Coordinate Geometry Practice Questions — ACT — Page 9

Question 161Question →

A straight line graphed on a coordinate grid has a slope of $1.25$ and passes through the point $(4, 11)$ . If the point $(k, -4)$ also lies on this line, what is the value of $k$ ?

Show answer & explanation

Answer: -8

Answer

The value of

k

is

-8

.

The correct answer is

-8

. By using the point-slope formula of a line, we determine the equation to be

y - 11 = 1.25(x - 4)

. Substituting the coordinates of the point

(k, -4)

yields the equation

-4 - 11 = 1.25(k - 4)

. Solving this equation results in

k = -8

.

Step-by-Step Solution

1

Use the point-slope formula of a line,

y - y_1 = m(x - x_1)

, to write the equation of the line.

y - 11 = 1.25(x - 4)

We are given the slope

m = 1.25

and a point

(4, 11)

on the line.

2

Substitute the coordinates of the second point,

(k, -4)

, into the line's equation.

-4 - 11 = 1.25(k - 4)

Since the point

(k, -4)

lies on the line, its coordinates must satisfy the line's equation.

3

Solve the equation for

k

.

k = -8

Simplify the left side to get

-15 = 1.25(k - 4)

, divide both sides by

1.25

to get

-12 = k - 4

, and add

4

to both sides to find

k = -8

.

Key Concept

Writing and solving linear equations using point-slope form.

Question 162Question →

In the standard $(x, y)$ coordinate plane, the line segment $PQ$ has endpoints $P(-3, 5)$ and $Q(1, -1)$ . Segment $PQ$ is reflected across the line $y = -x$ , and the resulting image is then translated $3$ units to the right and $2$ units down to form segment $P'Q'$ . What is the midpoint of segment $P'Q'$ ?

(-5, 3)

(0, -2)

(1, -1)

(-5, -1)

(1, 3)

Show answer & explanation

Answer: $(1, -1)$

Answer

(1, -1)

The correct answer is the coordinate point

(1, -1)

. First, the midpoint of the original segment

PQ

is found using the midpoint formula:

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

. Reflecting this midpoint across the line

y = -x

swaps and negates the coordinates, transforming

(-1, 2)

to

(-2, 1)

. Finally, translating this point

3

units to the right and

2

units down results in the coordinates

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

.

Step-by-Step Solution

1

Find the midpoint of the original segment

PQ

.

The midpoint is

M(-1, 2)

.

Since the midpoint is invariant under linear coordinate transformations, we can find the midpoint of the original segment and transform it directly rather than transforming both endpoints and finding the new midpoint.

2

Reflect the midpoint across the line

y = -x

.

The reflected point is

M_1(-2, 1)

.

Reflecting a point across the line

y = -x

maps the coordinate

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

.

3

Translate the reflected point

3

units to the right and

2

units down.

The final point is

M'(1, -1)

.

Translating a point

3

units right adds

3

to the

x

-coordinate, and translating

2

units down subtracts

2

from the

y

-coordinate:

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

.

Key Concept

Transformations in the Coordinate Plane (Reflection and Translation)

Alternative Method

Instead of transforming the midpoint directly, you can apply the transformations to the endpoints

P(-3, 5)

and

Q(1, -1)

individually. Reflecting both across the line

y = -x

gives

P_1(-5, 3)

and

Q_1(1, -1)

. Translating both points

3

units right and

2

units down gives

P'(-2, 1)

and

Q'(4, -3)

. The midpoint of

P'Q'

is then calculated as

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

.

Estimated Time:1m 30s

Question 163Question →

A line, $L_1$ , is perpendicular to a second line whose equation is $5x - 2y = 15$ . The line $L_1$ intersects the $y$ -axis at $(0, 12)$ and passes through the midpoint of a line segment with endpoints at $(3, k)$ and $(7, 6)$ in the standard $(x, y)$ coordinate plane. What is the value of $k$ ?

Show answer & explanation

Answer: 14

Answer

The correct value of

k

is 14.

The slope of the line

5x - 2y = 15

is found by solving for

y

, yielding

y = \frac{5}{2}x - 7.5

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

\frac{5}{2}

, which is

-\frac{2}{5}

. Given the

y

-intercept

(0, 12)

, the equation of the perpendicular line

L_1

is

y = -\frac{2}{5}x + 12

. The midpoint of the segment with endpoints

(3, k)

and

(7, 6)

is calculated as

\left(\frac{3+7}{2}, \frac{k+6}{2}\right) = \left(5, \frac{k+6}{2}\right)

. Since the midpoint lies on

L_1

, substituting

x = 5

into the line equation gives

y = -\frac{2}{5}(5) + 12 = 10

. Equating this to the midpoint's

y

-coordinate expression gives

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

, which solves to

k = 14

.

Step-by-Step Solution

1

Find the slope of the line

5x - 2y = 15

by converting it to slope-intercept form (

y = mx + b

).

The slope of the line is

\frac{5}{2}

.

Rewriting the equation as

-2y = -5x + 15

and dividing by

-2

isolates

y

and reveals the slope.

2

Determine the perpendicular slope for line

L_1

.

The slope of

L_1

is

-\frac{2}{5}

.

Perpendicular lines have slopes that are negative reciprocals of one another.

3

Formulate the equation of line

L_1

using its slope and the given

y

-intercept

(0, 12)

.

The equation of

L_1

is

y = -\frac{2}{5}x + 12

.

The slope-intercept form is

y = mx + b

, where

m

is the slope and

b

is the

y

-coordinate of the

y

-intercept.

4

Find the midpoint of the line segment with endpoints

(3, k)

and

(7, 6)

in terms of

k

.

The midpoint is

\left(5, \frac{k+6}{2}\right)

.

The midpoint formula calculates the average of the

x

-coordinates and the average of the

y

-coordinates.

5

Substitute the midpoint coordinates into the equation of

L_1

and solve for

k

.

k = 14

Since the midpoint lies on line

L_1

, substituting its

x

and

y

values into the equation must satisfy the equality.

Key Concept

Using perpendicular slopes and the midpoint formula to determine unknown coordinate values.

Question 164Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices at $L(-2, -1)$ , $M(2, 10)$ , and $N(6, 4)$ . A line segment is drawn from vertex $L$ to the midpoint of side $MN$ . What is the length of this line segment?

14

2\sqrt{7}

\sqrt{65}

10

\sqrt{40}

Show answer & explanation

Answer: $10$

Answer

The length of the line segment is

10

.

First, find the midpoint

P

of side

MN

with vertices

M(2, 10)

and

N(6, 4)

using the midpoint formula:

P = \left(\frac{2 + 6}{2}, \frac{10 + 4}{2}\right) = (4, 7)

. Then, calculate the distance between vertex

L(-2, -1)

and the midpoint

P(4, 7)

using the distance formula:

d = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10

.

Step-by-Step Solution

1

Find the coordinates of the midpoint of side

MN

.

The midpoint is

P(4, 7)

.

Apply the midpoint formula

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

to the vertices

M(2, 10)

and

N(6, 4)

, yielding

P = \left(\frac{2 + 6}{2}, \frac{10 + 4}{2}\right) = (4, 7)

.

2

Calculate the distance between vertex

L(-2, -1)

and the midpoint

P(4, 7)

.

The distance is

10

.

Apply the distance formula

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

to points

L(-2, -1)

and

P(4, 7)

, yielding

d = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10

.

Key Concept

Distance and Midpoint Formulas

Question 165Question →

In the standard $(x, y)$ coordinate plane, the line $3x - 4y = 24$ intersects the $x$ -axis at point $A$ and the $y$ -axis at point $B$ . What is the length of the segment $AB$ ?

10

2\sqrt{7}

14

2

\frac{4}{3}

Show answer & explanation

Answer: $10$

Answer

The length of the segment

AB

is

10

.

To find the length of segment

AB

, we first determine the coordinates of points

A

and

B

. The

x

-intercept,

A

, is found by setting

y = 0

, giving

3x = 24 \Rightarrow x = 8

, so

A = (8, 0)

. The

y

-intercept,

B

, is found by setting

x = 0

, giving

-4y = 24 \Rightarrow y = -6

, so

B = (0, -6)

. Using the distance formula, the distance between

(8, 0)

and

(0, -6)

is

\sqrt{(8 - 0)^2 + (0 - (-6))^2} = \sqrt{64 + 36} = \sqrt{100} = 10

. This matches the correct option.

Step-by-Step Solution

1

Find the coordinates of point

A

, the

x

-intercept of the line.

Point

A

is

(8, 0)

.

Set

y = 0

in the equation

3x - 4y = 24

, which gives

3x = 24

, so

x = 8

.

2

Find the coordinates of point

B

, the

y

-intercept of the line.

Point

B

is

(0, -6)

.

Set

x = 0

in the equation

3x - 4y = 24

, which gives

-4y = 24

, so

y = -6

.

3

Calculate the distance between point

A(8, 0)

and point

B(0, -6)

using the distance formula.

The distance is

10

.

The distance formula is

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

. Substituting the coordinates gives

d = \sqrt{(0 - 8)^2 + (-6 - 0)^2} = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10

.

Key Concept

Finding intercepts of a linear equation and calculating the distance between two points on the coordinate plane.

Question 166Question →

A point $P$ in a coordinate plane undergoes a translation of 4 units to the left and 5 units up, followed by a reflection across the $y$ -axis. If the coordinates of the image point after both transformations are $(2, -3)$ , what are the coordinates of the original point $P$ ?

(-6, 2)

(6, -2)

(2, -8)

(-6, -8)

(-2, 8)

Show answer & explanation

Answer: $(2, -8)$

Answer

The coordinates of the original point

P

are

(2, -8)

.

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

(2, -3)

by reversing each transformation in reverse order. First, we reverse the reflection across the

y

-axis. Reflecting across the

y

-axis negates the x-coordinate, so negating the x-coordinate of

(2, -3)

gives the intermediate point

(-2, -3)

. Second, we reverse the translation of 4 units left and 5 units up by translating the intermediate point 4 units right and 5 units down. This gives

x = -2 + 4 = 2

and

y = -3 - 5 = -8

. Thus, the original coordinates of point

P

are

(2, -8)

.

Step-by-Step Solution

1

Reverse the reflection across the

y

-axis by applying it to the final image point

(2, -3)

.

The intermediate point is

(-2, -3)

.

Reflecting a point across the

y

-axis negates its x-coordinate while keeping its y-coordinate the same. Reversing this reflection also negates the x-coordinate.

2

Reverse the translation of 4 units left and 5 units up by translating the intermediate point

(-2, -3)

4 units right and 5 units down.

The original point

P

is

(2, -8)

.

To undo a translation, apply the opposite operations: add 4 to the x-coordinate and subtract 5 from the y-coordinate.

Key Concept

Reversing composite transformations in the coordinate plane

Question 167Question →

A line in the standard $(x, y)$ coordinate plane has a $y$ -intercept that is $4$ units greater than its slope. If the line passes through the point $(5, -2)$ , what is the $y$ -intercept of the line?

Show answer & explanation

Answer: 3

Answer

The correct answer is 3.

The

y

-intercept is found by setting up the equation using the given point and the relationship between the slope and the

y

-intercept, yielding

b = 3

.

Step-by-Step Solution

1

Define the relationship between the slope

m

and the

y

-intercept

b

.

m = b - 4

The problem states the

y

-intercept is

4

units greater than the slope, so

b = m + 4

.

2

Substitute the point

(5, -2)

and the slope expression into the slope-intercept equation

y = mx + b

.

-2 = (b - 4)(5) + b

The line passes through the point

(5, -2)

, so its coordinates must satisfy the line's equation.

3

Solve the algebraic equation for

b

.

b = 3

Expanding and simplifying the equation yields

-2 = 6b - 20

, which gives

6b = 18

, and dividing by

6

results in

b = 3

.

Key Concept

Linear Equations and Graphing

Question 168Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices at $A(-2, -3)$ , $B(6, 1)$ , and $C(1, 6)$ . The length of the altitude from vertex $C$ to side $AB$ can be written in the simplified radical form $a\sqrt{b}$ , where $a$ and $b$ are positive integers. What is the value of $a + b$ ?

Show answer & explanation

Answer: 8

Answer

The final answer is 8.

The altitude length of the triangle is

3\sqrt{5}

. In this simplified radical form, the coefficient

a

is 3 and the radicand

b

is 5. Summing these values gives

3 + 5 = 8

.

Step-by-Step Solution

1

Calculate the area of triangle

ABC

using the shoelace formula or by bounding the triangle in a rectangle.

Area = 30

The area is needed to determine the altitude length using the area formula of a triangle.

2

Calculate the length of the base side

AB

using the distance formula.

AB = 4\sqrt{5}

The base length is required to solve for the height perpendicular to it.

3

Set up the triangle area formula

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

to find the height

h

.

h = 3\sqrt{5}

This solves for the length of the altitude from vertex

C

to side

AB

.

4

Identify

a

and

b

from the simplified radical form

a\sqrt{b} = 3\sqrt{5}

and calculate

a + b

.

8

To provide the final requested sum.

Key Concept

Calculating the altitude of a triangle in the coordinate plane by relating its area and side lengths.

Question 169Question →

A circle in the standard $(x, y)$ coordinate plane has a diameter with one endpoint at $(-3, 5)$ and its center at $(2, -7)$ . What is the $y$ -coordinate of the other endpoint of the diameter?

Show answer & explanation

Answer: -19

Answer

The y-coordinate of the other endpoint of the diameter is -19.

Because the center of a circle is the midpoint of any diameter, the midpoint formula applies. For the y-coordinate, the equation is

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

. Substituting the y-coordinate of the given endpoint (

5

) and the center (

-7

) yields

-7 = \frac{5 + y_2}{2}

. Multiplying both sides by 2 gives

-14 = 5 + y_2

. Subtracting 5 from both sides results in

-19

.

Step-by-Step Solution

1

Identify the relationship between the circle's center and its diameter.

The center of the circle,

(2, -7)

, is the midpoint of the diameter.

By definition, the center of a circle bisects any diameter, making it the midpoint of the diameter's endpoints.

2

Apply the midpoint formula to set up an equation for the y-coordinate.

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

The y-coordinate of a midpoint is the average of the y-coordinates of the two endpoints.

3

Substitute the given values into the formula and solve for the unknown y-coordinate.

y_2 = -19

Substituting the given y-coordinates yields

-7 = \frac{5 + y_2}{2}

. Multiplying by 2 gives

-14 = 5 + y_2

, and subtracting 5 from both sides yields

-19

.

Key Concept

Finding a missing endpoint given the midpoint and one endpoint

Estimated Time:1m 0s

Question 170Question →

A geometric shape on a coordinate grid has a vertex at $T(2, -5)$ . The shape is translated $3$ units to the left and $4$ units up, and then the resulting image is reflected across the $y$ -axis. What are the coordinates of the final image of vertex $T$ ?

(-5, -1)

(-6, -8)

(1, -1)

(1, -9)

(-5, -9)

Show answer & explanation

Answer: $(1, -1)$

Answer

The correct coordinates of the final image of vertex

T

are

(1, -1)

.

To find the coordinates of the final image, we apply the transformations in the specified order. First, translating the point

(2, -5)

by

3

units to the left subtracts

3

from the

x

-coordinate, and translating it

4

units up adds

4

to the

y

-coordinate:

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

. Second, reflecting the point

(-1, -1)

across the

y

-axis negates the

x

-coordinate while keeping the

y

-coordinate the same:

(-(-1), -1) = (1, -1)

.

Step-by-Step Solution

1

Apply the horizontal and vertical translations to the pre-image point

T(2, -5)

.

Translating

3

units left subtracts

3

from the

x

-coordinate (

2 - 3 = -1

). Translating

4

units up adds

4

to the

y

-coordinate (

-5 + 4 = -1

). The intermediate point is

(-1, -1)

.

A translation of

h

units horizontally and

k

units vertically maps

(x, y)

to

(x + h, y + k)

.

2

Reflect the intermediate point

(-1, -1)

across the

y

-axis.

Reflecting across the

y

-axis negates the

x

-coordinate while keeping the

y

-coordinate the same:

(-(-1), -1) = (1, -1)

.

A reflection across the

y

-axis maps any point

(x, y)

to

(-x, y)

.

Key Concept

Transformations in the Coordinate Plane

Question 171Question →

A line in the standard $(x, y)$ coordinate plane has a slope of $\frac{3}{4}$ and contains the points $P(a, 3)$ and $Q(5, b)$ . If the distance between $P$ and $Q$ is $10$ units, what is one possible value of $a + b$ ?

-6

2

6

12

22

Show answer & explanation

Answer: $6$

Answer

The correct answer is

6

.

The correct answer is

6

. Using the slope definition

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

, we write the slope equation as

\frac{b - 3}{5 - a} = \frac{3}{4}

, which gives

b - 3 = \frac{3}{4}(5 - a)

. Substituting this expression into the distance formula

d = \sqrt{(5 - a)^2 + (b - 3)^2} = 10

yields

10 = \sqrt{(5 - a)^2 + \left(\frac{3}{4}(5 - a)\right)^2} = \frac{5}{4}|5 - a|

. Solving for the horizontal change gives

|5 - a| = 8

, which means

5 - a = \pm 8

. If

5 - a = 8

, then

a = -3

and

b - 3 = 6 \Rightarrow b = 9

. The sum of these coordinates is

-3 + 9 = 6

.

Step-by-Step Solution

1

Set up the slope equation using the definition of slope:

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

.

\frac{b - 3}{5 - a} = \frac{3}{4}

, which simplifies to

b - 3 = \frac{3}{4}(5 - a)

.

This establishes a relationship between the coordinate differences of points

P

and

Q

based on the given slope.

2

Set up the distance equation using the distance formula:

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

.

10 = \sqrt{(5 - a)^2 + (b - 3)^2}

.

This uses the given distance of

10

units to constrain the coordinate differences.

3

Substitute the slope relationship

b - 3 = \frac{3}{4}(5 - a)

into the distance equation and solve for

(5 - a)

.

10 = \sqrt{(5 - a)^2 + \left(\frac{3}{4}(5 - a)\right)^2} \Rightarrow 10 = \sqrt{(5 - a)^2 \left(1 + \frac{9}{16}\right)} \Rightarrow 10 = \frac{5}{4}|5 - a| \Rightarrow |5 - a| = 8

. Thus,

5 - a = 8

or

5 - a = -8

.

Substituting one variable simplifies the system to a single quadratic equation in terms of the horizontal change.

4

Solve for the two possible values of

a

and calculate the corresponding values of

b

.

Case 1: If

5 - a = 8

, then

a = -3

. Substituting this back gives

b - 3 = \frac{3}{4}(8) = 6 \Rightarrow b = 9

. The sum is

a + b = -3 + 9 = 6

. Case 2: If

5 - a = -8

, then

a = 13

. Substituting this back gives

b - 3 = \frac{3}{4}(-8) = -6 \Rightarrow b = -3

. The sum is

a + b = 13 + (-3) = 10

.

Evaluating both branches of the absolute value equation yields the two valid coordinate combinations that satisfy both the slope and distance criteria.

Key Concept

Calculating the slope of a line and using it in conjunction with the distance formula to find unknown coordinates on the coordinate plane.

Question 172Question →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 - 10x - 4y = 0$ . The line $y = x$ intersects the circle at the origin $(0, 0)$ and at a second point $P$ . What is the $x$ -coordinate of point $P$ ?

Show answer & explanation

Answer: 7

Answer

The correct answer is 7.

To find the points of intersection between the circle and the line, we substitute the line's equation

y = x

into the circle's equation

x^2 + y^2 - 10x - 4y = 0

. This substitution yields

2x^2 - 14x = 0

. Factoring out

2x

gives

2x(x - 7) = 0

, which has solutions

x = 0

and

x = 7

. The solution

x = 0

corresponds to the origin

(0, 0)

. Therefore, the

x

-coordinate of the second point

P

is

7

.

Step-by-Step Solution

1

Substitute

y = x

into the equation of the circle.

x^2 + x^2 - 10x - 4x = 0

Since the line is

y = x

, any point of intersection must satisfy both equations.

2

Combine like terms to simplify the quadratic equation.

2x^2 - 14x = 0

Grouping the quadratic and linear terms yields a single equation in terms of

x

.

3

Factor the quadratic equation.

2x(x - 7) = 0

Factoring allows us to identify the individual roots of the equation.

4

Solve for

x

and identify the value for point

P

.

x = 7

The solution

x = 0

corresponds to the origin. Thus, the

x

-coordinate of the second point

P

must be

7

.

Key Concept

Finding points of intersection between a line and a circle in the coordinate plane.

Question 173Question →

In the standard $(x, y)$ coordinate plane, point $P(2, -3)$ is an endpoint of line segment $PQ$ , and $M(-1, 5)$ is the midpoint of $PQ$ . What is the distance from endpoint $Q$ to the point $(0, 10)$ ?

2

\sqrt{7}

5

\sqrt{26}

7

Show answer & explanation

Answer: 5

Answer

5

To solve this problem, first determine the coordinates of endpoint

Q(x, y)

using the midpoint formula. Since

M(-1, 5)

is the midpoint of segment

PQ

and

P(2, -3)

is one endpoint, the horizontal coordinate satisfies

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

, which yields

x = -4

. The vertical coordinate satisfies

5 = \frac{-3 + y}{2}

, which yields

y = 13

. Thus, the coordinates of

Q

are

(-4, 13)

. Next, calculate the distance between

Q(-4, 13)

and the target point

(0, 10)

using the distance formula:

d = \sqrt{(0 - (-4))^2 + (10 - 13)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5

. This matches the correct value.

Step-by-Step Solution

1

Set up equations using the midpoint formula to determine the coordinates of endpoint

Q(x_Q, y_Q)

.

-1 = \frac{2 + x_Q}{2}

and

5 = \frac{-3 + y_Q}{2}

The coordinates of a midpoint are the averages of the coordinates of the two endpoints.

2

Solve the midpoint equations for the coordinates of

Q

.

x_Q = -4

and

y_Q = 13

, so

Q = (-4, 13)

Multiplying by two and subtracting the coordinates of the first endpoint isolates the variables.

3

Apply the distance formula to find the distance between

Q(-4, 13)

and the target point

(0, 10)

.

d = \sqrt{(0 - (-4))^2 + (10 - 13)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

The distance formula calculates the straight line distance between two points in a coordinate plane.

Key Concept

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

Estimated Time:1m 30s

Question 174Question →

In the standard $(x, y)$ coordinate plane, point $P$ is rotated $90^\circ$ counterclockwise about the origin, and then translated $3$ units to the left and $5$ units down. If the coordinates of the resulting image point $P'$ are $(-6, 2)$ , what are the coordinates of the original point $P$ ?

(-3, 9)

(-7, -3)

(7, -3)

(7, 3)

(3, -9)

Show answer & explanation

Answer: $(7, 3)$

Answer

(7, 3)

The correct answer is

(7, 3)

. Working backward from the final image point

(-6, 2)

, we first reverse the translation of

3

units left and

5

units down by translating the point

3

units right and

5

units up, yielding

(-6+3, 2+5) = (-3, 7)

. We then reverse the

90^\circ

counterclockwise rotation by performing a

90^\circ

clockwise rotation about the origin. The rule for a

90^\circ

clockwise rotation is

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

, which maps

(-3, 7)

to

(7, 3)

.

Step-by-Step Solution

1

Identify the sequence of transformations and set up the backward process.

The forward process is: Point

P \rightarrow

Rotate

90^\circ

counterclockwise

\rightarrow

Translate left

3

and down

5 \rightarrow P'(-6, 2)

. To find the original point

P

, we must reverse these steps starting from

P'

: Translate right

3

and up

5 \rightarrow

Rotate

90^\circ

clockwise

\rightarrow

Point

P

.

Working backward requires reversing both the order and the direction of each transformation in the sequence.

2

Apply the reverse translation to the image point

P'(-6, 2)

.

Translating

3

units to the right adds

3

to the

x

-coordinate:

-6 + 3 = -3

. Translating

5

units up adds

5

to the

y

-coordinate:

2 + 5 = 7

. This gives the intermediate point

P_1(-3, 7)

.

Reversing a translation of

3

units left and

5

units down requires adding

3

to the

x

-coordinate and

5

to the

y

-coordinate.

3

Apply the reverse rotation to the intermediate point

P_1(-3, 7)

.

The reverse of a

90^\circ

counterclockwise rotation is a

90^\circ

clockwise rotation. The coordinate rule for a

90^\circ

clockwise rotation about the origin is

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

. Applying this rule to

P_1(-3, 7)

yields

(7, -(-3)) = (7, 3)

.

Rotating a point

90^\circ

clockwise about the origin swaps the coordinates and negates the new

y

-coordinate.

Key Concept

Reversing composite transformations in the coordinate plane

Question 175Question →

Line $A$ passes through the points $(-3, 4)$ and $(5, 8)$ in a coordinate plane. Line $B$ is perpendicular to Line $A$ and passes through the point $(2, -1)$ . What is the $y$ -coordinate of the point where Line $B$ crosses the $y$ -axis?

Show answer & explanation

Answer: 3

Answer

The correct answer is 3.

The correct answer is 3. Finding the slope of the first line gives 0.5. The negative reciprocal of 0.5 is -2, which is the slope of the perpendicular line. Setting up the slope equation between the point (2, -1) and the y-intercept (0, 3) confirms the y-coordinate is indeed 3.

Step-by-Step Solution

1

Calculate the slope of Line A.

The slope of Line A is 0.5.

Using the slope formula

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

for points

(-3, 4)

and

(5, 8)

gives

m_A = \frac{8 - 4}{5 - (-3)} = \frac{4}{8} = 0.5

.

2

Find the slope of Line B.

The slope of Line B is -2.

Because Line B is perpendicular to Line A, its slope must be the negative reciprocal of the slope of Line A:

m_B = -\frac{1}{0.5} = -2

.

3

Determine the y-intercept of Line B.

The y-coordinate of the y-intercept is 3.

Line B passes through

(2, -1)

and

(0, b)

. Using the slope formula:

\frac{b - (-1)}{0 - 2} = -2 \implies \frac{b + 1}{-2} = -2 \implies b + 1 = 4 \implies b = 3

.

Key Concept

The slope of a line measures its steepness, calculated as the change in

y

divided by the change in

x

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

Alternative Method

Instead of using the slope formula directly to solve for the y-intercept, write the equation of Line B in point-slope form:

y - y_1 = m(x - x_1)

. Substituting the point

(2, -1)

and slope

-2

gives

y - (-1) = -2(x - 2)

, which simplifies to

y + 1 = -2x + 4

, or

y = -2x + 3

. To find the y-intercept, set

x = 0

, which yields

y = 3

.

Estimated Time:1m 30s

Question 176Question →

In the standard $(x, y)$ coordinate plane, a line passes through the point $(3, 1)$ and has a slope of $\frac{2}{5}$ . If the line intersects the $y$ -axis at the point $(0, b)$ , what is the value of $b$ ?

-\frac{13}{2}

-\frac{4}{5}

\frac{11}{5}

-\frac{1}{5}

\frac{5}{4}

Show answer & explanation

Answer: $-\frac{1}{5}$

Answer

The value of the

y

-coordinate of the

y

-intercept is

-\frac{1}{5}

.

To find the

y

-intercept of the line, we can use the point-slope equation of a line:

y - y_1 = m(x - x_1)

, where

m

is the slope and

(x_1, y_1)

is a point on the line. Substituting the given point

(3, 1)

and slope

\frac{2}{5}

yields

y - 1 = \frac{2}{5}(x - 3)

. Since the

y

-intercept occurs at

x = 0

, we substitute

0

for

x

to find the corresponding

y

-value:

y - 1 = \frac{2}{5}(0 - 3)

, which simplifies to

y - 1 = -\frac{6}{5}

. Adding

1

to both sides gives

y = -\frac{1}{5}

. Therefore, the value of the

y

-coordinate of the

y

-intercept is

-\frac{1}{5}

.

Step-by-Step Solution

1

Identify the point-slope form of a linear equation.

y - y_1 = m(x - x_1)

This formula allows us to write the equation of a line when we know its slope (

m

) and a point on the line (

(x_1, y_1)

).

2

Substitute the given slope

m = \frac{2}{5}

and point

(3, 1)

into the formula.

y - 1 = \frac{2}{5}(x - 3)

This establishes the equation for the specific line described in the problem.

3

Substitute

x = 0

to find the

y

-intercept of the line.

y - 1 = \frac{2}{5}(0 - 3) \implies y - 1 = -\frac{6}{5}

The

y

-intercept of any graph is the point where it crosses the

y

-axis, which always has an

x

-coordinate of

0

.

4

Solve for

y

to determine the value of

b

.

y = 1 - \frac{6}{5} = -\frac{1}{5}

, so

b = -\frac{1}{5}

Isolating

y

gives the

y

-coordinate of the

y

-intercept, which is

b

since the point is defined as

(0, b)

.

Key Concept

Writing and evaluating linear equations in point-slope form to find intercepts.

Question 177Question →

In the standard $(x, y)$ coordinate plane, the equation $x^2 + y^2 + 8x - 12y + c = 0$ defines a circle. If this circle is tangent to the $y$ -axis, what is the value of the constant $c$ ?

16

36

48

52

68

Show answer & explanation

Answer: $36$

Answer

The value of the constant

c

is

36

.

By completing the square on

x^2 + y^2 + 8x - 12y + c = 0

, we get the standard form equation

(x+4)^2 + (y-6)^2 = 52-c

. This shows the circle's center is at

(-4, 6)

and its radius squared is

r^2 = 52-c

. Because the circle is tangent to the

y

-axis, the radius is the horizontal distance from the center to the

y

-axis, which is

4

units. Therefore, the radius squared is

16

. Equating the two expressions for the radius squared gives

52-c = 16

, which solves to

c = 36

.

Step-by-Step Solution

1

Group the

x

and

y

terms and move the constant to the right side of the equation.

(x^2 + 8x) + (y^2 - 12y) = -c

To prepare the equation for completing the square.

2

Complete the square for both the

x

and

y

groups by adding the square of half the coefficient of the linear terms to both sides.

(x^2 + 8x + 16) + (y^2 - 12y + 36) = -c + 16 + 36

, which simplifies to

(x+4)^2 + (y-6)^2 = 52-c

.

To rewrite the equation in standard circle form

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

.

3

Determine the radius of the circle based on the geometric condition of being tangent to the

y

-axis.

The center of the circle is

(-4, 6)

. Since the circle is tangent to the

y

-axis (the line

x=0

), the radius

r

is the horizontal distance from the center to the

y

-axis, which is

|-4| = 4

.

A circle tangent to a vertical line has its radius equal to the horizontal distance from its center to that line.

4

Set the radius squared equal to the right side of the standard equation and solve for

c

.

r^2 = 4^2 = 16

. Setting

52 - c = 16

gives

c = 36

.

In standard form, the right-hand side represents

r^2

.

Key Concept

Converting the general form of a circle equation to standard form

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

by completing the square, and using the geometric definition of tangency to determine the radius.

Question 178Question →

In the standard $(x, y)$ coordinate plane, a line is defined by the equation $2x + 3y = 22$ . A second line passes through the origin $(0, 0)$ and intersects the first line at a point where $x = 5$ . What is the slope of this second line?

Show answer & explanation

Answer: 0.8

Answer

The slope of the second line is

0.8

(or

\frac{4}{5}

).

The intersection point has an

x

-coordinate of

5

. Substituting this into

2x + 3y = 22

gives

2(5) + 3y = 22

, which simplifies to

3y = 12

, or

y = 4

. Thus, the intersection point is

(5, 4)

. The second line passes through

(0, 0)

and

(5, 4)

. Using the slope formula, the slope is

\frac{4 - 0}{5 - 0} = 0.8

.

Step-by-Step Solution

1

Substitute

x = 5

into the equation of the first line to find the

y

-coordinate of the intersection point.

The intersection point is

(5, 4)

.

Since the two lines intersect at

x = 5

, the intersection point must satisfy the equation of the first line.

2

Use the slope formula to find the slope of the line connecting

(0, 0)

and

(5, 4)

.

The slope is

0.8

.

The second line passes through the origin

(0, 0)

and the intersection point

(5, 4)

, so its slope is the ratio of the change in

y

to the change in

x

.

Key Concept

Finding the slope of a line given two points on the coordinate plane, where one point is determined by the intersection of two linear paths.

Alternative Method

Since the second line passes through the origin

(0, 0)

, its equation is of the form

y = mx

, where

m

is the slope. At the intersection point

(5, y)

, we have

y = 5m

. We can substitute this directly into the first line's equation:

2x + 3y = 22 \Rightarrow 2(5) + 3(5m) = 22 \Rightarrow 10 + 15m = 22 \Rightarrow 15m = 12 \Rightarrow m = \frac{12}{15} = 0.8

.

Estimated Time:1m 30s

Question 179Question →

A coordinate grid is laid over a map of a city. A subway station is located at $S(-1, 9)$ and a bus terminal is located at $B(7, -1)$ . A passenger transfer center is built at the midpoint of the line segment connecting the subway station and the bus terminal. What is the distance, in grid units, between the transfer center and a parking garage located at $(7, 7)$ ?

Show answer & explanation

Answer: 5

Answer

The distance between the passenger transfer center and the parking garage is 5 grid units.

First, the midpoint is determined by averaging the

x

-coordinates and

y

-coordinates of the endpoints:

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

and

y = \frac{9 - 1}{2} = 4

, giving the transfer center coordinates of

(3, 4)

. Second, the distance between

(3, 4)

and

(7, 7)

is found using the distance formula:

d = \sqrt{(7 - 3)^2 + (7 - 4)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Calculate the coordinates of the midpoint of the line segment connecting the subway station

S(-1, 9)

and the bus terminal

B(7, -1)

.

The midpoint is

M(3, 4)

.

The transfer center is located exactly halfway between the two endpoints, which corresponds to the midpoint.

2

Calculate the distance between the midpoint

M(3, 4)

and the parking garage at

(7, 7)

.

The distance is 5.

Applying the distance formula to

M(3, 4)

and

(7, 7)

yields the straight-line distance.

Key Concept

Distance and Midpoint Formulas

Alternative Method

Instead of using the distance formula, one can draw a right triangle on the grid with vertices at

(3, 4)

,

(7, 7)

, and

(7, 4)

. The horizontal leg has a length of

|7 - 3| = 4

, and the vertical leg has a length of

|7 - 4| = 3

. Using the Pythagorean theorem, the hypotenuse (distance) is

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

.

Estimated Time:1m 30s

Question 180Question →

In the standard $(x, y)$ coordinate plane, the circle defined by the equation $x^2 + y^2 - 12x - 4y + 31 = 0$ is translated $3$ units to the right and $5$ units down. A line passing through the origin with a non-zero slope $m$ is tangent to this translated circle. What is the value of $m$ ?

Show answer & explanation

Answer: -0.75

Answer

The correct answer is -0.75.

The correct answer is -0.75. By completing the square on the original equation, we find the circle

(x-6)^2 + (y-2)^2 = 9

with center

(6, 2)

and radius

3

. Translating the circle shifts the center to

(9, -3)

. A line passing through the origin with slope

m

has the equation

mx - y = 0

. For this line to be tangent to the circle, its perpendicular distance from the center

(9, -3)

must equal the radius

3

. Using the distance formula, we get the equation

\frac{|9m + 3|}{\sqrt{m^2 + 1}} = 3

. Simplifying gives

|3m + 1| = \sqrt{m^2 + 1}

. Squaring both sides yields

9m^2 + 6m + 1 = m^2 + 1

, which simplifies to

8m^2 + 6m = 0

. The non-zero solution is

m = -0.75

.

Step-by-Step Solution

1

Complete the square for the original circle equation

x^2 + y^2 - 12x - 4y + 31 = 0

.

(x - 6)^2 + (y - 2)^2 = 9

.

Grouping the

x

and

y

terms and completing the square yields the standard form of a circle, showing that the original center is

(6, 2)

and the radius

r = \sqrt{9} = 3

.

2

Apply the translation of

3

units right and

5

units down to the center

(6, 2)

.

New center is

(6 + 3, 2 - 5) = (9, -3)

.

Translating a point in the coordinate plane shifts the coordinates directly:

x

-coordinate increases by

3

and

y

-coordinate decreases by

5

. The radius remains

3

.

3

Set up the distance from the new center

(9, -3)

to the tangent line

y = mx

.

Distance equation:

\frac{|9m - (-3)|}{\sqrt{m^2 + 1}} = 3

.

The line

y = mx

can be rewritten as

mx - y = 0

. The perpendicular distance from the center to a tangent line must equal the radius of the circle, which is

3

.

4

Simplify the distance equation and solve for

m

.

|3m + 1| = \sqrt{m^2 + 1}

.

Factoring out

3

from the numerator and dividing both sides by

3

simplifies the equation before squaring.

5

Square both sides of the equation and solve the resulting quadratic equation.

2m(4m + 3) = 0 \implies m = 0

or

m = -0.75

.

Squaring both sides yields

(3m + 1)^2 = m^2 + 1 \implies 9m^2 + 6m + 1 = m^2 + 1 \implies 8m^2 + 6m = 0

. Since we are looking for a non-zero slope, the only valid solution is

m = -0.75

.

Key Concept

Using completing the square, coordinate translations, and the point-to-line distance formula to solve circle tangency problems

Coordinate Geometry

Answer

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Key Concept

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