Coordinate Geometry Practice Questions — ACT — Page 10

Question 181Question →

A triangle has vertices $K(-2, 3)$ , $L(-2, 1)$ , and $M(2, 1)$ in the standard $(x, y)$ coordinate plane. The triangle is first reflected across the line $y = x$ , and then the reflected image is translated $3$ units to the right and $4$ units down. What are the coordinates of the final image of vertex $K$ ?

(0, 2)

(1, -1)

(6, -6)

(0, -2)

(5, -1)

Show answer & explanation

Answer: $(6, -6)$

Answer

(6, -6)

The correct answer is

(6, -6)

. To find this, start with the coordinates of vertex

K(-2, 3)

. First, apply the reflection across the line

y = x

, which maps

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

. This gives the intermediate coordinates

K'(3, -2)

. Second, apply the translation

3

units to the right and

4

units down, which maps

(x, y) \rightarrow (x + 3, y - 4)

. Adding

3

to the

x

-coordinate and subtracting

4

from the

y

-coordinate of the intermediate point yields the final coordinates

(3 + 3, -2 - 4) = (6, -6)

.

Step-by-Step Solution

1

Apply the reflection rule to the pre-image vertex

K(-2, 3)

K'(3, -2)

Reflecting a point

(x, y)

across the line

y = x

swaps its coordinates, mapping it to

(y, x)

.

2

Apply the translation rule to the reflected point

K'(3, -2)

K''(6, -6)

Translating a point

3

units to the right and

4

units down adds

3

to the

x

-coordinate and subtracts

4

from the

y

-coordinate.

Key Concept

Composite transformations in the coordinate plane

Question 182Question →

A custom printing company uses a linear relationship to determine the total cost of printing t-shirts. The table below displays the total cost, $C$ , in dollars, for printing $x$ t-shirts:

Number of Shirts ( $x$ )	Total Cost ( $C$ )
15	\$162.50
25	\$237.50
35	\$312.50

If the total cost consists of a one-time setup fee plus a constant charge per t-shirt, what is the one-time setup fee?

\$7.50

\$10.83

\$12.50

\$50.00

\$112.50

Show answer & explanation

Answer: \$50.00

Answer

The one-time setup fee is \$50.00.

The correct answer of \

50.00 is found by first determining the constant rate of change (slope) from the table, which is \$7.50 per shirt. Substituting this rate and the data point for 15 shirts into the linear model

C = mx + b

gives

162.50 = 7.50(15) + b

. Solving for the y-intercept

b

yields

50.00, which represents the one-time setup fee.

Step-by-Step Solution

1

Calculate the constant charge per t-shirt (the slope,

m

) by finding the change in cost divided by the change in the number of shirts between two data points.

m = \frac{237.50 - 162.50}{25 - 15} = \frac{75.00}{10} = 7.50

Since the relationship is linear, the rate of change (cost per shirt) remains constant.

2

Write the linear equation representing the total cost:

C = 7.50x + b

, where

b

represents the one-time setup fee.

C = 7.50x + b

This sets up the slope-intercept form of the linear relationship.

3

Substitute one of the points from the table, such as

(15, 162.50)

, into the equation to solve for

b

.

162.50 = 7.50(15) + b

162.50 = 112.50 + b

b = 50.00

Using a known coordinate point allows us to isolate and solve for the unknown y-intercept (the setup fee).

Key Concept

Linear Equations and Graphing

Question 183Question →

In the standard $(x, y)$ coordinate plane, the center of a circle is located at the midpoint of the line segment with endpoints $A\left(\frac{1}{2}, 9\right)$ and $B\left(\frac{9}{2}, 3\right)$ . If the circle passes through the point $P\left(\frac{13}{2}, 9\right)$ , what is the radius of the circle?

6

7

5

25

2\sqrt{13}

Show answer & explanation

Answer: 5

Answer

The radius of the circle is 5.

The center of the circle is the midpoint of segment AB. Calculating the midpoint of

A(0.5, 9)

and

B(4.5, 3)

gives

M(2.5, 6)

. The radius is the distance from this center

M(2.5, 6)

to the given point

P(6.5, 9)

on the circle. The distance is

\sqrt{(6.5 - 2.5)^2 + (9 - 6)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Find the coordinates of the center of the circle by calculating the midpoint of segment AB.

Center

M = \left(\frac{1/2 + 9/2}{2}, \frac{9 + 3}{2}\right) = \left(\frac{5}{2}, 6\right) = (2.5, 6)

.

The center of the circle is located at the midpoint of the line segment AB.

2

Use the distance formula to find the distance between the center

M(2.5, 6)

and the point

P(6.5, 9)

on the circle.

r = \sqrt{(6.5 - 2.5)^2 + (9 - 6)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

.

The radius of a circle is the distance from its center to any point on its boundary.

Key Concept

Using the midpoint formula to locate the center of a circle, and then applying the distance formula between the center and a boundary point to find the radius.

Alternative Method

You can visualize the horizontal and vertical distances on a coordinate plane. The horizontal distance between the x-coordinates of the center

(2.5, 6)

and the point

(6.5, 9)

is

6.5 - 2.5 = 4

. The vertical distance is

9 - 6 = 3

. Recognizing the 3-4-5 right triangle triplet immediately gives the distance (hypotenuse) as 5.

Estimated Time:1m 15s

Question 184Question →

A line segment with endpoints $C(2, 3)$ and $D(6, 3)$ is plotted in the standard $(x, y)$ coordinate plane. If this segment is rotated $90^\circ$ clockwise about the origin and then translated $4$ units up, what are the coordinates of the midpoint of the resulting segment?

(3, -8)

(7, -4)

(3, 0)

(-3, 8)

(4, 7)

Show answer & explanation

Answer: $(3, 0)$

Answer

(3, 0)

The correct answer is obtained by first calculating the midpoint of the original segment

CD

, which yields

(4, 3)

. Applying a

90^\circ

clockwise rotation about the origin maps the point

(x, y)

to

(y, -x)

, transforming

(4, 3)

to

(3, -4)

. Translating this intermediate point 4 units up adds 4 to its y-coordinate, resulting in

(3, 0)

.

Step-by-Step Solution

1

Find the midpoint of the original line segment

CD

using the midpoint formula

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

.

M = \left(\frac{2 + 6}{2}, \frac{3 + 3}{2}\right) = (4, 3)

Since translations and rotations are rigid transformations, the midpoint of the transformed segment is the image of the midpoint of the original segment.

2

Apply the rotation of

90^\circ

clockwise about the origin to the midpoint

M(4, 3)

. The rule for a clockwise rotation of

90^\circ

is

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

.

M'(3, -4)

Rotating a point

(x, y)

by

90^\circ

clockwise maps it to

(y, -x)

.

3

Translate the point

M'(3, -4)

by

4

units up. The rule for translating a point

d

units up is

(x, y) \rightarrow (x, y + d)

.

M''(3, -4 + 4) = (3, 0)

A vertical translation upward increases the y-coordinate of the point by the given number of units.

Key Concept

Transformations in the Coordinate Plane

Question 185Question →

The circle defined by the equation $x^2 + y^2 - 6x + 8y - 11 = 0$ is plotted in the standard $(x, y)$ coordinate plane. A second circle has a center that is the reflection of the first circle's center across the line $y = x$ . If the radius of the second circle is twice the radius of the first circle, which of the following is the equation of the second circle?

(x - 4)^2 + (y + 3)^2 = 144

(x + 4)^2 + (y - 3)^2 = 12

(x - 3)^2 + (y + 4)^2 = 144

(x + 4)^2 + (y - 3)^2 = 144

(x + 4)^2 + (y - 3)^2 = 72

Show answer & explanation

Answer: $(x + 4)^2 + (y - 3)^2 = 144$

Answer

(x + 4)^2 + (y - 3)^2 = 144

Completing the square on

x^2 + y^2 - 6x + 8y - 11 = 0

gives

(x - 3)^2 + (y + 4)^2 = 36

. The center is

(3, -4)

and the radius is

6

. Reflecting

(3, -4)

across the line

y = x

swaps the coordinates, yielding the new center

(-4, 3)

. Doubling the radius yields

12

. The equation of the second circle is therefore

(x - (-4))^2 + (y - 3)^2 = 12^2

, which simplifies to

(x + 4)^2 + (y - 3)^2 = 144

.

Step-by-Step Solution

1

Group the

x

and

y

terms of the given equation

x^2 + y^2 - 6x + 8y - 11 = 0

and add 11 to both sides.

(x^2 - 6x) + (y^2 + 8y) = 11

To prepare the terms for completing the square.

2

Complete the square for both the

x

and

y

groups by adding

(6/2)^2 = 9

and

(8/2)^2 = 16

to both sides.

(x - 3)^2 + (y + 4)^2 = 36

To express the circle's equation in standard form

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

.

3

Identify the center and radius of the first circle from the standard form.

Center is

(3, -4)

and radius is

r_1 = \sqrt{36} = 6

.

These characteristics are needed to find the properties of the second circle.

4

Reflect the center

(3, -4)

across the line

y = x

by swapping the

x

and

y

coordinates, and double the radius.

New center is

(-4, 3)

and new radius is

r_2 = 2 \times 6 = 12

.

To satisfy the geometric transformations specified in the problem statement.

5

Write the standard equation of the second circle using the new center

(-4, 3)

and new radius

12

.

(x + 4)^2 + (y - 3)^2 = 144

Substituting

h = -4

,

k = 3

, and

r = 12

into the standard equation

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

.

Key Concept

Completing the square to find standard circle characteristics, and applying reflections across the line

y = x

.

Estimated Time:2m 0s

Question 186Question →

In the standard $(x, y)$ coordinate plane, where $1$ unit represents $1$ kilometer, a playground is located at $(1, 10)$ and a picnic shelter is located at $(5, -2)$ . A water fountain is to be placed exactly halfway between the playground and the picnic shelter. What is the distance, in kilometers, from the origin $(0, 0)$ to the water fountain?

Show answer & explanation

Answer: 5

Answer

The distance from the origin to the water fountain is

5

kilometers.

The midpoint of the segment representing the water fountain is determined to be

(3, 4)

. The distance from the origin

(0, 0)

to

(3, 4)

is found using the Pythagorean relation, yielding

\sqrt{3^2 + 4^2} = 5

kilometers.

Step-by-Step Solution

1

Find the coordinates of the midpoint between

(1, 10)

and

(5, -2)

.

The midpoint is

(3, 4)

.

The midpoint formula is

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

. Substituting the given endpoints gives

M = \left(\frac{1 + 5}{2}, \frac{10 + (-2)}{2}\right) = (3, 4)

.

2

Calculate the distance from the origin

(0, 0)

to the midpoint

(3, 4)

.

The distance is

5

kilometers.

The distance formula from the origin to any point

(x, y)

is

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

. Substituting the coordinates

(3, 4)

yields

d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

.

Key Concept

Using the midpoint formula to find the center of a line segment, followed by the distance formula from the origin to that point.

Question 187Question →

In the standard $(x, y)$ coordinate plane, a circle is tangent to the $x$ -axis at the point $(5, 0)$ . If the circle is also tangent to the line $y = \frac{4}{3}x$ and its center lies in the first quadrant, what is the $y$ -coordinate of the center of the circle?

Show answer & explanation

Answer: 2.5

Answer

The

y

-coordinate of the center of the circle is

2.5

.

A circle tangent to the

x

-axis at

(5, 0)

has a center along the vertical line

x = 5

. Since the center is in the first quadrant, its coordinates can be represented as

(5, k)

where

k > 0

, and the radius is

k

. The distance from the center

(5, k)

to the line

4x - 3y = 0

must also equal the radius

k

. Using the point-to-line distance formula, we write

\frac{|4(5) - 3k|}{\sqrt{4^2 + (-3)^2}} = k

, which simplifies to

|20 - 3k| = 5k

. Since the center must lie in the first quadrant (

k > 0

), we solve

20 - 3k = 5k

to find

k = 2.5

. The alternative case

20 - 3k = -5k

gives

k = -10

, which lies in the fourth quadrant and is thus excluded.

Step-by-Step Solution

1

Identify the coordinates of the center and the radius of the circle based on the

x

-axis tangency.

Center:

(5, k)

and Radius:

r = k

(where

k > 0

).

A circle tangent to the

x

-axis at

(5, 0)

has its center directly above or below this point on the line

x = 5

. Since the center is in the first quadrant, its

y

-coordinate

k

must be positive, making the radius

k

.

2

Use the distance from the center to the second tangent line to write an equation for

k

.

\frac{|4(5) - 3k|}{\sqrt{4^2 + (-3)^2}} = k

The distance from the center

(5, k)

to the tangent line

4x - 3y = 0

must equal the radius of the circle.

3

Simplify the equation and solve for

k

.

|20 - 3k| = 5k

, leading to

k = 2.5

or

k = -10

.

Simplifying the denominator yields

5

. Multiplying both sides by

5

gives the absolute value equation, which resolves to

20 - 3k = 5k

or

20 - 3k = -5k

.

4

Select the valid solution using the quadrant constraint.

k = 2.5

The center must lie in the first quadrant, which requires

k > 0

. Thus,

k = -10

is discarded, and the correct value is

2.5

.

Key Concept

The relationship between a circle's center, its radius, and its tangent lines in the coordinate plane.

Question 188Question →

On a map of a town laid out on a standard coordinate grid, a straight path has an $x$ -intercept of $-6$ and a $y$ -intercept of $4$ . A landmark is located on this path at a point where the $x$ -coordinate is $9$ . What is the $y$ -coordinate of the landmark?

Show answer & explanation

Answer: 10

Answer

The

y

-coordinate of the landmark is 10.

The correct answer is 10. The linear path passes through the points

(-6, 0)

and

(0, 4)

. Using the slope formula, the slope is

m = \frac{4 - 0}{0 - (-6)} = \frac{2}{3}

. The slope-intercept equation of the line is

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

. Substituting the landmark's

x

-coordinate of

9

into the equation gives

y = \frac{2}{3}(9) + 4 = 10

.

Step-by-Step Solution

1

Identify the coordinates of the points representing the intercepts on the coordinate grid.

The points are

(-6, 0)

and

(0, 4)

.

An

x

-intercept of

-6

corresponds to the coordinate point

(-6, 0)

, and a

y

-intercept of

4

corresponds to the coordinate point

(0, 4)

.

2

Calculate the slope of the linear path.

The slope is

\frac{2}{3}

.

Using the slope formula

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

, substitute the points

(-6, 0)

and

(0, 4)

to get

m = \frac{4 - 0}{0 - (-6)} = \frac{4}{6} = \frac{2}{3}

.

3

Write the linear equation representing the path.

The equation is

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

.

With a slope of

m = \frac{2}{3}

and a

y

-intercept of

b = 4

, the equation in slope-intercept form (

y = mx + b

) is

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

.

4

Find the

y

-coordinate of the landmark by substituting its

x

-coordinate.

The

y

-coordinate is 10.

Substitute

x = 9

into the equation:

y = \frac{2}{3}(9) + 4 = 6 + 4 = 10

.

Key Concept

Linear Equations and Graphing

Question 189Question →

Points $K(-2, 5)$ and $L(6, -1)$ are the endpoints of segment $KL$ on a coordinate plane. What is the distance from the midpoint of segment $KL$ to the point $P(5, 7)$ ?

\sqrt{10}

\sqrt{34}

8

10

\sqrt{130}

Show answer & explanation

Answer: $\sqrt{34}$

Answer

The correct answer is the square root of 34.

To find the distance from the midpoint of segment

KL

to the point

P(5, 7)

, we first calculate the midpoint

M

of

KL

. Using the midpoint formula, the coordinates are

x = \frac{-2 + 6}{2} = 2

and

y = \frac{5 + (-1)}{2} = 2

, giving

M(2, 2)

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

M(2, 2)

and

P(5, 7)

, which is

d = \sqrt{(5 - 2)^2 + (7 - 2)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}

. This confirms the correct distance is the square root of 34.

Step-by-Step Solution

1

Calculate the coordinates of the midpoint of segment

KL

using the midpoint formula:

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

.

The midpoint is

M(2, 2)

since

x = \frac{-2 + 6}{2} = 2

and

y = \frac{5 + (-1)}{2} = 2

.

Finding the midpoint of the segment is required before finding the distance from it to point

P

.

2

Calculate the distance between the midpoint

M(2, 2)

and the point

P(5, 7)

using the distance formula:

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

.

The distance is

\sqrt{34}

since

d = \sqrt{(5 - 2)^2 + (7 - 2)^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}

.

This determines the final distance required by the question.

Key Concept

Applying both the midpoint and distance formulas sequentially to solve coordinate geometry problems.

Question 190Question →

A square in the standard $(x, y)$ coordinate plane has vertices at $A(1, 1)$ , $B(3, 1)$ , $C(3, 3)$ , and $D(1, 3)$ . The square is dilated by a scale factor of $3$ with the center of dilation at point $A(1, 1)$ . The resulting image is then translated $2$ units left and $4$ units up. What are the coordinates of the final image of vertex $C$ ?

(7, 13)

(9, 3)

(1, 19)

(5, 11)

(11, 13)

Show answer & explanation

Answer: $(5, 11)$

Answer

The final coordinates of the image of vertex

C

are

(5, 11)

.

To find the coordinates of the image of vertex

C(3, 3)

after the dilation, we calculate the horizontal and vertical distances from the center of dilation

A(1, 1)

to

C(3, 3)

. Both distances are

2

units. Since the scale factor is

3

, these distances are tripled to

6

units. Adding these to the coordinates of the center

A(1, 1)

gives the dilated point

(1 + 6, 1 + 6) = (7, 7)

. Next, we apply the translation of

2

units left (subtracting

2

from the

x

-coordinate) and

4

units up (adding

4

to the

y

-coordinate) to

(7, 7)

, which yields the final coordinates

(7 - 2, 7 + 4) = (5, 11)

.

Step-by-Step Solution

1

Find the coordinates of the image of vertex

C(3, 3)

after a dilation with scale factor

3

centered at

A(1, 1)

.

The coordinates after dilation are

(7, 7)

.

The horizontal and vertical distances from the center

A(1, 1)

to

C(3, 3)

are both

2

units (

3 - 1 = 2

). Multiplying these distances by the scale factor of

3

yields

6

units. Adding these new distances to the coordinates of the center

A(1, 1)

gives

(1 + 6, 1 + 6) = (7, 7)

.

2

Apply a translation of

2

units left and

4

units up to the point

(7, 7)

.

The final coordinates are

(5, 11)

.

Translating

2

units left subtracts

2

from the

x

-coordinate (

7 - 2 = 5

). Translating

4

units up adds

4

to the

y

-coordinate (

7 + 4 = 11

).

Key Concept

Transformations in the Coordinate Plane

Estimated Time:1m 30s

Question 191Question →

The table below shows coordinate points $(x, y)$ that lie on a line $L$ in the standard $(x, y)$ coordinate system:

$x$	$y$
$-2$	$-7$
$1$	$2$
$4$	$11$

What is the $x$ -intercept of line $L$ ?

\frac{1}{3}

-\frac{1}{3}

-1

-5

\frac{5}{3}

Show answer & explanation

Answer: $\frac{1}{3}$

Answer

The correct answer is

\frac{1}{3}

The correct answer is

\frac{1}{3}

. First, find the slope of the line using the points

(-2, -7)

and

(1, 2)

from the table:

m = \frac{2 - (-7)}{1 - (-2)} = 3

. Next, substitute the slope and the coordinate point

(1, 2)

into the point-slope equation to get

y - 2 = 3(x - 1)

, which simplifies to

y = 3x - 1

. To find the

x

-intercept, set

y = 0

and solve for

x

:

0 = 3x - 1

, which gives

x = \frac{1}{3}

.

Step-by-Step Solution

1

Calculate the slope (

m

) of line

L

using the coordinate points

(-2, -7)

and

(1, 2)

from the table.

The slope is

m = \frac{2 - (-7)}{1 - (-2)} = \frac{9}{3} = 3

.

The slope of a line represents its constant rate of change and is calculated using the formula

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

.

2

Write the linear equation in slope-intercept form using the slope

m = 3

and the point

(1, 2)

.

Using the point-slope form:

y - 2 = 3(x - 1) \Rightarrow y - 2 = 3x - 3 \Rightarrow y = 3x - 1

.

Using point-slope form

y - y_1 = m(x - x_1)

allows us to easily simplify the relationship into slope-intercept form

y = mx + b

.

3

Find the

x

-intercept of the line by setting

y = 0

and solving the equation for

x

.

Setting

y = 0

gives

0 = 3x - 1 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}

.

The

x

-intercept is the coordinate value where the graph intersects the

x

-axis, which mathematically occurs when

y = 0

.

Key Concept

Determining the equation of a line from tabular coordinates and calculating its intercepts

Question 192Question →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 + y^2 - 6x + 4y - 12 = 0$ . A line $L$ passes through the center of this circle and is perpendicular to the line with equation $3x - 4y = 12$ . What is the $y$ -coordinate of the intersection point of line $L$ and the circle that has a positive $y$ -value?

2

6

-6

1

5

Show answer & explanation

Answer: 2

Answer

The y-coordinate of the intersection point with a positive y-value is 2.

Completing the square of the circle equation

x^2 + y^2 - 6x + 4y - 12 = 0

yields

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

, giving center

(3, -2)

and radius

r = 5

. The given line

3x - 4y = 12

has a slope of

3/4

, so a perpendicular line

L

has a slope of

-4/3

. Since line

L

passes through the center

(3, -2)

and has a slope of

-4/3

, moving a distance of 5 units along this line (which corresponds to a horizontal change of

\pm 3

and vertical change of

\mp 4

) gives the intersection points

(6, -6)

and

(0, 2)

. The y-coordinate of the point with a positive y-value is 2.

Step-by-Step Solution

1

Complete the square for the circle's equation to find its center and radius.

The equation is rewritten as

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

, which represents a circle with center

(3, -2)

and radius

r = 5

.

Converting to standard form

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

gives the circle's center and radius.

2

Find the slope of the given line

3x - 4y = 12

and determine the slope of the perpendicular line

L

.

The slope of the given line is

3/4

. The slope of line

L

is the negative reciprocal:

-4/3

.

Perpendicular lines have slopes that are negative reciprocals.

3

Find the intersection points of line

L

and the circle.

Since line

L

passes through the center

(3, -2)

and has a slope of

-4/3

, points on the line at a distance of the radius

r=5

are found by moving 3 units horizontally and 4 units vertically. This yields the points

(3 + 3, -2 - 4) = (6, -6)

and

(3 - 3, -2 + 4) = (0, 2)

.

The intersection points lie exactly one radius away from the center along the line.

4

Identify the y-coordinate with a positive value.

Between the points

(6, -6)

and

(0, 2)

, the point with a positive y-value is

(0, 2)

, so the y-coordinate is 2.

The question specifies the intersection point must have a positive y-value.

Key Concept

Circle standard equations and perpendicular lines on the coordinate plane

Question 193Question →

In the standard $(x, y)$ coordinate plane, a point $P$ is reflected across the $y$ -axis and then translated 4 units down and 3 units right. The coordinates of the resulting image point, $P'$ , are $(1, -2)$ . What are the coordinates of the original point $P$ ?

(-2, -2)

(-2, 2)

(2, -6)

(2, -2)

(2, 2)

Show answer & explanation

Answer: $(2, 2)$

Answer

The coordinates of the original point

P

are

(2, 2)

.

To find the coordinates of the original point

P(x, y)

, we apply the inverse transformations in reverse order to the final image point

P'(1, -2)

. First, we undo the translation of 4 units down and 3 units right by translating

P'

4 units up and 3 units left. This results in the intermediate coordinates

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

. Next, we undo the reflection across the

y

-axis by reflecting

(-2, 2)

across the

y

-axis again (since a reflection is its own inverse). Changing the sign of the

x

-coordinate gives

(2, 2)

. Alternatively, setting up the equations

-x + 3 = 1

and

y - 4 = -2

and solving them yields

x = 2

and

y = 2

, which corresponds to the point

(2, 2)

.

Step-by-Step Solution

1

Set up equations to express the composite transformations of the point

P(x, y)

to

P'(1, -2)

.

Reflecting

P(x, y)

across the

y

-axis changes the sign of the

x

-coordinate, yielding

(-x, y)

. Translating this point 4 units down and 3 units right yields

(-x + 3, y - 4)

.

Establishing the mathematical relationship for each transformation is necessary to work backward to find the coordinates of the pre-image.

2

Equate the coordinates of the transformed point to the coordinates of the final image point

P'(1, -2)

.

-x + 3 = 1

and

y - 4 = -2

This sets up two independent linear equations that can be solved for the original coordinates

x

and

y

.

3

Solve the equations for

x

and

y

.

From

-x + 3 = 1

, we get

-x = -2

, which means

x = 2

. From

y - 4 = -2

, we get

y = 2

. Therefore, the coordinates of

P

are

(2, 2)

.

Solving these equations gives the exact coordinates of the original pre-image point.

Key Concept

Working backward with composite transformations in the coordinate plane

Estimated Time:1m 15s

Question 194Question →

A triangle has a vertex at $P(-4, 3)$ in the standard $(x, y)$ coordinate plane. If the triangle is reflected across the line $y = -x$ and then translated $3$ units to the right and $4$ units down, what are the coordinates of the image of vertex $P$ after both transformations?

(0, 0)

(6, -8)

(-6, 8)

(7, -1)

(-1, -7)

Show answer & explanation

Answer: $(0, 0)$

Answer

(0, 0)

The coordinate

(0, 0)

is correct because reflecting the point

P(-4, 3)

across the line

y = -x

swaps and negates the coordinates, transforming it to

(-3, 4)

. Then, translating this point

3

units to the right and

4

units down is calculated as

(-3 + 3, 4 - 4)

, which simplifies to

(0, 0)

.

Step-by-Step Solution

1

Apply the reflection across the line

y = -x

to the coordinate

P(-4, 3)

.

The rule for reflection across the line

y = -x

is

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

. Applying this rule to

P(-4, 3)

yields

(-3, 4)

.

To find the coordinates of the vertex after the first transformation in the composite sequence.

2

Apply the translation of

3

units right and

4

units down to the intermediate point

(-3, 4)

.

Translating

3

units to the right adds

3

to the

x

-coordinate, and translating

4

units down subtracts

4

from the

y

-coordinate:

(-3 + 3, 4 - 4) = (0, 0)

.

To find the final position after the second transformation in the composite sequence.

Key Concept

Applying composite transformations in the coordinate plane, specifically a reflection across the line

y = -x

followed by a translation.

Question 195Question →

In the standard $(x, y)$ coordinate plane, line $T$ is perpendicular to the line with the equation $y = 3x - 5$ . If line $T$ passes through the point $(6, 2)$ , what is the $y$ -intercept of line $T$ ?

-16

-10

4

20

0

Show answer & explanation

Answer: 4

Answer

The correct answer is 4

The slope of the line given by the equation

y = 3x - 5

is

3

. Since line

T

is perpendicular to this line, its slope must be the negative reciprocal of

3

, which is

-\frac{1}{3}

. Using the point-slope form with the point

(6, 2)

, the equation of line

T

is

y - 2 = -\frac{1}{3}(x - 6)

. To find the

y

-intercept of line

T

, we set

x = 0

and solve for

y

:

y - 2 = -\frac{1}{3}(0 - 6) \implies y - 2 = 2 \implies y = 4

.

Step-by-Step Solution

1

Identify the slope of the given line from its equation.

The slope of the line

y = 3x - 5

is

3

.

The equation is in slope-intercept form (

y = mx + b

), where

m

is the slope.

2

Determine the slope of the perpendicular line

T

.

The slope of line

T

is

-\frac{1}{3}

.

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

3

Use the point-slope form to write the equation of line

T

with the point

(6, 2)

.

The equation of line

T

is

y - 2 = -\frac{1}{3}(x - 6)

.

The point-slope form is

y - y_1 = m(x - x_1)

where

(x_1, y_1)

is a point on the line and

m

is the slope.

4

Find the

y

-intercept of line

T

by setting

x = 0

.

y = 4

.

The

y

-intercept is the value of

y

where the line crosses the

y

-axis (when

x = 0

).

Key Concept

Finding the equation and y-intercept of a perpendicular line using negative reciprocal slopes

Question 196Question →

In the standard $(x, y)$ coordinate plane, the midpoint of the line segment with endpoints $A(1, 3)$ and $B\left(7, \frac{x}{2}\right)$ is $M$ . If the distance from the origin $(0, 0)$ to $M$ is $\sqrt{\frac{89}{4}}$ units, and $x$ is a positive number, what is the value of $x$ ?

2

4

8

14

16

Show answer & explanation

Answer: 4

Answer

The correct value of

x

is

4

.

The midpoint

M

of

A(1, 3)

and

B\left(7, \frac{x}{2}\right)

has coordinates

\left(\frac{1+7}{2}, \frac{3 + \frac{x}{2}}{2}\right) = \left(4, \frac{x+6}{4}\right)

. The distance from

(0, 0)

to

M

is

\sqrt{4^2 + \left(\frac{x+6}{4}\right)^2} = \sqrt{\frac{89}{4}}

. Squaring both sides yields

16 + \frac{(x+6)^2}{16} = \frac{89}{4}

. Multiplying through by 16 gives

256 + (x+6)^2 = 356

, which simplifies to

(x+6)^2 = 100

. Taking the square root of both sides gives

x+6 = 10

(since

x

is positive), which simplifies to

x = 4

.

Step-by-Step Solution

1

Find the coordinates of the midpoint

M(x_M, y_M)

of the segment

AB

.

M = \left(4, \frac{x+6}{4}\right)

By the midpoint formula,

x_M = \frac{1 + 7}{2} = 4

and

y_M = \frac{3 + \frac{x}{2}}{2} = \frac{\frac{6+x}{2}}{2} = \frac{x+6}{4}

.

2

Set up the distance equation from the origin

(0, 0)

to

M\left(4, \frac{x+6}{4}\right)

using the distance formula.

\sqrt{4^2 + \left(\frac{x+6}{4}\right)^2} = \sqrt{\frac{89}{4}}

The distance between the origin

(0, 0)

and any point

(x, y)

is given by

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

.

3

Solve the distance equation for the positive variable

x

.

x = 4

Squaring both sides gives

16 + \frac{(x+6)^2}{16} = \frac{89}{4}

. Multiplying by 16 yields

256 + (x+6)^2 = 356

, which simplifies to

(x+6)^2 = 100

. Since

x

is positive,

x+6 = 10 \implies x = 4

.

Key Concept

Solving coordinate geometry problems by combining the midpoint formula and the distance formula.

Question 197Question →

In the standard $(x, y)$ coordinate plane, line $L_1$ passes through the points $(2, 5)$ and $(6, -3)$ . A second line, $L_2$ , is perpendicular to $L_1$ and intersects $L_1$ at its $y$ -intercept. What is the $x$ -coordinate of the $x$ -intercept of $L_2$ ?

Show answer & explanation

Answer: -18

Answer

The

x

-coordinate of the

x

-intercept of

L_2

is

-18

.

First, the slope of

L_1

is calculated as

-2

using the slope formula. Substituting one of the points into the slope-intercept form gives the

y

-intercept of

L_1

as

(0, 9)

. Since

L_2

is perpendicular to

L_1

, its slope is the negative reciprocal of

-2

, which is

\frac{1}{2}

. Since

L_2

shares the

y

-intercept

(0, 9)

, its equation is

y = \frac{1}{2}x + 9

. Setting

y = 0

to find the

x

-intercept gives

x = -18

.

Step-by-Step Solution

1

Calculate the slope of line

L_1

.

The slope of

L_1

is

-2

.

Using the slope formula

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

with the given points

(2, 5)

and

(6, -3)

yields

m_1 = \frac{-3 - 5}{6 - 2} = -2

.

2

Find the

y

-intercept of

L_1

.

The

y

-intercept is

(0, 9)

.

Substituting

m = -2

and the coordinates of

(2, 5)

into the slope-intercept equation

y = mx + b

gives

5 = -2(2) + b

, which simplifies to

b = 9

.

3

Find the slope of the perpendicular line,

L_2

.

The slope of

L_2

is

\frac{1}{2}

.

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

-2

is

\frac{1}{2}

.

4

Determine the equation of

L_2

and calculate its

x

-intercept.

The

x

-coordinate of the

x

-intercept of

L_2

is

-18

.

Since

L_2

passes through the

y

-intercept

(0, 9)

, its equation is

y = \frac{1}{2}x + 9

. Setting

y = 0

to find the

x

-intercept yields

0 = \frac{1}{2}x + 9

, which solves to

x = -18

.

Key Concept

Determining the equation and intercepts of a line perpendicular to a given line that passes through a specific shared point.

Estimated Time:1m 30s

Question 198Question →

In the standard $(x, y)$ coordinate plane, a line segment has endpoints $A(-3, k)$ and $B(5, 3)$ . If the midpoint of segment $AB$ lies on the $x$ -axis, what is the length of segment $AB$ ?

Show answer & explanation

Answer: 10

Answer

The length of segment

AB

is 10.

The midpoint of segment

AB

with endpoints

A(-3, k)

and

B(5, 3)

is

\left(1, \frac{k + 3}{2}\right)

. Since the midpoint lies on the

x

-axis, its

y

-coordinate must be

0

. Solving

\frac{k + 3}{2} = 0

gives

k = -3

. This means the endpoints are

A(-3, -3)

and

B(5, 3)

. The distance between these two points is

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

.

Step-by-Step Solution

1

Set up the equation for the

y

-coordinate of the midpoint.

\frac{k + 3}{2} = 0

The midpoint of a segment with endpoints

(x_1, y_1)

and

(x_2, y_2)

has a

y

-coordinate of

\frac{y_1 + y_2}{2}

. Since the midpoint lies on the

x

-axis, its

y

-coordinate must be

0

.

2

Solve the equation for

k

.

k = -3

Multiply both sides of the equation by

2

to get

k + 3 = 0

, then subtract

3

from both sides.

3

Substitute

k = -3

to find the coordinates of point

A

.

A(-3, -3)

This provides both complete endpoints,

A(-3, -3)

and

B(5, 3)

, which are needed to find the distance.

4

Apply the distance formula to find the length of segment

AB

.

AB = 10

The distance formula is

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

. Substituting the coordinates gives

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

.

Key Concept

Distance and Midpoint Formulas

Question 199Question →

The endpoint $A$ of segment $AB$ is located at $(-3, 5)$ in the standard $(x, y)$ coordinate plane. Segment $AB$ is translated 4 units right and 2 units down, and then reflected across the line $y = -x$ . What are the coordinates of the image of endpoint $A$ after these transformations?

(-7, 7)

(-7, -1)

(-3, -1)

(3, 1)

(-1, -3)

Show answer & explanation

Answer: $(-3, -1)$

Answer

The coordinates of the image of endpoint

A

are

(-3, -1)

.

Translating the point

A(-3, 5)

4 units right and 2 units down gives

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

. Reflecting the point

(1, 3)

across the line

y = -x

swaps and negates the coordinates, giving

(-3, -1)

.

Step-by-Step Solution

1

Apply the translation of 4 units right and 2 units down to the initial coordinates of point

A(-3, 5)

.

The x-coordinate changes by

+4

and the y-coordinate changes by

-2

:

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

.

Translation shifts the coordinates directly by adding the horizontal change to

x

and subtracting the vertical change from

y

.

2

Apply the reflection across the line

y = -x

to the translated point

(1, 3)

.

Swapping and negating both coordinates of

(1, 3)

yields

(-3, -1)

.

The reflection rule across the line

y = -x

maps any point

(x, y)

to

(-y, -x)

.

Key Concept

Composite transformations in the coordinate plane involving translation and reflection across the line

y = -x

.

Question 200Question →

In the standard $(x, y)$ coordinate plane, a triangle has a vertex at $T(-1, 4)$ . The triangle is reflected across the $y$ -axis, and then dilated by a scale factor of $2$ with the center of dilation at the point $(3, 1)$ . What are the coordinates of the image of vertex $T$ after this sequence of transformations?

(5, 7)

(2, 8)

(-1, 7)

(-5, -9)

(7, -5)

Show answer & explanation

Answer: $(-1, 7)$

Answer

The correct coordinates of the image of vertex

T

are

(-1, 7)

.

The correct coordinates are found by first reflecting the vertex across the

y

-axis to get

(1, 4)

, and then applying the dilation formula relative to the center

(3, 1)

with a scale factor of

2

, which yields

(-1, 7)

.

Step-by-Step Solution

1

Reflect the vertex

T(-1, 4)

across the

y

-axis.

T'(1, 4)

Reflecting a point

(x, y)

across the

y

-axis maps it to

(-x, y)

, which changes the sign of the x-coordinate of

T(-1, 4)

to get

(1, 4)

.

2

Apply the dilation with a scale factor of

2

centered at

(3, 1)

to the reflected point

T'(1, 4)

.

T''(-1, 7)

The dilation formula for a point

(x, y)

with scale factor

k

and center

(h, k_c)

is

(h + k(x - h), k_c + k(y - k_c))

. Substituting

x = 1

,

y = 4

,

k = 2

,

h = 3

, and

k_c = 1

yields the x-coordinate

3 + 2(1 - 3) = -1

and the y-coordinate

1 + 2(4 - 1) = 7

.

Key Concept

Composite transformations in the coordinate plane involving reflections and non-origin-centered dilations.

Alternative Method

Instead of calculating algebraically, you can sketch the transformations on a grid. Plot

T(-1, 4)

, reflect it across the y-axis to find

T'(1, 4)

. Mark the center of dilation at

P(3, 1)

. Find the vector from

P

to

T'

, which is

2

units left and

3

units up. Double this vector to

4

units left and

6

units up from

P(3, 1)

, which lands at

(-1, 7)

.

Estimated Time:1m 30s

Coordinate Geometry

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