Coordinate Geometry Practice Questions — ACT — Page 3

Question 41Question →

In the standard $(x,y)$ coordinate plane, the point $P(3, -4)$ is translated $5$ units to the left and $2$ units up to map onto point $P'$ . What are the coordinates of $P'$ ?

(-2, -6)

(8, -2)

(-2, -2)

(8, -6)

(1, 1)

Show answer & explanation

Answer: $(-2, -2)$

Answer

The coordinates of point

P'

are

(-2, -2)

.

To find the coordinates of the image point

P'

after translation, apply the shifts to the coordinates of the pre-image point

P(3, -4)

. Translating

5

units left subtracts

5

from the x-coordinate:

3 - 5 = -2

. Translating

2

units up adds

2

to the y-coordinate:

-4 + 2 = -2

. Thus, the coordinates of

P'

are

(-2, -2)

.

Step-by-Step Solution

1

Determine the effect of translating

5

units to the left on the x-coordinate.

Subtract

5

from the x-coordinate:

3 - 5 = -2

.

Horizontal translations to the left decrease the x-value.

2

Determine the effect of translating

2

units up on the y-coordinate.

Add

2

to the y-coordinate:

-4 + 2 = -2

.

Vertical translations upward increase the y-value.

3

Combine the new coordinates to find the image point

P'

.

P' = (-2, -2)

The translation maps the original point

P(3, -4)

to the new coordinates

P'(-2, -2)

.

Key Concept

Translating a point

(x, y)

horizontally by

h

units and vertically by

k

units yields the image point

(x + h, y + k)

. Shifts to the left and down correspond to negative values for

h

and

k

, while shifts to the right and up correspond to positive values.

Question 42Question →

In the standard $(x, y)$ coordinate plane, what is the distance, in coordinate units, between the points $(1, 2)$ and $(4, 6)$ ?

\frac{4}{3}

5

\sqrt{7}

7

25

Show answer & explanation

Answer: $5$

Answer

The distance is

5

coordinate units.

The correct answer is

5

because applying the distance formula

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

to the points

(1, 2)

and

(4, 6)

yields

\sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Identify the coordinates of the two given points.

Let

(x_1, y_1) = (1, 2)

and

(x_2, y_2) = (4, 6)

.

This establishes the coordinate values for the distance formula.

2

Calculate the difference between the

x

-coordinates and the difference between the

y

-coordinates.

x_2 - x_1 = 4 - 1 = 3

and

y_2 - y_1 = 6 - 2 = 4

.

These differences represent the horizontal and vertical side lengths of the right triangle formed by the two points.

3

Apply the distance formula

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

.

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

.

Squaring the coordinate differences, adding them, and taking the square root yields the straight-line distance.

Key Concept

Using the distance formula to find the straight-line distance between two points in the coordinate plane.

Question 43Question →

Two vertices of an equilateral triangle are located at the points $(\frac{1}{2}, 2)$ and $(\frac{5}{2}, 2)$ in the standard $(x,y)$ coordinate plane. If the third vertex is located above the given line segment, what are its coordinates?

(\frac{3}{4}, 2 + \sqrt{3})

(\frac{3}{2}, 2 + \sqrt{3})

(\frac{7}{4}, 2 + \sqrt{3})

(\frac{3}{2}, \sqrt{7})

(\frac{3}{2}, 4)

Show answer & explanation

Answer: $(\frac{3}{2}, 2 + \sqrt{3})$

Answer

(\frac{3}{2}, 2 + \sqrt{3})

The midpoint of the base segment is at

(\frac{3}{2}, 2)

and the vertical height of the equilateral triangle is

\sqrt{3}

. Since the vertex lies above the segment, we add the height to the

y

-coordinate of the midpoint, giving the coordinates

(\frac{3}{2}, 2 + \sqrt{3})

.

Step-by-Step Solution

1

Find the midpoint and length of the segment connecting the two given vertices

(\frac{1}{2}, 2)

and

(\frac{5}{2}, 2)

.

Since both points lie on the horizontal line

y = 2

, the distance between them is

\frac{5}{2} - \frac{1}{2} = 2

. The

x

-coordinate of the midpoint is

\frac{\frac{1}{2} + \frac{5}{2}}{2} = \frac{3}{2}

, so the midpoint is at

(\frac{3}{2}, 2)

.

The third vertex of an equilateral triangle lies on the perpendicular bisector of the opposite side, which passes through its midpoint.

2

Calculate the height of the equilateral triangle using the Pythagorean theorem.

The side length of the triangle is

2

. The distance from a vertex to the midpoint of the opposite side is

1

. The height

h

satisfies

1^2 + h^2 = 2^2 \Rightarrow 1 + h^2 = 4 \Rightarrow h = \sqrt{3}

.

The height of an equilateral triangle divides it into two

30^\circ\text{-}60^\circ\text{-}90^\circ

right triangles.

3

Determine the coordinates of the third vertex by applying the height to the midpoint.

Since the base is horizontal, the altitude is vertical. Thus, the third vertex has the same

x

-coordinate as the midpoint,

\frac{3}{2}

. The

y

-coordinate is the

y

-coordinate of the midpoint plus the height, which is

2 + \sqrt{3}

.

The vertex must be located above the segment, so we add the height to the

y

-coordinate of the midpoint.

Key Concept

Distance and Midpoint Formulas

Question 44Question →

A circle in the standard $(x, y)$ coordinate plane has its center at $(2, -1)$ and passes through the point $(5, 3)$ . What is the diameter, in coordinate units, of the circle?

5

10

14

25

50

Show answer & explanation

Answer: 10

Answer

10

The distance between the center of the circle at

(2, -1)

and the point on the circle

(5, 3)

represents the radius (

r

). Applying the distance formula:

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

. Since the diameter of a circle is twice its radius, the diameter is

2 \times 5 = 10

.

Step-by-Step Solution

1

Use the distance formula to find the radius of the circle, which is the distance between the center

(2, -1)

and the point

(5, 3)

.

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

The radius of a circle is the distance between its center and any point on its boundary.

2

Multiply the radius by 2 to find the diameter of the circle.

d = 2r = 2(5) = 10

The diameter of a circle is always twice the length of its radius.

Key Concept

Calculating the radius of a circle using the distance formula and doubling it to find the diameter.

Question 45Question →

An ellipse in the standard $(x, y)$ coordinate plane is defined by the equation $25x^2 + 9y^2 - 100x + 54y - 44 = 0$ . Which of the following points is a focus of this ellipse?

(-2, 7)

(2, -1)

(2, 1)

(6, -3)

(2, 19)

Show answer & explanation

Answer: $(2, 1)$

Answer

(2, 1)

The correct answer is the point

(2, 1)

because rewriting the general equation of the ellipse in standard form gives

\frac{(x-2)^2}{9} + \frac{(y+3)^2}{25} = 1

. The center is

(2, -3)

and the major axis is vertical with a focal distance of

c = \sqrt{25 - 9} = 4

. Adding this distance to the

y

-coordinate of the center yields the focus

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

.

Step-by-Step Solution

1

Group the

x

and

y

terms and factor out the coefficients.

25(x^2 - 4x) + 9(y^2 + 6y) = 44

This prepares the algebraic equation for completing the square.

2

Complete the square for both the

x

and

y

groups.

25(x-2)^2 + 9(y+3)^2 = 225

Adding

25 \times 4 = 100

and

9 \times 9 = 81

to both sides maintains equality while converting the quadratic expressions into perfect square trinomials.

3

Divide both sides by

225

to write the equation in standard form.

\frac{(x-2)^2}{9} + \frac{(y+3)^2}{25} = 1

The standard form of an ellipse equation is

\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1

(for a vertical major axis), which directly reveals the center

(h, k)

and the axis parameters.

4

Identify the center, axis lengths, and calculate the focal distance

c

.

Center is

(2, -3)

,

a^2 = 25

,

b^2 = 9

. Thus,

c = \sqrt{25 - 9} = 4

.

The focal distance

c

for an ellipse is determined by the relation

c = \sqrt{a^2 - b^2}

.

5

Determine the coordinates of the foci.

Foci are

(2, -3 \pm 4)

, which simplifies to

(2, 1)

and

(2, -7)

.

Since

a^2 = 25

is under the

y

-term, the ellipse is vertically oriented, meaning the foci lie on the vertical line passing through the center.

Key Concept

Rewriting the general equation of an ellipse to standard form and finding its foci.

Estimated Time:2m 30s

Question 46Question →

On a coordinate plane, triangle $PQR$ has vertices $P(2, 3)$ , $Q(5, 3)$ , and $R(2, 7)$ . The triangle undergoes a sequence of three transformations:

1. A dilation centered at the point $C(1, 1)$ with a scale factor of $2$ .
2. A reflection across the line $y = -x$ .
3. A translation of $3$ units to the left and $4$ units up.

What are the coordinates of the final image of vertex $Q$ after this sequence of transformations?

(-9, -6)

(2, 13)

(-2, -13)

(-8, -5)

(-7, -4)

Show answer & explanation

Answer: $(-8, -5)$

Answer

The final image of vertex

Q

is located at the coordinates

(-8, -5)

.

Applying the transformations sequentially yields the correct result. First, the vector from the dilation center at

(1, 1)

to

(5, 3)

is

(4, 2)

. Scaling this vector by

2

gives

(8, 4)

, which when added back to

(1, 1)

places the intermediate image at

(9, 5)

. Second, reflecting

(9, 5)

across the line

y = -x

yields

(-5, -9)

. Third, translating this point

3

units left and

4

units up results in

(-5 - 3, -9 + 4) = (-8, -5)

.

Step-by-Step Solution

1

Find the vector from the center of dilation

C(1, 1)

to the point

Q(5, 3)

.

Vector

\vec{CQ} = (5 - 1, 3 - 1) = (4, 2)

.

To perform a dilation centered at a point other than the origin, we must first find the displacement of the target point relative to that center.

2

Multiply the vector by the scale factor of

2

and add it back to the coordinates of the center

C(1, 1)

.

First intermediate point

Q' = (1, 1) + 2(4, 2) = (1 + 8, 1 + 4) = (9, 5)

.

This scales the distance from the center of dilation by

2

and finds the coordinate of the image point

Q'

.

3

Apply the reflection rule for the line

y = -x

to the point

Q'(9, 5)

.

Second intermediate point

Q'' = (-5, -9)

.

Reflecting a coordinate

(x, y)

across the line

y = -x

swaps and negates both coordinates, mapping

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

.

4

Translate the point

Q''(-5, -9)

by subtracting

3

from the x-coordinate and adding

4

to the y-coordinate.

Final point

Q''' = (-5 - 3, -9 + 4) = (-8, -5)

.

Translating left reduces the x-value, and translating up increases the y-value.

Key Concept

Composite transformations in the coordinate plane including non-origin dilations, reflections, and translations.

Question 47Question →

The table below shows corresponding values of $x$ and $y$ for a linear relationship in the standard $(x, y)$ coordinate plane.

$x$	$y$
$-3$	$13$
$1$	$5$
$5$	$a$
$b$	$-7$

What is the value of $a + b$ ?

4

8

-8

28

-2

Show answer & explanation

Answer: 4

Answer

4

The correct answer is

4

. The slope of the line is determined to be

-2

using the points

(-3, 13)

and

(1, 5)

. This gives the equation of the line as

y = -2x + 7

. Substituting

x = 5

yields

a = -3

, and substituting

y = -7

yields

b = 7

. Adding these values together gives

4

.

Step-by-Step Solution

1

Calculate the slope of the linear relationship using the points

(-3, 13)

and

(1, 5)

.

Slope

m = -2

Since the relationship is linear, the slope is constant and is given by

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

.

2

Determine the equation of the line using the point-slope formula with point

(1, 5)

and slope

-2

.

Equation is

y = -2x + 7

Using

y - y_1 = m(x - x_1)

allows us to find the relation between any

x

and

y

coordinate on this line.

3

Find the value of

a

by substituting

x = 5

into the linear equation.

a = -3

The table indicates that when

x = 5

, the

y

-value is

a

.

4

Find the value of

b

by substituting

y = -7

into the linear equation and solving for

x

.

b = 7

The table indicates that when

y = -7

, the

x

-value is

b

.

5

Sum the values of

a

and

b

.

a + b = 4

The question asks for the sum

a + b

.

Key Concept

Linear Equations and Graphing

Estimated Time:1m 30s

Question 48Question →

In the standard $(x, y)$ coordinate plane, a line segment has endpoints at $(3, -4)$ and $(9, 8)$ . What is the $y$ -coordinate of the midpoint of this line segment?

Show answer & explanation

Answer: 2

Answer

The

y

-coordinate of the midpoint is

2

.

The

y

-coordinate of the midpoint is calculated by finding the average of the

y

-coordinates of the endpoints:

\frac{y_1 + y_2}{2}

. Substituting

y_1 = -4

and

y_2 = 8

gives

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

.

Step-by-Step Solution

1

Identify the

y

-coordinates of the two given endpoints

(3, -4)

and

(9, 8)

.

y_1 = -4

and

y_2 = 8

The midpoint formula relies on the coordinates of the endpoints.

2

Calculate the average of the

y

-coordinates using the formula

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

.

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

The

y

-coordinate of a midpoint is the arithmetic mean of the

y

-coordinates of the endpoints.

Key Concept

Midpoint Formula

Estimated Time:45s

Question 49Question →

A hyperbola in the standard $(x, y)$ coordinate plane is defined by the equation:

16x^2 - 9y^2 - 32x + 36y - 164 = 0

What are the coordinates of the foci of this hyperbola?

(1, 7)

and

(1, -3)

(6, -2)

and

(-4, -2)

(6, 2)

and

(-4, 2)

(4, -2)

and

(-6, -2)

(1 + \sqrt{7}, 2)

and

(1 - \sqrt{7}, 2)

Show answer & explanation

Answer: $(6, 2)$ and $(-4, 2)$

Answer

The foci of the hyperbola are

(6, 2)

and

(-4, 2)

.

The correct answer is the set of coordinates

(6, 2)

and

(-4, 2)

. After rewriting the hyperbola's equation in standard form by completing the square, we obtain

\frac{(x - 1)^2}{9} - \frac{(y - 2)^2}{16} = 1

. The center of the hyperbola is

(1, 2)

. Since the

x^2

term is positive, it has a horizontal transverse axis. The distance from the center to each focus,

c

, satisfies

c^2 = a^2 + b^2 = 9 + 16 = 25

, so

c = 5

. Adding and subtracting this focal distance from the

x

-coordinate of the center yields the foci at

(1 + 5, 2) = (6, 2)

and

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

.

Step-by-Step Solution

1

Group the

x

-terms and

y

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

16(x^2 - 2x) - 9(y^2 - 4y) = 164

Grouping the terms allows us to prepare for completing the square for both variables.

2

Complete the square for

x^2 - 2x

by adding

1

inside the parentheses, and for

y^2 - 4y

by adding

4

inside the parentheses. Add the corresponding balanced quantities to the right side:

16(1) = 16

and

-9(4) = -36

.

16(x^2 - 2x + 1) - 9(y^2 - 4y + 4) = 164 + 16 - 36

16(x - 1)^2 - 9(y - 2)^2 = 144

This rewrites the quadratic expressions into perfect square binomials.

3

Divide both sides of the equation by

144

to express it in standard form.

\frac{(x - 1)^2}{9} - \frac{(y - 2)^2}{16} = 1

The standard form of a horizontal hyperbola is

\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1

, which lets us identify the center,

a

, and

b

directly.

4

Identify the center

(h, k)

,

a^2

, and

b^2

, then calculate the focal distance

c

using the relation

c^2 = a^2 + b^2

.

Center is

(1, 2)

.

a^2 = 9

and

b^2 = 16

.

c^2 = 9 + 16 = 25 \implies c = 5

.

Foci are located at a distance of

c

from the center along the transverse axis.

5

Determine the coordinates of the foci by shifting the

x

-coordinate of the center by

\pm c

since the transverse axis is horizontal.

Foci coordinates are

(1 \pm 5, 2)

, which gives

(6, 2)

and

(-4, 2)

.

Adding and subtracting

c

from the center's

x

-coordinate gives the locations of the two foci.

Key Concept

Rewriting a hyperbola equation in standard form by completing the square and finding its foci.

Question 50Question →

In the standard $(x,y)$ coordinate plane, the set of all points equidistant from the line $3x - 4y = 8$ and the line $5x + 12y = 13$ consists of two perpendicular lines. What is the slope of the line in this set that has a positive slope?

-8

\frac{2}{19}

\frac{4}{7}

\frac{1}{8}

8

Show answer & explanation

Answer: $\frac{1}{8}$

Answer

The slope of the line with a positive slope is 1/8.

By using the point-to-line distance formula, the equidistant relationship

\frac{|3x - 4y - 8|}{5} = \frac{|5x + 12y - 13|}{13}

yields two linear equations:

14x - 112y - 39 = 0

and

64x + 8y - 169 = 0

. The slopes of these lines are 1/8 and -8, respectively. The line with the positive slope has a slope of 1/8.

Step-by-Step Solution

1

Set up the distance formula from a point

(x,y)

to both given lines.

The distance

d_1

to the first line is

\frac{|3x - 4y - 8|}{\sqrt{3^2 + (-4)^2}} = \frac{|3x - 4y - 8|}{5}

. The distance

d_2

to the second line is

\frac{|5x + 12y - 13|}{\sqrt{5^2 + 12^2}} = \frac{|5x + 12y - 13|}{13}

.

Points equidistant from both lines must satisfy

d_1 = d_2

.

2

Equate the two distance expressions to represent the equidistant relationship.

\frac{|3x - 4y - 8|}{5} = \frac{|5x + 12y - 13|}{13}

This represents the geometric condition of being equidistant from both lines.

3

Solve for Case 1 where the expressions inside the absolute values have the same sign.

13(3x - 4y - 8) = 5(5x + 12y - 13) \Rightarrow 39x - 52y - 104 = 25x + 60y - 65 \Rightarrow 14x - 112y - 39 = 0

.

One of the two bisecting lines is found when the signs match.

4

Solve for Case 2 where the expressions inside the absolute values have opposite signs.

13(3x - 4y - 8) = -5(5x + 12y - 13) \Rightarrow 39x - 52y - 104 = -25x - 60y + 65 \Rightarrow 64x + 8y - 169 = 0

.

The other bisecting line is found when the signs are opposite.

5

Find the slope of each resulting linear equation to identify the positive one.

For

14x - 112y - 39 = 0

, the slope is

-\frac{14}{-112} = \frac{1}{8}

. For

64x + 8y - 169 = 0

, the slope is

-\frac{64}{8} = -8

.

The slope of a line in the standard form

Ax + By + C = 0

is given by

-\frac{A}{B}

.

Key Concept

The set of points equidistant from two intersecting lines forms two perpendicular lines representing the angle bisectors of the original lines, which can be determined by equating their point-to-line distance formulas.

Alternative Method

Instead of using the distance formula, one could find the angle of inclination of each line using trigonometry:

\theta_1 = \arctan(3/4)

and

\theta_2 = \arctan(-5/12)

. The angle bisectors have inclinations at the average of these two angles plus or minus 90 degrees. Converting back to slopes using

m = \tan(\theta)

yields the same results.

Estimated Time:3m 0s

Question 51Question →

A hyperbola in the standard $(x, y)$ coordinate plane is defined by the equation $9x^2 - 16y^2 - 36x - 32y - 124 = 0$ . One of the foci of this hyperbola is located at the point $(f, -1)$ , where $f > 0$ . What is the value of $f$ ?

Show answer & explanation

Answer: 7

Answer

7

Completing the square transforms the equation into the standard form of a horizontal hyperbola,

\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1

, which has its center at

(2, -1)

with

a^2 = 16

and

b^2 = 9

. The distance to the foci is

c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = 5

. Since the transverse axis is horizontal, the foci are located at

(2 \pm 5, -1)

, which are

(-3, -1)

and

(7, -1)

. Given the constraint that

f > 0

, the positive x-coordinate of the focus is 7.

Step-by-Step Solution

1

Group the terms and prepare to complete the square.

9(x^2 - 4x) - 16(y^2 + 2y) = 124

Grouping the variables helps isolate the quadratic expressions for completing the square.

2

Complete the square for both the

x

and

y

terms.

9(x - 2)^2 - 16(y + 1)^2 = 144

To complete the square for

x^2 - 4x

, add 4 inside the first parentheses, adding

9 \times 4 = 36

to the right side. To complete the square for

y^2 + 2y

, add 1 inside the second parentheses, which subtracts

16 \times 1 = 16

from the right side because of the leading negative coefficient. This leaves the right side as

124 + 36 - 16 = 144

.

3

Divide both sides of the equation by the constant to find the standard form.

\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1

Dividing by 144 puts the equation in the standard horizontal hyperbola form:

\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1

.

4

Find the distance

c

from the center to the foci.

c = 5

For a hyperbola, the relationship between the semi-axes and the focal distance is

c = \sqrt{a^2 + b^2}

. Substituting

a^2 = 16

and

b^2 = 9

gives

c = \sqrt{16 + 9} = 5

.

5

Determine the coordinates of the foci and extract the value of

f

.

f = 7

The center of the hyperbola is

(h, k) = (2, -1)

. The foci are located at

(h \pm c, k) = (2 \pm 5, -1)

, which corresponds to the points

(-3, -1)

and

(7, -1)

. Since the problem states

f > 0

, the target focus must be

(7, -1)

, meaning

f = 7

.

Key Concept

Converting a general hyperbola equation into standard form to calculate focal points

Question 52Question →

A triangle has a vertex located at the point with coordinates $(7, -2)$ . If the triangle is translated $4$ units to the left and $5$ units up, what are the coordinates of this vertex after the translation?

(3, 3)

(11, 3)

(3, -7)

(11, -7)

(-6, 12)

Show answer & explanation

Answer: $(3, 3)$

Answer

(3, 3)

Applying a translation of

4

units left means subtracting

4

from the

x

-coordinate of the point

(7, -2)

, resulting in

7 - 4 = 3

. Translating

5

units up means adding

5

to the

y

-coordinate, resulting in

-2 + 5 = 3

. Therefore, the new coordinates of the vertex are

(3, 3)

.

Step-by-Step Solution

1

Identify the horizontal translation and apply it to the

x

-coordinate.

Since the translation is

4

units to the left, we subtract

4

from the initial

x

-coordinate of

7

:

7 - 4 = 3

.

Moving left along the

x

-axis decreases the coordinate value.

2

Identify the vertical translation and apply it to the

y

-coordinate.

Since the translation is

5

units up, we add

5

to the initial

y

-coordinate of

-2

:

-2 + 5 = 3

.

Moving up along the

y

-axis increases the coordinate value.

3

Combine the new coordinates into an ordered pair.

The final coordinates are

(3, 3)

.

The coordinates

(x', y')

form the final position of the vertex.

Key Concept

Translating points in the coordinate plane by adding or subtracting values from their coordinates.

Estimated Time:45s

Question 53Question →

A local gym charges a monthly membership fee of $12.50$ dollars plus a one-time registration fee of $b$ dollars. The total cost, $y$ , in dollars, for $x$ months is given by the linear equation $y = 12.50x + b$ . If a member paid a total of $100$ dollars for $6$ months of membership, what is the registration fee, $b$ , in dollars?

75.00

81.50

25.00

175.00

37.50

Show answer & explanation

Answer: $25.00$

Answer

25.00

The correct answer is found by substituting the total cost of

100

for

y

and the number of months,

6

, for

x

in the equation

y = 12.50x + b

, giving

100 = 12.50(6) + b

. Multiplying

12.50

by

6

yields

75

, and subtracting

75

from

100

gives the registration fee

b = 25

.

Step-by-Step Solution

1

Identify and substitute the given values into the linear equation.

Substitute

y = 100

(total cost) and

x = 6

(number of months) into

y = 12.50x + b

to get

100 = 12.50(6) + b

.

This sets up the equation with one variable, allowing us to solve for the unknown registration fee

b

.

2

Calculate the total cost of the monthly fees.

12.50 \times 6 = 75

. The equation becomes

100 = 75 + b

.

We multiply the monthly fee rate by the number of months to find the portion of the total cost spent on monthly fees.

3

Solve for the registration fee

b

.

b = 100 - 75 = 25

.

Subtracting the total monthly fees from the total cost isolates

b

to find the registration fee.

Key Concept

Solving linear equations in slope-intercept form by substituting known values to find the y-intercept.

Question 54Question →

In the standard $(x, y)$ coordinate plane, the line defined by the equation $3x - 4y = 24$ intersects the $x$ -axis at a certain point. What is the $x$ -coordinate of this intersection point?

Show answer & explanation

Answer: 8

Answer

The

x

-coordinate of the intersection point is

8

.

To find the

x

-intercept, set

y = 0

in the equation

3x - 4y = 24

. This simplifies to

3x = 24

. Dividing both sides of the equation by

3

gives

x = 8

.

Step-by-Step Solution

1

Set

y = 0

in the equation.

3x - 4(0) = 24 \implies 3x = 24

The intersection of any graph with the

x

-axis occurs where the

y

-coordinate is

0

.

2

Solve for

x

.

x = 8

Dividing both sides of

3x = 24

by

3

isolates the variable

x

.

Key Concept

Finding the

x

-intercept of a line by setting

y = 0

Alternative Method

Convert the standard form equation

3x - 4y = 24

into slope-intercept form:

-4y = -3x + 24 \implies y = \frac{3}{4}x - 6

. To find the

x

-intercept, set

y = 0

and solve the equation

0 = \frac{3}{4}x - 6 \implies 6 = \frac{3}{4}x \implies x = 8

.

Estimated Time:45s

Question 55Question →

A digital graphic designer positions a control point of a logo at the coordinates $(-4, 1)$ on a computer screen's coordinate grid. The designer then translates the logo so that the control point moves $3$ units to the left and $6$ units up. What are the coordinates of the control point in its new position?

(-1, 7)

(-7, 7)

(-7, -5)

(-1, -5)

(-10, 4)

Show answer & explanation

Answer: $(-7, 7)$

Answer

The correct coordinates of the control point in its new position are

(-7, 7)

.

To find the coordinates of the image after a translation, adjust the original coordinates based on the direction of movement. Since the point starting at

(-4, 1)

is translated

3

units to the left, subtract

3

from the

x

-coordinate:

-4 - 3 = -7

. Since it is translated

6

units up, add

6

to the

y

-coordinate:

1 + 6 = 7

. This gives the final coordinates of

(-7, 7)

.

Step-by-Step Solution

1

Identify the initial coordinates of the point and the translation instructions.

Initial coordinates are

(x, y) = (-4, 1)

. The translation is

3

units to the left and

6

units up.

This establishes the starting point and the transformation rules that need to be applied.

2

Apply the horizontal translation to the

x

-coordinate.

x' = -4 - 3 = -7

Translating a point to the left on the coordinate plane decreases its

x

-value, so we subtract

3

from the initial

x

-coordinate.

3

Apply the vertical translation to the

y

-coordinate.

y' = 1 + 6 = 7

Translating a point upward on the coordinate plane increases its

y

-value, so we add

6

to the initial

y

-coordinate.

Key Concept

Applying translations to coordinate points by adding or subtracting units from the

x

- and

y

-coordinates depending on the direction of movement.

Question 56Question →

In the standard $(x, y)$ coordinate plane, a line segment $AB$ has midpoint $M(3, 4)$ . If endpoint $A$ lies on the line $y = 2x - 7$ , and the length of segment $AB$ is $10$ units, what is the product of all possible $x$ -coordinates of endpoint $A$ ?

Show answer & explanation

Answer: 21

Answer

The product of all possible

x

-coordinates of endpoint

A

is

21

.

By determining that the distance from endpoint

A(x, 2x - 7)

to the midpoint

M(3, 4)

is half of the segment length

AB

(which is

5

), we set up the distance formula equation:

(x - 3)^2 + (2x - 11)^2 = 25

. Simplifying this equation leads to the quadratic expression

x^2 - 10x + 21 = 0

, which factors into

(x - 3)(x - 7) = 0

. The two possible

x

-coordinates are

3

and

7

. Multiplying these values yields the product

21

.

Step-by-Step Solution

1

Find the distance between endpoint

A

and midpoint

M

.

The distance

AM

is

5

.

Since

M

is the midpoint of segment

AB

of length

10

, the distance from either endpoint to the midpoint is half of the total length:

10 \div 2 = 5

.

2

Express the coordinates of endpoint

A

in terms of a single variable.

Endpoint

A

is represented as

(x, 2x - 7)

.

Endpoint

A

lies on the line

y = 2x - 7

.

3

Apply the distance formula to find the relationship for

x

.

(x - 3)^2 + (2x - 11)^2 = 25

The distance between

A(x, 2x - 7)

and

M(3, 4)

is

5

, so the square of the distance is

5^2 = 25

.

4

Simplify the quadratic equation.

x^2 - 10x + 21 = 0

Expanding

(x - 3)^2 + (2x - 11)^2 = 25

yields

x^2 - 6x + 9 + 4x^2 - 44x + 121 = 25

. Combining like terms gives

5x^2 - 50x + 130 = 25

. Subtracting

25

from both sides gives

5x^2 - 50x + 105 = 0

. Dividing the entire equation by

5

yields

x^2 - 10x + 21 = 0

.

5

Solve for the possible values of

x

.

x = 3

or

x = 7

Factoring the quadratic equation yields

(x - 3)(x - 7) = 0

, which gives the roots

x = 3

and

x = 7

.

6

Calculate the product of the possible

x

-coordinates.

21

The product of the two possible

x

-coordinates is

3 \times 7 = 21

.

Key Concept

Distance and Midpoint Formulas

Alternative Method

Instead of expanding the quadratic equation, one can use the geometric interpretation. The points

A

are the intersections of the circle

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

and the line

y = 2x - 7

. Substituting

y = 2x - 7

directly into the circle equation and simplifying to

x^2 - 10x + 21 = 0

is the most direct approach.

Estimated Time:2m 0s

Question 57Question →

In the standard $(x,y)$ coordinate plane, a triangle has vertices at $A(2, 4)$ , $B(6, 4)$ , and $C(2, 10)$ . The triangle undergoes a series of transformations: first, it is reflected across the line $y = x$ ; next, the resulting image is dilated by a scale factor of $\frac{1}{2}$ with the center of dilation at $(2, 2)$ ; finally, this image is translated $1$ unit to the left and $3$ units down. What are the coordinates of the final image of vertex $C$ ?

(4, -2)

(7, 5)

(5, -1)

(5, 5)

(-5, -3)

Show answer & explanation

Answer: $(5, -1)$

Answer

(5, -1)

The correct answer is

(5, -1)

because applying the three transformations sequentially yields the correct final coordinates: first, reflecting

C(2, 10)

across the line

y = x

gives

C'(10, 2)

; second, dilating

C'(10, 2)

by a scale factor of

\frac{1}{2}

centered at

(2, 2)

gives

C''(6, 2)

; and finally, translating

C''(6, 2)

by

1

unit left and

3

units down results in

(5, -1)

.

Step-by-Step Solution

1

Reflect the initial coordinates of vertex

C(2, 10)

across the line

y = x

.

C'(10, 2)

Reflecting a point

(x, y)

across the line

y = x

interchanges the coordinates, mapping

(x, y)

to

(y, x)

.

2

Dilate the point

C'(10, 2)

by a scale factor of

k = \frac{1}{2}

centered at

P(2, 2)

.

C''(6, 2)

The formula for a dilation centered at

(a, b)

is

(x'', y'') = (a + k(x' - a), b + k(y' - b))

. Substituting

a=2, b=2, k=\frac{1}{2}, x'=10, y'=2

gives

(2 + \frac{1}{2}(10-2), 2 + \frac{1}{2}(2-2)) = (2+4, 2+0) = (6, 2)

.

3

Translate the point

C''(6, 2)

by

1

unit to the left and

3

units down.

C'''(5, -1)

Translating

1

unit left subtracts

1

from the

x

-coordinate (

6 - 1 = 5

), and translating

3

units down subtracts

3

from the

y

-coordinate (

2 - 3 = -1

).

Key Concept

Composite transformations in the coordinate plane involving reflections, dilations with non-origin centers, and translations.

Estimated Time:2m 0s

Question 58Question →

In the standard $(x, y)$ coordinate plane, a line with a negative slope passes through the point $(4, 3)$ and has an $x$ -intercept that is twice its $y$ -intercept. If the equation of this line is written in the form $Ax + By = C$ , where $A$ , $B$ , and $C$ are integers with no common factor greater than 1, and $A > 0$ , what is the value of $C$ ?

Show answer & explanation

Answer: 10

Answer

The value of

C

is 10.

The correct answer is 10. By setting the intercepts as

(2b, 0)

and

(0, b)

, we find the slope is

m = -\frac{1}{2}

. Applying the point-slope formula with

(4, 3)

gives

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

, which simplifies to

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

. Converting this to standard form with

A > 0

yields

x + 2y = 10

, where

C = 10

.

Step-by-Step Solution

1

Represent the coordinates of the intercepts using a variable.

The

y

-intercept is

(0, b)

and the

x

-intercept is

(2b, 0)

.

The problem states that the

x

-intercept is twice the

y

-intercept.

2

Calculate the slope (

m

) of the line using the intercepts.

m = \frac{b - 0}{0 - 2b} = -\frac{1}{2}

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

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

.

3

Find the equation of the line using the point-slope form with point

(4, 3)

.

y - 3 = -\frac{1}{2}(x - 4) \Rightarrow y = -\frac{1}{2}x + 5

The point-slope form of a linear equation is

y - y_1 = m(x - x_1)

.

4

Convert the equation to the standard form

Ax + By = C

.

x + 2y = 10

Multiplying the equation by 2 and moving the

x

term to the left side results in integer coefficients where the coefficient of

x

is positive (

A = 1 > 0

).

5

Identify the value of

C

from the standard form equation.

C = 10

Comparing

x + 2y = 10

with

Ax + By = C

shows that

A=1

,

B=2

, and

C=10

. These integers share no common factors greater than 1.

Key Concept

Converting a linear equation to standard form using given geometric features and a coordinate point.

Alternative Method

Instead of using the point-slope formula, substitute the point

(4, 3)

directly into the intercept form of a linear equation,

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

. Since

a = 2b

, this becomes

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

. Simplifying this gives

\frac{2}{b} + \frac{3}{b} = 1 \Rightarrow \frac{5}{b} = 1 \Rightarrow b = 5

. Thus,

a = 10

. The intercept form is

\frac{x}{10} + \frac{y}{5} = 1

. Multiplying by 10 to clear denominators gives

x + 2y = 10

, where

C = 10

.

Estimated Time:1m 30s

Question 59Question →

In the standard $(x, y)$ coordinate plane, the perpendicular bisector of the line segment with endpoints $(1, 2)$ and $(5, 10)$ intersects the curve $y = x^2$ at two points. If one of these intersection points lies in the second quadrant, what is the $y$ -coordinate of this point?

Show answer & explanation

Answer: 9

Answer

The correct answer is

9

.

The perpendicular bisector of the segment connecting

(1, 2)

and

(5, 10)

passes through their midpoint

(3, 6)

and has a slope of

-\frac{1}{2}

(the negative reciprocal of

2

). The equation of this line is

x + 2y = 15

. Substituting

y = x^2

yields the quadratic equation

2x^2 + x - 15 = 0

, which factors into

(2x - 5)(x + 3) = 0

. The solution in the second quadrant corresponds to the negative

x

-coordinate,

x = -3

. Squaring this value gives a

y

-coordinate of

9

.

Step-by-Step Solution

1

Find the midpoint of the segment with endpoints

(1, 2)

and

(5, 10)

.

The midpoint is

(3, 6)

.

The perpendicular bisector of a segment passes through its midpoint.

2

Calculate the slope of the segment and the slope of the perpendicular bisector.

The slope of the segment is

2

, and the perpendicular slope is

-\frac{1}{2}

.

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

3

Write the equation of the perpendicular bisector.

The equation of the line is

x + 2y = 15

.

Using the point-slope form with point

(3, 6)

and slope

-\frac{1}{2}

yields

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

, which simplifies to

x + 2y = 15

.

4

Find the intersection points of the line and the curve

y = x^2

.

The

x

-coordinates of the intersection points are

2.5

and

-3

.

Substituting

y = x^2

into

x + 2y = 15

gives the quadratic equation

2x^2 + x - 15 = 0

, which factors as

(2x - 5)(x + 3) = 0

.

5

Identify the point in the second quadrant and find its

y

-coordinate.

The point is

(-3, 9)

, so the

y

-coordinate is

9

.

A point in the second quadrant must have a negative

x

-coordinate (

x = -3

) and a positive

y

-coordinate (

y = 9

).

Key Concept

Perpendicular bisectors and systems of linear-quadratic equations in coordinate geometry

Question 60Question →

In the standard $(x,y)$ coordinate plane, the point $A(-4, 3)$ is reflected across the $x$ -axis and then translated $5$ units to the right to map onto point $A'$ . What are the coordinates of $A'$ ?

(1, -3)

(-9, -3)

(1, 3)

(9, 3)

(-9, 3)

Show answer & explanation

Answer: $(1, -3)$

Answer

(1, -3)

Reflecting the point

A(-4, 3)

across the

x

-axis negates its

y

-coordinate, resulting in

(-4, -3)

. Then, translating this point

5

units to the right increases its

x

-coordinate by

5

, which yields

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

.

Step-by-Step Solution

1

Reflect the point

A(-4, 3)

across the

x

-axis.

The rule for reflection across the

x

-axis is

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

. Applying this rule to

A(-4, 3)

yields the intermediate point

(-4, -3)

.

A reflection across the

x

-axis negates the

y

-coordinate of the point while keeping the

x

-coordinate the same.

2

Translate the reflected point

(-4, -3)

by

5

units to the right.

The rule for translating a point

h

units to the right is

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

. Adding

5

to the

x

-coordinate yields

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

.

Translating a point to the right increases its

x

-coordinate by the translation distance.

Key Concept

Applying a sequence of coordinate transformations (reflection followed by translation) to a point.

Estimated Time:45s

Coordinate Geometry

Answer

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