Coordinate Geometry Practice Questions — ACT — Page 6

Question 101Question →

A water tank is being drained at a constant rate. After $2$ hours of draining, the tank contains $140$ gallons of water, and after $5$ hours, it contains $80$ gallons. When the relationship between the time spent draining, $t$ , and the volume of water, $v$ , is plotted on a standard coordinate plane with $t$ on the horizontal axis, what is the slope of the resulting line?

-60

-20

-\frac{1}{20}

20

60

Show answer & explanation

Answer: $-20$

Answer

The slope of the line is

-20

.

The correct answer is

-20

. The slope represents the rate of change of the volume of water with respect to time. Using the points

(2, 140)

and

(5, 80)

, the slope is calculated as the change in volume divided by the change in time:

\frac{80 - 140}{5 - 2} = \frac{-60}{3} = -20

.

Step-by-Step Solution

1

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

The two points are

(2, 140)

and

(5, 80)

, where time

t

represents the horizontal coordinate and volume

v

represents the vertical coordinate.

The problem defines the horizontal axis as representing time

t

and the vertical axis as representing volume

v

.

2

Apply the slope formula

m = \frac{v_2 - v_1}{t_2 - t_1}

using the coordinates from Step 1.

m = \frac{80 - 140}{5 - 2} = \frac{-60}{3} = -20

.

Calculating the ratio of change in the vertical axis to the change in the horizontal axis gives the slope of the line.

Key Concept

Slope as a constant rate of change in a linear relationship

Estimated Time:1m 0s

Question 102Question →

In the standard $(x, y)$ coordinate plane, line $L$ has a $y$ -intercept of $4$ . When line $L$ is translated horizontally to the right by $3$ units, the $y$ -intercept of the resulting line is $6$ units less than the $y$ -intercept of line $L$ . What is the $x$ -intercept of line $L$ ?

2

-8

-2

-\frac{1}{2}

-6

Show answer & explanation

Answer: $-2$

Answer

The

x

-intercept of line

L

is

-2

.

The original line

L

can be written as

y = mx + 4

. Translating the line horizontally to the right by

3

units replaces

x

with

x - 3

, resulting in the equation

y = m(x - 3) + 4 = mx - 3m + 4

. The

y

-intercept of this new line is

4 - 3m

. We are given that this new

y

-intercept is

6

units less than the original

y

-intercept of

4

, which gives the equation

4 - 3m = 4 - 6

. Solving for

m

yields

-3m = -6 \implies m = 2

. With a slope of

2

and a

y

-intercept of

4

, the equation of the original line

L

is

y = 2x + 4

. To find the

x

-intercept, we set

y = 0

, which gives

2x + 4 = 0 \implies x = -2

.

Step-by-Step Solution

1

Write the general equation of line

L

in slope-intercept form.

y = mx + 4

, where

m

is the slope of line

L

.

The

y

-intercept of line

L

is given as

4

, which represents the constant term

b

in

y = mx + b

.

2

Apply the horizontal translation of

3

units to the right.

y = m(x - 3) + 4

, which simplifies to

y = mx - 3m + 4

.

Translating a function horizontally to the right by

h

units replaces

x

with

x - h

.

3

Determine the slope

m

using the change in

y

-intercept.

The new

y

-intercept is

4 - 3m

. Since this is

6

units less than the original

y

-intercept of

4

, we set up the equation

4 - 3m = 4 - 6 \implies -3m = -6 \implies m = 2

.

We relate the algebraic expression for the new

y

-intercept to the given numerical relationship.

4

Find the

x

-intercept of line

L

by setting

y = 0

.

2x + 4 = 0 \implies 2x = -4 \implies x = -2

.

The

x

-intercept is the value of

x

when the line crosses the

x

-axis, meaning

y = 0

.

Key Concept

Linear Equations and Graphing

Question 103Question →

A circle in the first quadrant of the standard $(x,y)$ coordinate plane is tangent to the $x$ -axis and is also tangent to the line $y = \frac{4}{3}x$ . If the center of the circle lies on the line with equation $y = 3x - 10$ , what is the radius of the circle?

Show answer & explanation

Answer: 2

Answer

The radius of the circle is 2.

The correct answer is obtained by determining that the center of a circle tangent to the

x

-axis in the first quadrant has the form

(h, R)

where

R

is the radius. Using the distance from this point to the line

4x - 3y = 0

gives

h = 2R

to ensure the center remains in the first quadrant. Substituting

(2R, R)

into the line

y = 3x - 10

yields

R = 3(2R) - 10

, which solves to

R = 2

.

Step-by-Step Solution

1

Determine the relation between the circle's center coordinates and its radius.

The center is

(h, R)

where

R

is the radius.

Because the circle is tangent to the

x

-axis and lies in the first quadrant, the y-coordinate of its center must equal its radius.

2

Apply the point-to-line distance formula from the center to the line

y = \frac{4}{3}x

.

The relation is

|4h - 3R| = 5R

.

The distance from the center

(h, R)

to the line

4x - 3y = 0

must equal the radius

R

.

3

Solve the absolute value equation for

h

in terms of

R

.

Since

h > 0

, we find

h = 2R

.

The positive case

4h - 3R = 5R

gives

h = 2R

, whereas the negative case

4h - 3R = -5R

gives a negative

h

which violates the first-quadrant condition.

4

Substitute the center coordinates

(2R, R)

into the line

y = 3x - 10

and solve for

R

.

R = 2

.

The center lies on this line, so its coordinates must satisfy the line's equation.

Key Concept

Equations and Graphs of Circles

Question 104Question →

In the standard $(x,y)$ coordinate plane, a rectangle has vertices at $(-2, -3)$ , $(4, -3)$ , $(4, 2)$ , and $(-2, 2)$ . What is the area, in square units, of this rectangle?

Show answer & explanation

Answer: 30

Answer

The area of the rectangle is 30 square units.

The area of a rectangle is the product of its length and width. By finding the difference between the x-coordinates of the horizontal vertices (

4 - (-2) = 6

) and the difference between the y-coordinates of the vertical vertices (

2 - (-3) = 5

), we find the dimensions to be 6 and 5. Multiplying these gives

6 \times 5 = 30

.

Step-by-Step Solution

1

Determine the length of the horizontal sides of the rectangle.

The horizontal sides have a length of 6 units.

The horizontal sides connect vertices with the same y-coordinates, such as

(-2, -3)

and

(4, -3)

. The distance is the difference in their x-coordinates:

4 - (-2) = 6

.

2

Determine the length of the vertical sides of the rectangle.

The vertical sides have a length of 5 units.

The vertical sides connect vertices with the same x-coordinates, such as

(4, -3)

and

(4, 2)

. The distance is the difference in their y-coordinates:

2 - (-3) = 5

.

3

Calculate the area of the rectangle.

The area of the rectangle is 30 square units.

The area of a rectangle is found by multiplying its length by its width:

\text{Area} = 6 \times 5 = 30

.

Key Concept

Finding the area of a rectangle on the coordinate plane by calculating the lengths of its horizontal and vertical sides.

Question 105Question →

On a coordinate grid, line $d$ is represented by the equation $y = -\frac{3}{4}x + 5$ . Line $e$ is perpendicular to line $d$ and passes through the point $(1, 2)$ . Which of the following is the equation of line $e$ ?

y = -\frac{4}{3}x + \frac{10}{3}

y = -\frac{3}{4}x + \frac{11}{4}

y = \frac{4}{3}x + \frac{2}{3}

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

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

Show answer & explanation

Answer: $y = \frac{4}{3}x + \frac{2}{3}$

Answer

The equation of line

e

is

y = \frac{4}{3}x + \frac{2}{3}

.

The correct equation has a slope of

\frac{4}{3}

and a y-intercept of

\frac{2}{3}

. The slope of the given line is

-\frac{3}{4}

, meaning any line perpendicular to it must have a slope that is the negative reciprocal, which is

\frac{4}{3}

. Using the slope-intercept form

y = mx + b

with the point

(1, 2)

allows us to solve for

b

by calculating

2 = \frac{4}{3}(1) + b

, which simplifies to

b = 2 - \frac{4}{3} = \frac{2}{3}

. Writing this together in slope-intercept form yields

y = \frac{4}{3}x + \frac{2}{3}

.

Step-by-Step Solution

1

Identify the slope of the given line

d

.

The slope of line

d

is

-\frac{3}{4}

.

The equation

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

is in slope-intercept form (

y = mx + b

), where the coefficient of

x

represents the slope.

2

Determine the slope of the perpendicular line

e

.

The slope of line

e

is

\frac{4}{3}

.

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

-\frac{3}{4}

is

\frac{4}{3}

.

3

Substitute the perpendicular slope and the given point

(1, 2)

into the slope-intercept equation to solve for the y-intercept

b

.

b = \frac{2}{3}

Plugging the values into

y = mx + b

gives

2 = \frac{4}{3}(1) + b

. Solving for

b

requires subtracting

\frac{4}{3}

from

2

, which yields

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

.

4

Write the final equation of line

e

in slope-intercept form.

y = \frac{4}{3}x + \frac{2}{3}

Substituting the slope

m = \frac{4}{3}

and y-intercept

b = \frac{2}{3}

into the standard slope-intercept form equation.

Key Concept

Finding the equation of a line perpendicular to a given line through a given point using negative reciprocal slopes.

Estimated Time:1m 0s

Question 106Question →

A straight hiking trail ascends a hill at a constant incline. On a coordinate grid where the units represent meters, the path of the trail is a straight line starting at the coordinate point $(10, 150)$ and ending at the coordinate point $(90, 190)$ . What is the slope of this trail?

Show answer & explanation

Answer: 0.5

Answer

The correct answer is

0.5

(or

\frac{1}{2}

).

The slope of a line is defined as the change in the

y

-coordinates divided by the change in the

x

-coordinates:

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

. By substituting the coordinates of the start of the trail

(10, 150)

and the end of the trail

(90, 190)

, we calculate

\frac{190 - 150}{90 - 10} = \frac{40}{80} = 0.5

.

Step-by-Step Solution

1

Identify the coordinates from the problem statement.

(x_1, y_1) = (10, 150)

and

(x_2, y_2) = (90, 190)

To calculate the slope between two points, we first need to define their coordinates.

2

Recall the formula for the slope of a line passing through two points.

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

Slope represents the vertical change (rise) divided by the horizontal change (run).

3

Substitute the coordinate values into the slope formula and simplify.

m = \frac{190 - 150}{90 - 10} = \frac{40}{80} = 0.5

Plugging the values into the formula yields the constant rate of change (slope) of the trail.

Key Concept

Slope of a Line

Question 107Question →

A triangle in the standard $(x, y)$ coordinate plane has vertices at $A(1, 1)$ , $B(10, 16)$ , and $C(5, 9)$ . A line passes through the point $P(4, 7)$ on the side $AC$ and intersects the side $AB$ at a point $Q$ . If this line divides the triangle into two regions of equal area, what is the length of the line segment $PQ$ ?

Show answer & explanation

Answer: 5

Answer

5

The total area of the triangle is 6. To divide the triangle into two equal-area regions, each region must have an area of 3. Since point

P(4,7)

lies

\frac{3}{4}

of the way from

A

to

C

, the area of the sub-triangle

\triangle PBC

is only

\frac{1}{4} \times 6 = 1.5

. Thus, the dividing line must intersect side

AB

at a point

Q

to form

\triangle APQ

with an area of 3. Using the area ratio formula,

\text{Area}(\triangle APQ) = \frac{AP}{AC} \times \frac{AQ}{AB} \times \text{Area}(\triangle ABC) \implies 3 = \frac{3}{4} \times \frac{AQ}{AB} \times 6

, which gives

\frac{AQ}{AB} = \frac{2}{3}

. Using the section formula, the coordinates of

Q

are

A + \frac{2}{3}(B - A) = (1, 1) + \frac{2}{3}(9, 15) = (7, 11)

. Finally, the length of

PQ

is

\sqrt{(7-4)^2 + (11-7)^2} = \sqrt{3^2 + 4^2} = 5

.

Step-by-Step Solution

1

Calculate the area of the entire triangle

ABC

using the Shoelace formula.

Area of

\triangle ABC = 6

.

Establishing the total area of the triangle is necessary to determine the target area of 3 for each of the two equal-area regions.

2

Determine which side of the triangle the dividing line intersects by comparing the area of

\triangle PBC

to the target area of 3.

The line must intersect side

AB

at a point

Q

.

Since

P(4,7)

lies

\frac{3}{4}

of the way along

AC

, the base

PC

is

\frac{1}{4}

of

AC

. The area of

\triangle PBC

is

\frac{1}{4} \times 6 = 1.5

. Since this is less than 3, the dividing line cannot intersect side

BC

and must intersect side

AB

instead.

3

Set up the area ratio equation for

\triangle APQ

to find the ratio

\frac{AQ}{AB}

.

\frac{AQ}{AB} = \frac{2}{3}

.

The area of

\triangle APQ

is given by

\text{Area}(\triangle APQ) = \frac{AP}{AC} \times \frac{AQ}{AB} \times \text{Area}(\triangle ABC) \implies 3 = \frac{3}{4} \times \frac{AQ}{AB} \times 6 \implies \frac{AQ}{AB} = \frac{2}{3}

.

4

Find the coordinates of

Q

using the section formula along segment

AB

from

A(1,1)

to

B(10,16)

.

Q(7, 11)

.

Applying

Q = A + \frac{2}{3}(B - A) = (1, 1) + \frac{2}{3}(9, 15) = (7, 11)

yields the exact coordinates of

Q

.

5

Calculate the length of segment

PQ

using the distance formula between

P(4,7)

and

Q(7,11)

.

PQ = 5

.

The question asks for the length of the segment

PQ

, which is the distance between these two points.

Key Concept

Using coordinate geometry formulas and area ratios to solve problems involving geometric figures on the coordinate plane.

Question 108Question →

In the standard $(x, y)$ coordinate plane, the point $M$ is the midpoint of the line segment with endpoints $P(-4, -2)$ and $Q(4, 2)$ . A second line segment is drawn from $M$ to a point $R(x, y)$ such that the length of the segment $MR$ is $8$ . If the midpoint of the segment $MR$ lies on the line $3x - 4y + 12 = 0$ , what is the smallest possible value of $x$ ?

Show answer & explanation

Answer: -8

Answer

The smallest possible value of

x

is

-8

.

By finding the midpoint

M(0,0)

of

PQ

and writing the midpoint of

MR

as

\left(\frac{x}{2}, \frac{y}{2}\right)

, we substitute this into the line equation to find

y = \frac{3}{4}x + 6

. We then substitute this into the distance formula equation

x^2 + y^2 = 64

to get the quadratic equation

25x^2 + 144x - 448 = 0

, which yields the solutions

x = -8

and

x = 2.24

. The smallest possible value is

-8

.

Step-by-Step Solution

1

Calculate the coordinates of the midpoint

M

of segment

PQ

.

M = (0, 0)

The midpoint formula states that the midpoint of a segment with endpoints

(x_1, y_1)

and

(x_2, y_2)

is

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

.

2

Express the midpoint

N

of segment

MR

in terms of

R(x, y)

.

N = \left(\frac{x}{2}, \frac{y}{2}\right)

The midpoint of

M(0, 0)

and

R(x, y)

is found by averaging their coordinates.

3

Substitute the coordinates of

N

into the equation of the line

3x - 4y + 12 = 0

.

y = \frac{3}{4}x + 6

Since the midpoint

N

lies on the line, its coordinates must satisfy the line's equation, which gives a linear relationship between

x

and

y

.

4

Set up the equation for the distance

MR = 8

.

x^2 + y^2 = 64

The distance formula between

M(0, 0)

and

R(x, y)

is

d = \sqrt{x^2 + y^2}

, and squaring both sides gives

x^2 + y^2 = d^2

.

5

Substitute

y = \frac{3}{4}x + 6

into the distance equation and solve the quadratic equation.

x = -8

and

x = 2.24

Substituting the linear relationship into the quadratic circle equation gives a single quadratic equation in terms of

x

, which can be solved using the quadratic formula.

6

Determine the smallest value of

x

from the two possible solutions.

-8

Comparing the two solutions,

-8

is smaller than

2.24

.

Key Concept

Distance and Midpoint Formulas

Alternative Method

Instead of solving algebraically, one can scale the line

3x - 4y + 12 = 0

by a factor of 2 centered at the origin

M(0,0)

to directly obtain the line equation on which

R

lies:

3x - 4y + 24 = 0

. Then, find the intersection of this line with the circle

x^2 + y^2 = 64

using substitution.

Estimated Time:2m 30s

Question 109Question →

In the standard $(x, y)$ coordinate plane, the points $(k, -3)$ , $(1, k)$ , and $(4, 9)$ lie on the same straight line, where $k$ is a constant. Which of the following is a possible value for the slope of this line?

-\frac{3}{4}

\frac{4}{3}

-\frac{4}{3}

4

\frac{22}{3}

Show answer & explanation

Answer: $-\frac{4}{3}$

Answer

The correct answer is

-\frac{4}{3}

.

The correct answer is

-\frac{4}{3}

. If three points are collinear, the slope computed between any two pairs must be equal. Setting the slope between

(k, -3)

and

(1, k)

equal to the slope between

(1, k)

and

(4, 9)

gives the equation

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

. Cross-multiplying yields

3(k + 3) = (9 - k)(1 - k)

, which simplifies to

k^2 - 13k = 0

. Solving this gives

k = 0

or

k = 13

. Substituting

k = 13

back into the coordinates gives the points

(13, -3)

,

(1, 13)

, and

(4, 9)

. The slope of the line passing through these points is

\frac{9 - 13}{4 - 1} = -\frac{4}{3}

.

Step-by-Step Solution

1

Set up the collinearity condition using the slope formula

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

.

The slope of the line segment between

(k, -3)

and

(1, k)

must equal the slope of the line segment between

(1, k)

and

(4, 9)

.

Since all three points lie on the same straight line, the slope between any two pairs of points must be equal.

2

Write the algebraic expressions for the slopes and set them equal to each other.

\frac{k - (-3)}{1 - k} = \frac{9 - k}{4 - 1} \Rightarrow \frac{k + 3}{1 - k} = \frac{9 - k}{3}

This establishes a rational equation containing the variable

k

.

3

Cross-multiply to eliminate the fractions and simplify.

3(k + 3) = (9 - k)(1 - k) \Rightarrow 3k + 9 = 9 - 10k + k^2

Cross-multiplying allows us to convert the rational equation into a polynomial equation.

4

Rearrange the terms into standard quadratic form and solve for

k

.

k^2 - 13k = 0 \Rightarrow k(k - 13) = 0 \Rightarrow k = 0 \text{ or } k = 13

Setting the quadratic expression to zero allows us to find the two possible values for the constant

k

by factoring.

5

Calculate the slope of the line for each possible value of

k

.

If

k = 0

, the slope is

m = \frac{9 - 0}{4 - 1} = 3

. If

k = 13

, the slope is

m = \frac{9 - 13}{4 - 1} = -\frac{4}{3}

.

We must check which of the calculated slopes matches one of the given choices.

Key Concept

Finding the slope of a line and using the condition of collinearity to solve for missing coordinates.

Estimated Time:2m 0s

Question 110Question →

A rectangle has vertices at $A(-1, -2)$ , $B(3, -2)$ , $C(3, 1)$ , and $D(-1, 1)$ in the standard $(x,y)$ coordinate plane. What is the perimeter of this rectangle?

7

10

11

12

14

Show answer & explanation

Answer: 14

Answer

14

The horizontal length of the rectangle is

3 - (-1) = 4

units, and the vertical length is

1 - (-2) = 3

units. Using the formula for the perimeter of a rectangle,

P = 2(\text{width} + \text{height})

, we get

P = 2(4 + 3) = 14

units.

Step-by-Step Solution

1

Calculate the horizontal length (width) of the rectangle.

width = 4

Since vertices

A(-1, -2)

and

B(3, -2)

share the same y-coordinate, the horizontal distance between them is the difference in their x-coordinates:

3 - (-1) = 4

.

2

Calculate the vertical length (height) of the rectangle.

height = 3

Since vertices

B(3, -2)

and

C(3, 1)

share the same x-coordinate, the vertical distance between them is the difference in their y-coordinates:

1 - (-2) = 3

.

3

Use the perimeter formula for a rectangle to find the total perimeter.

perimeter = 14

The perimeter is given by

P = 2(\text{width} + \text{height}) = 2(4 + 3) = 2(7) = 14

.

Key Concept

Perimeter of geometric figures on the coordinate plane

Estimated Time:45s

Question 111Question →

A graphic designer uses a coordinate plane to design a logo containing a triangular shape. The final image of the triangle has vertices at $(3, -1)$ , $(5, -3)$ , and $(2, -5)$ . The designer created this final image by performing a $90^\circ$ counterclockwise rotation of the original triangle about the origin, followed by a translation of $4$ units to the right and $3$ units down. What are the coordinates of the vertices of the original triangle?

(2, 1)

,

(0, -1)

, and

(-2, 2)

(-2, -1)

,

(0, 1)

, and

(2, -2)

(-4, -7)

,

(-6, -9)

, and

(-8, -6)

(5, 0)

,

(7, 2)

, and

(9, -1)

(-1, -2)

,

(1, 0)

, and

(-2, 2)

Show answer & explanation

Answer: $(2, 1)$ , $(0, -1)$ , and $(-2, 2)$

Answer

The coordinates of the vertices of the original triangle are

(2, 1)

,

(0, -1)

, and

(-2, 2)

.

The correct answer is obtained by working backward from the final image. First, undo the translation of

4

units right and

3

units down by translating the final vertices

4

units left and

3

units up. This yields intermediate vertices

(-1, 2)

,

(1, 0)

, and

(-2, -2)

. Second, undo the

90^\circ

counterclockwise rotation by rotating these intermediate vertices

90^\circ

clockwise about the origin. The algebraic rule for a

90^\circ

clockwise rotation is

(x, y) \rightarrow (y, -x)

. Applying this rule to

(-1, 2)

,

(1, 0)

, and

(-2, -2)

gives the original coordinates

(2, 1)

,

(0, -1)

, and

(-2, 2)

.

Step-by-Step Solution

1

Identify the two transformations in reverse order and define their inverse operations.

The final transformation was a translation of

4

units to the right and

3

units down, so the first step in working backward is a translation of

4

units to the left and

3

units up. The initial transformation was a

90^\circ

counterclockwise rotation about the origin, so the second step in working backward is a

90^\circ

clockwise rotation about the origin.

To find the pre-image, we must apply the inverse of each transformation in the reverse order of how they were originally applied.

2

Apply the inverse translation of

4

units left and

3

units up to the final image vertices:

(3, -1)

,

(5, -3)

, and

(2, -5)

.

The intermediate vertices are:

A'(3, -1) \rightarrow A_1(3 - 4, -1 + 3) = (-1, 2)

;

B'(5, -3) \rightarrow B_1(5 - 4, -3 + 3) = (1, 0)

;

C'(2, -5) \rightarrow C_1(2 - 4, -5 + 3) = (-2, -2)

.

Undoing a translation of

(+4, -3)

requires subtracting

4

from the

x

-coordinates and adding

3

to the

y

-coordinates.

3

Apply the inverse rotation, a

90^\circ

clockwise rotation about the origin, to the intermediate vertices.

The rule for a

90^\circ

clockwise rotation about the origin is

(x_1, y_1) \rightarrow (y_1, -x_1)

. Applying this to the intermediate vertices yields:

A_1(-1, 2) \rightarrow A(2, 1)

;

B_1(1, 0) \rightarrow B(0, -1)

;

C_1(-2, -2) \rightarrow C(-2, 2)

.

A

90^\circ

clockwise rotation is the inverse of a

90^\circ

counterclockwise rotation, which reverses the coordinates and changes the sign of the new

y

-coordinate.

Key Concept

Finding the pre-image of a figure under composite transformations in the coordinate plane by applying inverse transformations in reverse order.

Question 112Question →

In the standard $(x, y)$ coordinate plane, triangle $PQR$ has vertices $P(2, 1)$ , $Q(5, 1)$ , and $R(2, 5)$ . The triangle is first rotated $90^\circ$ counterclockwise about the origin to form triangle $P'Q'R'$ . Next, triangle $P'Q'R'$ is reflected across the $y$ -axis to form triangle $P''Q''R''$ . What are the coordinates of the vertex $R''$ ?

(5, 2)

(-5, 2)

(5, -2)

(-5, -2)

(2, -5)

Show answer & explanation

Answer: $(5, 2)$

Answer

(5, 2)

The correct answer is the coordinate pair

(5, 2)

. First, rotating the point

R(2, 5)

counterclockwise by

90^\circ

about the origin uses the transformation rule

(x, y) \rightarrow (-y, x)

, which maps

R(2, 5)

to

R'(-5, 2)

. Next, reflecting the point

R'(-5, 2)

across the

y

-axis uses the transformation rule

(x, y) \rightarrow (-x, y)

, which maps

R'(-5, 2)

to

R''(5, 2)

.

Step-by-Step Solution

1

Apply the rotation of

90^\circ

counterclockwise about the origin to the vertex

R(2, 5)

.

The rule for a

90^\circ

counterclockwise rotation is

(x, y) \rightarrow (-y, x)

. Applying this to

R(2, 5)

yields

R'(-5, 2)

.

To find the coordinates after the first transformation step.

2

Apply the reflection across the

y

-axis to the intermediate point

R'(-5, 2)

.

The rule for reflection across the

y

-axis is

(x, y) \rightarrow (-x, y)

. Applying this to

R'(-5, 2)

yields

R''(5, 2)

.

To find the final coordinates after the second transformation step.

Key Concept

Applying composite transformations (rotation followed by reflection) to coordinates in the coordinate plane.

Question 113Question →

A line has a slope of $-3$ . If the sum of its $x$ -intercept and its $y$ -intercept is $12$ , what is the $y$ -intercept of the line?

Show answer & explanation

Answer: 9

Answer

The y-intercept of the line is 9.

The equation of the line is

y = -3x + b

, where

b

is the

y

-intercept. The

x

-intercept is found by setting

y=0

, which yields

x = \frac{b}{3}

. Setting their sum to

12

gives

\frac{b}{3} + b = 12

, which simplifies to

4b = 36

and results in

b = 9

.

Step-by-Step Solution

1

Express the equation of the line using the slope-intercept form.

The equation of the line is

y = -3x + b

, where

b

is the

y

-intercept.

We are given that the slope of the line is

-3

.

2

Find the

x

-intercept of the line in terms of

b

.

Setting

y = 0

gives

0 = -3x + b \implies 3x = b \implies x = \frac{b}{3}

. Thus, the

x

-intercept is

\frac{b}{3}

.

The

x

-intercept of a line is the

x

-coordinate where the line crosses the

x

-axis, which occurs when

y = 0

.

3

Set up an equation using the given sum of the intercepts.

\frac{b}{3} + b = 12

The problem states that the sum of the

x

-intercept and the

y

-intercept is

12

.

4

Solve the linear equation for the

y

-intercept

b

.

Multiply the entire equation by

3

to clear the denominator:

b + 3b = 36 \implies 4b = 36 \implies b = 9

.

Solving this equation yields the value of the

y

-intercept.

Key Concept

Linear equations, slope-intercept form, and finding intercepts.

Alternative Method

Alternatively, you can write the equation of the line in intercept form:

\frac{x}{a} + \frac{y}{b} = 1

, where

a

is the

x

-intercept and

b

is the

y

-intercept. The slope of this line is

-\frac{b}{a} = -3

, which means

b = 3a

. Since the sum of the intercepts is

12

, we write

a + b = 12

. Substituting

b = 3a

into this sum gives

a + 3a = 12 \implies 4a = 12 \implies a = 3

. Therefore, the

y

-intercept

b

is

3(3) = 9

.

Estimated Time:1m 30s

Question 114Question →

In the standard $(x, y)$ coordinate plane, line $p$ is perpendicular to the line represented by the equation $y = -0.2x + 9$ . What is the slope of line $p$ ?

Show answer & explanation

Answer: 5

Answer

The slope of line

p

is

5

.

The given line is in slope-intercept form

y = mx + b

with a slope of

-0.2

, which can be written as

-\frac{1}{5}

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

-\frac{1}{5}

, which is

5

.

Step-by-Step Solution

1

Identify the slope of the given line from its equation.

The slope of the line

y = -0.2x + 9

is

-0.2

(or

-\frac{1}{5}

).

The equation is in slope-intercept form

y = mx + b

, where the coefficient of

x

represents the slope

m

.

2

Calculate the slope of the perpendicular line.

The negative reciprocal of

-\frac{1}{5}

is

5

.

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

m_1 \cdot m_2 = -1

).

Key Concept

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

Estimated Time:45s

Question 115Question →

A line passes through the points $(-3, 2)$ and $(5, -4)$ in a coordinate plane. What is the slope of this line?

\frac{4}{3}

\frac{3}{4}

-\frac{3}{4}

-\frac{4}{3}

-\frac{1}{4}

Show answer & explanation

Answer: $-\frac{3}{4}$

Answer

-\frac{3}{4}

The correct answer shows a slope of

-\frac{3}{4}

. Applying the slope formula

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

to the points

(-3, 2)

and

(5, -4)

yields

\frac{-4 - 2}{5 - (-3)} = \frac{-6}{8}

, which simplifies to

-\frac{3}{4}

.

Step-by-Step Solution

1

Identify the coordinates of the two points and state the slope formula.

The points are

(x_1, y_1) = (-3, 2)

and

(x_2, y_2) = (5, -4)

. The slope formula is

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

.

To find the slope of a line, we calculate the ratio of vertical change (rise) to horizontal change (run).

2

Substitute the coordinate values into the slope formula.

m = \frac{-4 - 2}{5 - (-3)}

This sets up the subtraction of the

y

-coordinates in the numerator and the

x

-coordinates in the denominator.

3

Simplify the numerator and denominator to calculate the final slope.

m = \frac{-6}{8} = -\frac{3}{4}

Subtracting a negative number in the denominator is equivalent to addition:

5 - (-3) = 8

. Simplifying the fraction yields the slope of

-\frac{3}{4}

.

Key Concept

The slope

m

of a line passing through points

(x_1, y_1)

and

(x_2, y_2)

is defined as the change in

y

divided by the change in

x

:

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

.

Question 116Question →

An equation of a circle is given by $x^2 + y^2 - 8x + 6y + k = 0$ , where $k$ is a constant. A second circle has a center that is the reflection of the first circle's center across the line $y = x$ . If the second circle is tangent to the $x$ -axis and has the same radius as the first circle, what is the value of $k$ ?

9

16

21

34

41

Show answer & explanation

Answer: 9

Answer

The value of the constant

k

is 9.

The correct answer is 9. Completing the square for the first circle's equation gives

(x - 4)^2 + (y + 3)^2 = 25 - k

, which identifies the center as

(4, -3)

and the radius squared as

r^2 = 25 - k

. Reflecting

(4, -3)

across the line

y = x

swaps the coordinates to give the new center

(-3, 4)

. Because the second circle is tangent to the

x

-axis, its radius is the absolute value of the

y

-coordinate of its center, which is

|4| = 4

. Since both circles have the same radius, we set the radius squared equal to

4^2

:

25 - k = 16

, which yields

k = 9

.

Step-by-Step Solution

1

Complete the square for the first circle's equation

x^2 + y^2 - 8x + 6y + k = 0

to identify its center and radius.

(x - 4)^2 + (y + 3)^2 = 25 - k

, which represents a circle with center

(4, -3)

and radius squared

r^2 = 25 - k

.

This puts the equation into the standard form

(x - h)^2 + (y - k)^2 = r^2

to find the center and radius.

2

Reflect the center of the first circle across the line

y = x

.

Reflecting the point

(4, -3)

across the line

y = x

swaps the coordinates, yielding the new center

(-3, 4)

.

To find the center of the second circle.

3

Determine the radius of the second circle using the tangency condition.

Since the second circle is tangent to the

x

-axis, its radius is equal to the absolute value of the

y

-coordinate of its center, which is

|4| = 4

.

The distance from a circle's center

(h, k)

to the line of tangency

y = 0

(the

x

-axis) is equal to its radius.

4

Equate the radius squared of the first circle to the square of the radius of the second circle.

25 - k = 4^2 \implies 25 - k = 16 \implies k = 9

.

Both circles are stated to have the same radius.

Key Concept

Converting the general form of a circle's equation to standard form by completing the square, and using coordinate transformations and geometric tangency conditions to solve for unknowns.

Estimated Time:3m 0s

Question 117Question →

In the standard $(x, y)$ coordinate plane, line segment $PQ$ has endpoints $P(-2, 3)$ and $Q(4, 1)$ . First, segment $PQ$ is rotated $90^\circ$ counterclockwise about the origin to form segment $P'Q'$ . Next, segment $P'Q'$ is reflected across the line $y = x$ to form segment $P''Q''$ . What are the coordinates of the midpoint of segment $P''Q''$ ?

(1, -2)

(-1, 2)

(-2, 1)

(2, -1)

(-2, -1)

Show answer & explanation

Answer: $(1, -2)$

Answer

The coordinates of the midpoint of segment

P''Q''

are

(1, -2)

.

The midpoint of the original segment

PQ

is

(1, 2)

. Under a

90^\circ

counterclockwise rotation about the origin, the point

(x, y)

maps to

(-y, x)

, so

(1, 2)

maps to

(-2, 1)

. Under a reflection across the line

y = x

, the point

(x, y)

maps to

(y, x)

, so

(-2, 1)

maps to

(1, -2)

. Since rigid transformations preserve midpoints, the midpoint of the final segment is

(1, -2)

.

Step-by-Step Solution

1

Calculate the midpoint of the original segment

PQ

.

Midpoint

M = \left(\frac{-2+4}{2}, \frac{3+1}{2}\right) = (1, 2)

Since rotation and reflection are rigid transformations (isometries), the midpoint of the transformed segment is the transformed midpoint of the original segment.

2

Apply a

90^\circ

counterclockwise rotation about the origin to the midpoint coordinate

(1, 2)

.

Intermediate midpoint

M' = (-2, 1)

The coordinate rule for a

90^\circ

counterclockwise rotation about the origin is

(x, y) \rightarrow (-y, x)

.

3

Apply a reflection across the line

y = x

to the intermediate midpoint

(-2, 1)

.

Final midpoint

M'' = (1, -2)

The coordinate rule for reflection across the line

y = x

is

(x, y) \rightarrow (y, x)

.

Key Concept

Applying composite transformations (rotations and reflections) to geometric figures on the coordinate plane, utilizing the property that the midpoint of a transformed segment is the transformed midpoint of the original segment.

Estimated Time:1m 30s

Question 118Question →

In the standard $(x, y)$ coordinate plane, a rhombus $ABCD$ has vertices $A(-2, 3)$ and $C(4, 1)$ . If vertex $B$ lies on the line $y = 2x - 5$ , what is the area of the rhombus $ABCD$ ?

20

40

80

100

200

Show answer & explanation

Answer: 100

Answer

100

The correct answer is 100. By finding the midpoint of diagonal

AC

as

(1, 2)

and using the perpendicular slope of

3

, the line containing diagonal

BD

is

y = 3x - 1

. Intersecting this with

y = 2x - 5

yields vertex

B(-4, -13)

, which gives vertex

D(6, 17)

by midpoint symmetry. The diagonal lengths are

2\sqrt{10}

and

10\sqrt{10}

, and their half-product is 100.

Step-by-Step Solution

1

Find the midpoint

M

of diagonal

AC

.

M(1, 2)

Since

ABCD

is a rhombus, its diagonals bisect each other at their midpoint.

2

Calculate the slope of diagonal

AC

.

m_{AC} = -\frac{1}{3}

The slope is calculated using the formula

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

with vertices

A(-2, 3)

and

C(4, 1)

.

3

Find the equation of the line containing diagonal

BD

.

y = 3x - 1

The diagonals of a rhombus are perpendicular. The slope of

BD

is the negative reciprocal of

-\frac{1}{3}

, which is

3

. The line passes through

M(1, 2)

.

4

Find the coordinates of vertex

B

.

B(-4, -13)

Vertex

B

lies at the intersection of the diagonal line

y = 3x - 1

and the given line

y = 2x - 5

.

5

Find the coordinates of vertex

D

.

D(6, 17)

Since

M(1, 2)

is the midpoint of diagonal

BD

, we solve

\frac{-4 + x_D}{2} = 1

and

\frac{-13 + y_D}{2} = 2

.

6

Calculate the lengths of the diagonals

AC

and

BD

.

AC = 2\sqrt{10}

and

BD = 10\sqrt{10}

Use the distance formula

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

on the pairs of vertices.

7

Calculate the area of the rhombus.

100

The area of a rhombus is given by

\text{Area} = \frac{1}{2} \times d_1 \times d_2

.

Key Concept

Using perpendicular bisector properties of rhombus diagonals on a coordinate plane to find vertices and calculate area.

Question 119Question →

A straight road on a map is modeled by the linear equation $2x - 5y = 10$ in the standard $(x, y)$ coordinate plane. A second road, which is parallel to the first road, is being constructed. What is the slope of the second road?

2

-\frac{5}{2}

\frac{2}{5}

-\frac{2}{5}

\frac{5}{2}

Show answer & explanation

Answer: $\frac{2}{5}$

Answer

The slope of the second road is

\frac{2}{5}

Parallel lines have the same slope. Rewriting the line's equation

2x - 5y = 10

into slope-intercept form (

y = mx + b

) yields

y = \frac{2}{5}x - 2

. The slope of this line is the coefficient of

x

, which is

\frac{2}{5}

. Therefore, any line parallel to it must also have a slope of

\frac{2}{5}

.

Step-by-Step Solution

1

Determine the relationship between the slopes of parallel lines.

Parallel lines have identical slopes.

Since the second road is parallel to the first, its slope must be equal to the slope of the first road.

2

Convert the equation of the first road,

2x - 5y = 10

, into slope-intercept form (

y = mx + b

).

Subtract

2x

from both sides to get

-5y = -2x + 10

. Then, divide both sides by

-5

to get

y = \frac{2}{5}x - 2

.

In slope-intercept form, the coefficient of

x

(represented by

m

) is the slope of the line.

3

Identify the slope of the first road and match it to the parallel road.

The slope of the first road is

\frac{2}{5}

, so the slope of the parallel road is also

\frac{2}{5}

.

The slope is the coefficient of

x

, which is

\frac{2}{5}

.

Key Concept

Parallel lines in a coordinate plane have equal slopes.

Question 120Question →

In the standard $(x, y)$ coordinate plane, point $P$ undergoes two transformations. First, it is reflected across the $y$ -axis. Second, it is translated $3$ units down and $4$ units to the right, resulting in the image point $(1, -2)$ . What are the coordinates of the original point $P$ ?

(-3, -1)

(3, -5)

(-5, -5)

(3, 1)

(1, -3)

Show answer & explanation

Answer: $(3, 1)$

Answer

The coordinates of the original point

P

are

(3, 1)

.

To find the coordinates of the original point, we must work backward from the final image

(1, -2)

. First, we undo the translation (which was

3

units down and

4

units to the right) by performing the opposite actions: moving

3

units up and

4

units to the left. This shifts

(1, -2)

to

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

. Next, we undo the reflection across the

y

-axis. Reflecting a point across the

y

-axis negates its

x

-coordinate. The reflection of

(-3, 1)

across the

y

-axis yields the original point

(3, 1)

.

Step-by-Step Solution

1

Identify the transformations in reverse order to work backward from the final image

(1, -2)

to the original point

P

.

The final image is

(1, -2)

. We must first undo the translation (

3

units down,

4

units right) and then undo the reflection across the

y

-axis.

To find the pre-image, we apply the inverse transformations in the reverse order of the original operations.

2

Undo the translation by applying the opposite operations: move

3

units up and

4

units to the left.

The intermediate point is

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

.

Undoing a translation of

+4

in the

x

-direction and

-3

in the

y

-direction requires subtracting

4

from the

x

-coordinate and adding

3

to the

y

-coordinate.

3

Undo the reflection across the

y

-axis by reflecting the intermediate point

(-3, 1)

across the

y

-axis.

The original point

P

is

(3, 1)

.

Reflecting a point across the

y

-axis negates its

x

-coordinate. Since reflection is its own inverse, applying it again returns the original coordinates:

-(-3) = 3

.

Key Concept

Working backward through composite transformations in the coordinate plane.

Alternative Method

Instead of working backward step-by-step, write the transformation equations. If the original point is

P(x, y)

, the reflection across the

y

-axis gives

P'(-x, y)

. The subsequent translation of

3

units down and

4

units right gives

P''(-x + 4, y - 3)

. Set this expression equal to the final coordinates:

-x + 4 = 1 \implies x = 3

and

y - 3 = -2 \implies y = 1

, which gives the original point

(3, 1)

.

Estimated Time:1m 30s

Coordinate Geometry

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