated using the formula

m = \frac{y_2 - y_1}{x_2 - x_1}

for any two points

(x_1, y_1)

and

(x_2, y_2)

on the line.

Estimated Time:1m 0s

Question 215Question →

In the standard $(x, y)$ coordinate plane, three vertices of a rhombus are $(1, 2)$ , $(5, 5)$ , and $(9, 2)$ . If the $y$ -coordinate of the fourth vertex is less than 2, what are the coordinates of the fourth vertex?

(5, -1)

(13, 5)

(13, -1)

(6, -2)

(-3, 5)

Show answer & explanation

Answer: $(5, -1)$

Answer

The coordinates of the fourth vertex are

(5, -1)

.

The correct answer is the coordinate pair

(5, -1)

. Since opposite sides of a rhombus are parallel and congruent, the vector from

(5, 5)

to

(1, 2)

is

(-4, -3)

. Applying this vector to

(9, 2)

gives

(9 - 4, 2 - 3) = (5, -1)

. This point satisfies the requirement that the

y

-coordinate is less than 2, and all four side lengths are exactly 5.

Step-by-Step Solution

1

Calculate the vector translation between two adjacent vertices of the rhombus.

The vector from vertex

(5, 5)

to vertex

(1, 2)

is

(1 - 5, 2 - 5) = (-4, -3)

.

In a rhombus, opposite sides must be parallel and equal in length, meaning the translation from one vertex to another on one side must equal the translation on the opposite side.

2

Apply the translation vector to the third vertex to find the fourth vertex.

Applying the vector

(-4, -3)

to the vertex

(9, 2)

yields

(9 - 4, 2 - 3) = (5, -1)

.

This determines the coordinates of the fourth vertex that completes the parallelogram structure.

3

Verify that the resulting vertex satisfies all conditions of the problem.

The

y

-coordinate of

(5, -1)

is

-1

, which is less than 2. The side lengths are all equal to 5:

\sqrt{(5-1)^2 + (-1-2)^2} = 5

and

\sqrt{(5-9)^2 + (-1-2)^2} = 5

.

This confirms that the figure is a rhombus and satisfies the

y

-coordinate constraint.

Key Concept

Using vector translations and distance formulas to determine coordinates of geometric figures on the coordinate plane.

Question 216Question →

In the standard $(x, y)$ coordinate plane, line $L_1$ passes through the points $(k, 5)$ and $(4, -1)$ . Line $L_2$ is perpendicular to line $L_1$ and is represented by the equation $3x - 2y = 6$ . What is the value of $k$ ?

-2

0

-5

8

13

Show answer & explanation

Answer: -5

Answer

-5

The correct answer is

-5

. First, find the slope of line

L_2

by rewriting the equation

3x - 2y = 6

in slope-intercept form:

y = \frac{3}{2}x - 3

, which gives a slope of

\frac{3}{2}

. Because line

L_1

is perpendicular to line

L_2

, its slope must be the negative reciprocal of

\frac{3}{2}

, which is

-\frac{2}{3}

. Using the slope formula with the points

(k, 5)

and

(4, -1)

gives

\frac{-1 - 5}{4 - k} = -\frac{2}{3}

. Solving this equation gives

\frac{-6}{4 - k} = -\frac{2}{3}

, which simplifies to

18 = 8 - 2k

, leading to

k = -5

.

Step-by-Step Solution

1

Convert the equation of line

L_2

to slope-intercept form to find its slope.

3x - 2y = 6 \implies -2y = -3x + 6 \implies y = \frac{3}{2}x - 3

. Thus, the slope of

L_2

is

\frac{3}{2}

.

Converting to slope-intercept form (

y = mx + b

) isolates the slope as the coefficient of

x

.

2

Find the slope of line

L_1

using the relationship between perpendicular lines.

The slope of

L_1

is the negative reciprocal of

\frac{3}{2}

, which is

-\frac{2}{3}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

3

Use the slope formula with the points

(k, 5)

and

(4, -1)

to solve for

k

.

\frac{-1 - 5}{4 - k} = -\frac{2}{3} \implies \frac{-6}{4 - k} = -\frac{2}{3} \implies 18 = 2(4 - k) \implies 18 = 8 - 2k \implies 2k = -10 \implies k = -5

.

Setting the calculated slope of

L_1

equal to the slope formula expression allows us to solve for the missing coordinate parameter.

Key Concept

Finding a coordinate parameter by using the negative reciprocal relationship between the slopes of perpendicular lines.

Question 217Question →

In the standard $(x, y)$ coordinate plane, if a non-horizontal and non-vertical line has a slope of $m$ , then reflecting this line across the line $y = x$ produces a line with a slope of $\frac{1}{m}$ .

True

False

Show answer & explanation

Answer: True

Answer

True

Reflecting a point

(x, y)

across the line

y = x

swaps the coordinates to

(y, x)

. For any two points

(x_1, y_1)

and

(x_2, y_2)

on the original line, the slope is

m = \frac{y_2 - y_1}{x_2 - x_1}

. The corresponding points on the reflected line are

(y_1, x_1)

and

(y_2, x_2)

, yielding a slope of

m' = \frac{x_2 - x_1}{y_2 - y_1} = \frac{1}{m}

. Thus, the statement is true.

Step-by-Step Solution

1

Identify the coordinate transformation representing reflection across the line

y = x

.

Reflecting any point

(x, y)

across the line

y = x

yields the point

(y, x)

.

By definition of reflection across the identity line, the

x

-coordinate and

y

-coordinate of each point are swapped.

2

Set up the slope formula for the original line using two distinct points

(x_1, y_1)

and

(x_2, y_2)

.

The slope of the original line is

m = \frac{y_2 - y_1}{x_2 - x_1}

.

The slope of a line is the ratio of vertical change (rise) to horizontal change (run).

3

Determine the slope of the reflected line using the transformed points

(y_1, x_1)

and

(y_2, x_2)

.

The slope of the reflected line is

m' = \frac{x_2 - x_1}{y_2 - y_1}

.

Applying the slope formula to the reflected points swaps the numerator and denominator.

4

Relate the slope of the reflected line,

m'

, to the slope of the original line,

m

.

Since

m' = \frac{x_2 - x_1}{y_2 - y_1} = \frac{1}{\frac{y_2 - y_1}{x_2 - x_1}} = \frac{1}{m}

, the slope of the reflected line is

\frac{1}{m}

.

Taking the reciprocal of the original slope fraction yields the reflected slope fraction, which is valid since the line is non-horizontal (

m \neq 0

) and non-vertical (

m

is defined).

Key Concept

Reflection of a line across

y = x

swaps the rise and run of the line, inverting its slope.

Question 218Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices at $A(-3, -2)$ , $B(5, -2)$ , and $C(x, y)$ . If the area of the triangle is $24$ square units, which of the following could be the coordinates of vertex $C$ ?

(2, 1)

(2, 4)

(2, 6)

(2, 8)

(2, 10)

Show answer & explanation

Answer: $(2, 4)$

Answer

(2, 4)

The correct answer is the coordinate

(2, 4)

. The base of the triangle is the segment

AB

, which is horizontal since both vertices lie on the line

y = -2

. The length of this base is

5 - (-3) = 8

units. Using the formula for the area of a triangle,

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

, we have

24 = \frac{1}{2} \times 8 \times \text{height}

, which simplifies to

24 = 4 \times \text{height}

, giving a height of

6

units. Since the base lies along

y = -2

, the

y

-coordinate of the third vertex must be

6

units away from

-2

(either at

-2 + 6 = 4

or

-2 - 6 = -8

). The point

(2, 4)

satisfies this requirement.

Step-by-Step Solution

1

Calculate the length of the base of the triangle.

The base segment

AB

is horizontal because both

A(-3, -2)

and

B(5, -2)

have the same

y

-coordinate of

-2

. The length of the base is the difference in their

x

-coordinates:

5 - (-3) = 8

units.

To use the area formula of a triangle, we first need to determine the length of one of its sides to act as the base.

2

Use the area formula of a triangle to find its height.

The formula for the area of a triangle is

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

. Substituting the given area of

24

and the base length of

8

:

24 = \frac{1}{2} \times 8 \times \text{height} \Rightarrow 24 = 4 \times \text{height} \Rightarrow \text{height} = 6

units.

Knowing the area and the base allows us to find the vertical distance (height) from the base to the third vertex.

3

Determine the possible

y

-coordinates of vertex

C

.

Since the base lies on the horizontal line

y = -2

, the

y

-coordinate of vertex

C

must be exactly

6

units away from

-2

vertically. This means

y = -2 + 6 = 4

or

y = -2 - 6 = -8

. Thus,

C

can be any point with a

y

-coordinate of

4

or

-8

.

The height represents the vertical distance from the horizontal base line to the third vertex.

4

Match the calculated

y

-coordinates with the given options.

Among the options, only the coordinate

(2, 4)

has a

y

-coordinate of

4

.

We identify the option that provides a valid set of coordinates for vertex

C

.

Key Concept

Calculating coordinates of a geometric figure's vertex using the area formula and coordinate distances.

Question 219Question →

In the standard $(x, y)$ coordinate plane, a square has two opposite vertices at $(1, -3)$ and $(7, 5)$ . What is the area of the square, in square units?

Show answer & explanation

Answer: 50

Answer

The area of the square is 50 square units.

The length of the diagonal of the square is found using the distance formula between the two opposite vertices:

d = \sqrt{(7 - 1)^2 + (5 - (-3))^2} = \sqrt{6^2 + 8^2} = 10

. The area of a square can be calculated using its diagonal length

d

with the formula

\text{Area} = \frac{d^2}{2}

. Substituting

d = 10

gives

\text{Area} = \frac{10^2}{2} = 50

square units.

Step-by-Step Solution

1

Calculate the length of the diagonal of the square using the distance formula.

The diagonal length is

10

.

The distance between opposite vertices of a square represents the length of its diagonal.

2

Determine the area of the square using the diagonal length.

The area is

50

.

The area of a square with diagonal

d

is given by

\frac{d^2}{2}

.

Key Concept

Finding the area of a square on the coordinate plane using its diagonal.

Estimated Time:1m 30s

Question 220Question →

Two lines in a coordinate plane, $T_1$ and $T_2$ , are perpendicular to each other. Line $T_1$ has the equation $3x - 5y = 15$ , and line $T_2$ has the equation $ax + 9y = 20$ , where $a$ is a constant. What is the value of $a$ ?

Show answer & explanation

Answer: 15

Answer

The value of the constant

a

is 15.

The correct answer is 15. The slope of the line

3x - 5y = 15

is

\frac{3}{5}

. Since the two lines are perpendicular, the slope of the second line must be the negative reciprocal of

\frac{3}{5}

, which is

-\frac{5}{3}

. The slope of the line

ax + 9y = 20

is

-\frac{a}{9}

. Equating the two slopes yields

-\frac{a}{9} = -\frac{5}{3}

, which simplifies to

a = 15

.

Step-by-Step Solution

1

Find the slope of line

T_1

The slope of line

T_1

is

\frac{3}{5}

.

Writing the equation

3x - 5y = 15

in slope-intercept form

y = \frac{3}{5}x - 3

isolates the slope coefficient.

2

Find the perpendicular slope

The perpendicular slope is

-\frac{5}{3}

.

Perpendicular lines have slopes that are negative reciprocals of each other, so we invert the fraction and change the sign.

3

Determine the slope of line

T_2

in terms of

a

The slope of line

T_2

is

-\frac{a}{9}

.

Rewriting the equation

ax + 9y = 20

in slope-intercept form

y = -\frac{a}{9}x + \frac{20}{9}

isolates the slope coefficient.

4

Solve for

a

a = 15

Setting the slope of

T_2

equal to the perpendicular slope yields the equation

-\frac{a}{9} = -\frac{5}{3}

, which simplifies to

a = 15

.

Key Concept

The slopes of perpendicular lines are negative reciprocals of each other.

Coordinate Geometry

Answer

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