Coordinate Geometry Practice Questions — ACT — Page 8

Question 141Question →

A circle drawn in the standard $(x, y)$ coordinate plane has its center at the point $(3, -2)$ and passes through the point $(7, 1)$ . What is the radius of this circle?

Show answer & explanation

Answer: 5

Answer

The radius of the circle is 5.

The radius of a circle is defined as the distance from its center to any point on the boundary. Using the distance formula between the center

(3, -2)

and the point

(7, 1)

yields

r = \sqrt{(7 - 3)^2 + (1 - (-2))^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Identify the formula to find the distance between two points, which represents the radius of the circle.

The distance formula is

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

.

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

2

Substitute the coordinates of the center

(3, -2)

and the point

(7, 1)

into the formula.

r = \sqrt{(7 - 3)^2 + (1 - (-2))^2}

We substitute

x_1 = 3, y_1 = -2

and

x_2 = 7, y_2 = 1

into the distance formula.

3

Simplify the expression to determine the radius.

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

Calculate the squared differences, add them, and take the square root to find the final value.

Key Concept

Calculating the radius of a circle given its center and a point on the circle using the distance formula.

Question 142Question →

In the standard $(x, y)$ coordinate plane, the perpendicular bisector of the line segment with endpoints $D(-5, 8)$ and $E(3, 4)$ is represented by the equation $y = mx + b$ . What is the value of $m + b$ ?

2

\frac{7}{2}

5

7

10

Show answer & explanation

Answer: $10$

Answer

The sum of the slope and the y-intercept of the perpendicular bisector is

10

.

The slope of the segment

DE

is calculated as

m_{DE} = \frac{4 - 8}{3 - (-5)} = -\frac{1}{2}

. The perpendicular bisector has a slope

m

that is the negative reciprocal of this, which is

2

. The line must pass through the midpoint of

DE

, which is located at

\left(\frac{-5 + 3}{2}, \frac{8 + 4}{2}\right) = (-1, 6)

. Substituting

m = 2

and the point

(-1, 6)

into the slope-intercept form

y = mx + b

yields

6 = 2(-1) + b

, which simplifies to

b = 8

. Summing the slope and the

y

-intercept gives

m + b = 2 + 8 = 10

.

Step-by-Step Solution

1

Find the midpoint of the line segment

DE

.

Midpoint

M = \left(\frac{-5 + 3}{2}, \frac{8 + 4}{2}\right) = (-1, 6)

By definition, the perpendicular bisector must pass through the midpoint of the segment it bisects.

2

Calculate the slope of the line segment

DE

.

m_{DE} = \frac{4 - 8}{3 - (-5)} = \frac{-4}{8} = -\frac{1}{2}

The slope of the segment is needed to find the slope of any line perpendicular to it.

3

Determine the slope

m

of the perpendicular bisector.

m = -\frac{1}{m_{DE}} = -\frac{1}{-1/2} = 2

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

4

Find the

y

-intercept

b

of the perpendicular bisector.

Using the slope-intercept form

y = mx + b

with the slope

m = 2

and the midpoint

M(-1, 6)

:

6 = 2(-1) + b \implies 6 = -2 + b \implies b = 8

Substituting a known point on the line allows us to solve for the vertical intercept parameter.

5

Calculate the sum

m + b

.

m + b = 2 + 8 = 10

This is the value requested by the question.

Key Concept

The perpendicular bisector of a line segment passes through its midpoint at a right angle, meaning its slope is the negative reciprocal of the segment's slope.

Estimated Time:2m 0s

Question 143Question →

In the standard $(x,y)$ coordinate plane, a triangle has vertices at $A(1, 2)$ , $B(9, 2)$ , and $C(5, 8)$ . A horizontal line defined by the equation $y = k$ divides the area of the triangle into two regions of equal area. What is the value of $k$ ?

5

8 - 3\sqrt{2}

2 + 3\sqrt{2}

8 - 2\sqrt{3}

4

Show answer & explanation

Answer: $8 - 3\sqrt{2}$

Answer

8 - 3\sqrt{2}

The correct answer is

8 - 3\sqrt{2}

. Because the horizontal line

y=k

is parallel to the base of the triangle, it creates a smaller top triangle similar to the original one. The area of the original triangle is

24

and the area of the smaller triangle is

12

, yielding an area ratio of

\frac{1}{2}

. The height of the original triangle is

6

, and the height of the smaller triangle is

8-k

. Since the ratio of the areas of similar triangles is the square of the ratio of their heights, we write

\frac{1}{2} = \left(\frac{8-k}{6}\right)^2

. Solving this equation yields

k = 8 - 3\sqrt{2}

.

Step-by-Step Solution

1

Calculate the area of the entire triangle

ABC

.

The base

\overline{AB}

is horizontal along

y = 2

with a length of

9 - 1 = 8

. The height is the vertical distance from

y = 2

to the vertex

C(5, 8)

, which is

8 - 2 = 6

. The area of

\triangle ABC

is

\frac{1}{2} \times 8 \times 6 = 24

.

Finding the total area is necessary to determine the target area of the divided regions.

2

Find the area of the smaller triangle formed above the line

y = k

.

The line

y = k

divides the triangle into a smaller top triangle and a bottom trapezoid. Since both regions have equal areas, the area of the smaller top triangle is

\frac{24}{2} = 12

.

The line divides the original area of

24

into two equal parts of

12

each.

3

Set up the area ratio equation using the properties of similar triangles.

The smaller top triangle is similar to

\triangle ABC

because its base is parallel to

\overline{AB}

. The ratio of their areas is the square of the ratio of their heights:

\frac{12}{24} = \left(\frac{8 - k}{6}\right)^2 \implies \frac{1}{2} = \left(\frac{8 - k}{6}\right)^2

.

For similar figures, the area ratio is equal to the square of the scale factor.

4

Solve the ratio equation for

k

.

Taking the square root of both sides gives

\frac{1}{\sqrt{2}} = \frac{8 - k}{6}

, which simplifies to

\frac{\sqrt{2}}{2} = \frac{8 - k}{6}

. Multiplying both sides by

6

gives

3\sqrt{2} = 8 - k

, which yields

k = 8 - 3\sqrt{2}

.

This isolates

k

to find the exact vertical coordinate of the dividing line.

Key Concept

Using properties of similar figures to determine areas and coordinates on the coordinate plane.

Question 144Question →

A drone starts at position $P$ on a grid. The drone's path is programmed with a sequence of two movements: first, it is translated $6$ units to the right and $5$ units up, and then its position is reflected across the $y$ -axis. If the drone's final position is $(-4, 8)$ , what were the coordinates of its starting position $P$ ?

(10, 3)

(-2, 13)

(-2, 3)

(10, 13)

(-10, -13)

Show answer & explanation

Answer: $(-2, 3)$

Answer

The correct answer is

(-2, 3)

because working backward from the final position

(-4, 8)

by undoing the reflection across the

y

-axis gives

(4, 8)

, and then undoing the translation by subtracting

6

from the

x

-coordinate and

5

from the

y

-coordinate yields

(-2, 3)

.

The correct answer is

(-2, 3)

because we can find the starting position by working backward from the final position. The last transformation applied was a reflection across the

y

-axis. Undoing this reflection maps the final coordinates

(-4, 8)

to

(4, 8)

because reflecting across the

y

-axis changes the sign of the

x

-coordinate. The first transformation was a translation of

6

units to the right and

5

units up. To undo this translation, we subtract

6

from the

x

-coordinate and subtract

5

from the

y

-coordinate of

(4, 8)

, giving

(4 - 6, 8 - 5) = (-2, 3)

.

Step-by-Step Solution

1

Identify the final transformation and undo it. The final transformation is a reflection across the

y

-axis, which maps

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

.

The coordinates before reflection are

(4, 8)

.

Undoing a reflection across the

y

-axis on

(-4, 8)

means changing the sign of the

x

-coordinate:

-(-4) = 4

, while keeping the

y

-coordinate the same.

2

Identify the first transformation and undo it. The first transformation is a translation of

6

units right and

5

units up, which maps

(x, y) \rightarrow (x+6, y+5)

.

The starting coordinates

P

are

(-2, 3)

.

To undo the translation, subtract

6

from the

x

-coordinate and

5

from the

y

-coordinate of the intermediate point

(4, 8)

:

4 - 6 = -2

and

8 - 5 = 3

.

Key Concept

Transformations in the Coordinate Plane

Alternative Method

Instead of working backward, you can test the options by applying the transformations forward. For

(-2, 3)

, translating 6 units right and 5 units up gives

(-2+6, 3+5) = (4, 8)

. Reflecting across the

y

-axis then gives

(-4, 8)

, which matches the final position.

Estimated Time:1m 15s

Question 145Question →

In the standard $(x, y)$ coordinate plane, a line $L$ passes through the points $(a, a^2)$ and $(b, b^2)$ , where $a$ and $b$ are distinct real numbers. The slope of line $L$ is $8$ . If the midpoint of the line segment connecting these two points lies on the line $y = 5x - 1$ , what is the $y$ -coordinate of this midpoint?

Show answer & explanation

Answer: 19

Answer

The

y

-coordinate of the midpoint is

19

.

Applying the slope formula to the points

(a, a^2)

and

(b, b^2)

yields \frac{b^2-a^2}{b-a} = a+b = 8. The

x

-coordinate of the midpoint is \frac{a+b}{2} = 4. Substituting this value into the equation

y = 5x - 1

gives

y = 5(4) - 1 = 19

.

Step-by-Step Solution

1

Express the slope of line

L

in terms of

a

and

b

and set it equal to the given slope.

a + b = 8

The slope formula is

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

. For the points

(a, a^2)

and

(b, b^2)

, the slope is \frac{b^2 - a^2}{b - a}. Factoring the numerator gives \frac{(b-a)(b+a)}{b-a} = a + b. Since the slope is given as

8

, we establish

a + b = 8

.

2

Find the

x

-coordinate of the midpoint of the segment connecting the two points.

x_M = 4

The midpoint formula for the

x

-coordinate is

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

. For our points,

x_M = \frac{a+b}{2}

. Substituting the value

a+b = 8

gives

x_M = \frac{8}{2} = 4

.

3

Determine the

y

-coordinate of the midpoint using the line equation.

y_M = 19

The midpoint lies on the line

y = 5x - 1

. Substituting

x_M = 4

into this equation gives

y_M = 5(4) - 1 = 19

.

Key Concept

Slope of a Line

Alternative Method

Let

a

and

b

be the real roots of the quadratic equation

t^2 - 8t + 13 = 0

. By Vieta's formulas,

a+b = 8

and

ab = 13

. The

x

-coordinate of the midpoint is \frac{a+b}{2} = 4, and the

y

-coordinate is \frac{a^2+b^2}{2} = \frac{(a+b)^2 - 2ab}{2} = \frac{64 - 26}{2} = 19. Since the point

(4, 19)

satisfies

y = 5x - 1

, this confirms the existence of valid real coordinates

(a, a^2)

and

(b, b^2)

that produce the midpoint on the line.

Estimated Time:2m 0s

Question 146Question →

A municipal water reservoir is being drained at a constant rate. After $4$ hours of draining, the reservoir contains $18,000$ gallons of water. After $7$ hours of draining, it contains $13,500$ gallons of water. How many hours after the draining process begins will the reservoir be completely empty?

Show answer & explanation

Answer: 16

Answer

The reservoir will be completely empty

16

hours after draining begins.

The correct answer is

16

hours. The constant draining rate is calculated as the change in volume divided by the change in time:

\frac{13,500 - 18,000}{7 - 4} = -1,500

gallons per hour. Since there are

18,000

gallons left at the

4

-hour mark, it will take an additional

\frac{18,000}{1,500} = 12

hours to completely empty the reservoir. The total time from the start is

4 + 12 = 16

hours.

Step-by-Step Solution

1

Calculate the draining rate (the slope of the linear function).

The rate is

-1,500

gallons per hour.

The rate of change is the change in volume divided by the change in time:

\frac{13,500 - 18,000}{7 - 4} = \frac{-4,500}{3} = -1,500

gallons per hour.

2

Determine the remaining time needed to empty the reservoir after the

7

-hour mark.

It will take an additional

9

hours.

At

7

hours, the reservoir contains

13,500

gallons. Draining at a rate of

1,500

gallons per hour, the remaining time is

\frac{13,500}{1,500} = 9

hours.

3

Calculate the total elapsed time since the draining process began.

16

hours

Adding the initial

7

hours of draining to the additional

9

hours needed gives

7 + 9 = 16

hours.

Key Concept

Linear word problems and finding intercepts

Alternative Method

We can model the volume of water

W

as a linear function of time

t

using the slope-intercept form

W(t) = mt + b

. Substituting the rate

m = -1,500

and the point

(4, 18,000)

gives

18,000 = -1,500(4) + b

, which yields the

y

-intercept (initial volume)

b = 24,000

gallons. The linear model is

W(t) = -1,500t + 24,000

. Setting

W(t) = 0

to find when the reservoir is empty gives

0 = -1,500t + 24,000

, which simplifies to

t = \frac{24,000}{1,500} = 16

hours.

Estimated Time:1m 30s

Question 147Question →

In the standard $(x, y)$ coordinate plane, point $P$ is rotated $90^\circ$ clockwise about the origin, and then translated $3$ units to the left and $4$ units up to form the image point $P'(1, 2)$ . What are the coordinates of the pre-image point $P$ ?

(-1, 3)

(-6, -2)

(2, 4)

(-2, -4)

(1, -3)

Show answer & explanation

Answer: $(2, 4)$

Answer

The coordinates of the pre-image point

P

are

(2, 4)

.

To find the coordinates of the pre-image point, we must undo the transformations in reverse order. First, we undo the translation of

3

units left and

4

units up by translating the image point

(1, 2)

by

3

units right and

4

units down, yielding the intermediate point

(4, -2)

. Next, we undo the

90^\circ

clockwise rotation by rotating this intermediate point

90^\circ

counterclockwise about the origin. Using the rotation rule

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

on

(4, -2)

gives the pre-image point

(2, 4)

.

Step-by-Step Solution

1

Identify the sequence of forward transformations and set up the path to work backward from the final image point

P'(1, 2)

to the pre-image point

P

.

The forward sequence is:

P \xrightarrow{\text{Rotate } 90^\circ \text{ clockwise}} P_{\text{mid}} \xrightarrow{\text{Translate left 3, up 4}} P'(1, 2)

. To find

P

, we must apply the inverse transformations in reverse order:

P'(1, 2) \xrightarrow{\text{Translate right 3, down 4}} P_{\text{mid}} \xrightarrow{\text{Rotate } 90^\circ \text{ counterclockwise}} P

.

Working backward with inverse operations is the standard mathematical procedure to retrieve a pre-image from its transformed image.

2

Apply the inverse translation to the image point

P'(1, 2)

to find the coordinates of the intermediate point

P_{\text{mid}}

.

To undo a translation of

3

units left and

4

units up, translate

3

units right and

4

units down:

x = 1 + 3 = 4

, and

y = 2 - 4 = -2

. Thus,

P_{\text{mid}} = (4, -2)

.

Applying the opposite direction and magnitude to the coordinates reverses the translation effect.

3

Apply the inverse rotation to the intermediate point

P_{\text{mid}}(4, -2)

to find the coordinates of the pre-image point

P

.

The inverse of a

90^\circ

clockwise rotation is a

90^\circ

counterclockwise rotation about the origin. The rule for a

90^\circ

counterclockwise rotation is

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

. Applying this to

(4, -2)

gives

P(-(-2), 4) = P(2, 4)

.

Rotating

90^\circ

counterclockwise reverses the initial

90^\circ

clockwise rotation, returning the point to its original pre-image coordinates.

Key Concept

Determining the coordinates of a pre-image by reversing a composite transformation in the coordinate plane.

Question 148Question →

A right triangle is plotted in the standard $(x, y)$ coordinate plane. The vertices of the triangle are located at $(-3, -2)$ , $(5, -2)$ , and $(-3, 4)$ . What is the length of the hypotenuse of the triangle?

\frac{4}{3}

2\sqrt{7}

\frac{3}{5}

10

14

Show answer & explanation

Answer: $10$

Answer

10

The distance between the vertices

(-3, -2)

and

(5, -2)

along the horizontal line

y = -2

is

5 - (-3) = 8

units. The distance between the vertices

(-3, -2)

and

(-3, 4)

along the vertical line

x = -3

is

4 - (-2) = 6

units. Because horizontal and vertical segments meet at a right angle, they form the legs of a right triangle. Applying the Pythagorean theorem, the length of the hypotenuse is the square root of the sum of the squares of the legs:

\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10

.

Step-by-Step Solution

1

Determine the lengths of the two perpendicular legs of the right triangle on the coordinate plane.

The horizontal leg has a length of

5 - (-3) = 8

units, and the vertical leg has a length of

4 - (-2) = 6

units.

Because the segment between

(-3, -2)

and

(5, -2)

is horizontal (constant

y = -2

) and the segment between

(-3, -2)

and

(-3, 4)

is vertical (constant

x = -3

).

2

Apply the Pythagorean theorem to calculate the length of the hypotenuse.

The hypotenuse length is

\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10

units.

The Pythagorean theorem states that for any right triangle with legs

a

and

b

and hypotenuse

c

,

a^2 + b^2 = c^2

.

Key Concept

Finding the lengths of segments and applying the Pythagorean theorem on the coordinate plane.

Question 149Question →

A circle in the standard $(x,y)$ coordinate plane is defined by the equation $(x + 3)^2 + (y - 4)^2 = 25$ . What are the coordinates of the center and the length of the radius of this circle?

Center:

(3, -4)

; Radius:

5

Center:

(-3, 4)

; Radius:

25

Center:

(-3, 4)

; Radius:

5

Center:

(3, -4)

; Radius:

25

Center:

(-3, -4)

; Radius:

5

Show answer & explanation

Answer: Center: $(-3, 4)$ ; Radius: $5$

Answer

Center:

(-3, 4)

; Radius:

5

The standard form of a circle's equation is

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

, where the center is

(h, k)

and the radius is

r

. Matching

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

to this standard form, we find

h = -3

,

k = 4

, and

r^2 = 25

(which gives

r = 5

). Therefore, the center is

(-3, 4)

and the radius is

5

.

Step-by-Step Solution

1

Identify the standard form of a circle's equation.

The standard equation of a circle with center

(h, k)

and radius

r

is

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

.

This formula provides the template to match the given equation and extract the parameters.

2

Rewrite the given equation

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

to match the signs and exponents of the standard form.

The equation can be written as

(x - (-3))^2 + (y - 4)^2 = 5^2

.

Rewriting the terms helps to identify the exact values of

h

,

k

, and

r

without sign confusion.

3

Extract the center

(h, k)

and radius

r

from the rewritten equation.

Comparing the terms shows

h = -3

,

k = 4

, and

r = 5

, giving a center of

(-3, 4)

and a radius of

5

.

These extracted values are the final answer.

Key Concept

Equations and Graphs of Circles

Question 150Question →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $(x - 9)^2 + (y + 4)^2 = 81$ . What is the radius of this circle?

Show answer & explanation

Answer: 9

Answer

The radius of the circle is 9.

The standard equation of a circle in the coordinate plane is

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

, where

r

is the radius. For the equation

(x - 9)^2 + (y + 4)^2 = 81

, the right-hand side represents

r^2

, so

r^2 = 81

. Taking the positive square root of both sides gives

r = \sqrt{81} = 9

.

Step-by-Step Solution

1

Identify the standard form of the circle equation.

The standard equation of a circle is

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

, where

(h, k)

is the center and

r

is the radius.

This allows us to relate the terms of the given equation to the components of the circle.

2

Match the given equation to the standard form.

By comparing

(x - 9)^2 + (y + 4)^2 = 81

to

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

, we find that

r^2 = 81

.

The constant on the right side of the standard equation represents the square of the radius.

3

Solve for the radius

r

.

r = \sqrt{81} = 9

.

Taking the square root of

r^2

gives the actual radius of the circle.

Key Concept

Identifying the radius from the standard equation of a circle

Question 151Question →

In the standard $(x, y)$ coordinate plane, the midpoint of a line segment with endpoints $(1, -2)$ and $(7, 10)$ is the center of a circle. If the point $(8, 7)$ lies on the circle, what is the radius of the circle?

7

25

3\sqrt{5}

5

\sqrt{26}

Show answer & explanation

Answer: 5

Answer

The radius of the circle is 5.

The midpoint of the line segment with endpoints

(1, -2)

and

(7, 10)

is calculated as

(\frac{1+7}{2}, \frac{-2+10}{2}) = (4, 4)

, which is the center of the circle. The distance from the center

(4, 4)

to the point

(8, 7)

on the circle is the radius, which is

\sqrt{(8-4)^2 + (7-4)^2} = \sqrt{16+9} = 5

.

Step-by-Step Solution

1

Find the midpoint of the line segment with endpoints

(1, -2)

and

(7, 10)

to determine the center of the circle.

The center of the circle is

(4, 4)

.

The midpoint of a segment with endpoints

(x_1, y_1)

and

(x_2, y_2)

is given by

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

. Calculating this gives

(\frac{1+7}{2}, \frac{-2+10}{2}) = (4, 4)

.

2

Calculate the distance between the center of the circle

(4, 4)

and the point

(8, 7)

on the circle to find the radius.

The radius of the circle is 5.

The distance formula is

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

. Substituting the coordinates gives

\sqrt{(8-4)^2 + (7-4)^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5

.

Key Concept

Finding the midpoint of a line segment to determine a circle's center and using the distance formula to calculate its radius.

Question 152Question →

In the standard $(x,y)$ coordinate plane, a square has two adjacent vertices at $A(1, 4)$ and $B(4, 8)$ . If the square lies entirely in the first quadrant, what is the $x$ -coordinate of the vertex diagonally opposite to $A$ ?

Show answer & explanation

Answer: 8

Answer

The

x

-coordinate of the vertex diagonally opposite to

A

is

8

.

The vector from

A(1,4)

to

B(4,8)

is

\vec{AB} = (3,4)

with length

5

. Since the adjacent sides of a square must be perpendicular and equal in length, the vector translation must be either

(4, -3)

or

(-4, 3)

. Applying the translation

(4, -3)

to

B

and

A

gives

C(8, 5)

and

D(5, 1)

, both of which lie strictly in the first quadrant. The diagonally opposite vertex to

A

is

C

, and its

x

-coordinate is

8

.

Step-by-Step Solution

1

Calculate the vector representing side

AB

and its length.

\vec{AB} = (3, 4)

, which has a length of

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

.

This determines the length and orientation of the square's base.

2

Determine the perpendicular vectors of equal length to find the directions of the adjacent sides.

The perpendicular vectors of length

5

are

(4, -3)

and

(-4, 3)

.

The adjacent sides of a square are perpendicular and equal in length to side

AB

.

3

Find the coordinates of the other two vertices for both possible positions of the square.

Position 1 vertices:

C = B + (4, -3) = (8, 5)

and

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

. Position 2 vertices:

C' = B + (-4, 3) = (0, 11)

and

D' = A + (-4, 3) = (-3, 7)

.

Adding the perpendicular vectors to the coordinates of the base vertices yields the remaining vertices.

4

Determine which square lies entirely in the first quadrant and identify the target coordinate.

The square

ABCD

with vertices

C(8, 5)

and

D(5, 1)

lies entirely in the first quadrant because all

x

- and

y

-coordinates are positive. The vertex diagonally opposite to

A

is

C(8, 5)

, which has an

x

-coordinate of

8

.

The alternative square

ABC'D'

has a vertex at

D'(-3, 7)

, which lies in the second quadrant.

Key Concept

Using vector methods to determine the coordinates of a geometric figure on the coordinate plane under specific quadrant constraints.

Estimated Time:2m 0s

Question 153Question →

In the standard $(x, y)$ coordinate plane, a line $L$ passes through the point $(2, -3)$ and does not pass through the origin. The distance from the origin to the line's $x$ -intercept is twice the distance from the origin to the line's $y$ -intercept. Which of the following could be the slope of line $L$ ?

-2

-\frac{1}{2}

\frac{2}{3}

1

2

Show answer & explanation

Answer: $-\frac{1}{2}$

Answer

The slope of the line could be

-\frac{1}{2}

.

The correct answer is

-\frac{1}{2}

. By setting the

x

-intercept of the line to

(a, 0)

and the

y

-intercept to

(0, b)

, the distance condition gives

|a| = 2|b|

. Substituting the point

(2, -3)

into the intercept equation of a line,

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

, yields

\frac{2}{a} - \frac{3}{b} = 1

. Solving the two cases (

a = 2b

and

a = -2b

) gives the intercept pairs

(-4, -2)

and

(8, -4)

. Calculating the slope

m = -\frac{b}{a}

for both pairs results in

-\frac{1}{2}

and

\frac{1}{2}

. Since

-\frac{1}{2}

is one of these possible values, it is the correct choice.

Step-by-Step Solution

1

Represent the line

L

using its intercepts. Let the

x

-intercept of

L

be

(a, 0)

and the

y

-intercept be

(0, b)

. Since the line does not pass through the origin, we have

a \neq 0

and

b \neq 0

.

The equation of the line can be written in intercept form as:

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

Using intercept form allows us to directly relate the coordinates of the intercepts to the given point and the distance condition.

2

Translate the distance condition and the given point into equations. The distance from the origin to the

x

-intercept is

|a|

, and the distance to the

y

-intercept is

|b|

. The problem states that the distance to the

x

-intercept is twice the distance to the

y

-intercept, so

|a| = 2|b|

. Additionally, the line passes through

(2, -3)

, so we substitute

x = 2

and

y = -3

into the intercept equation.

|a| = 2|b|

and

\frac{2}{a} - \frac{3}{b} = 1

This sets up a system of equations to solve for the unknown intercepts

a

and

b

.

3

Solve the system of equations by analyzing the two cases for the absolute value:

a = 2b

and

a = -2b

.

Case 1: If

a = 2b

, then

\frac{2}{2b} - \frac{3}{b} = 1 \implies \frac{1}{b} - \frac{3}{b} = 1 \implies -\frac{2}{b} = 1 \implies b = -2

, which gives

a = -4

. Case 2: If

a = -2b

, then

\frac{2}{-2b} - \frac{3}{b} = 1 \implies -\frac{1}{b} - \frac{3}{b} = 1 \implies -\frac{4}{b} = 1 \implies b = -4

, which gives

a = 8

.

Resolving the absolute value yields the exact coordinates of the intercepts for both valid scenarios.

4

Calculate the slope

m

for both cases using the formula

m = -\frac{b}{a}

.

For Case 1,

m = -\frac{-2}{-4} = -\frac{1}{2}

. For Case 2,

m = -\frac{-4}{8} = \frac{1}{2}

.

The slope of a line with intercepts

(a, 0)

and

(0, b)

is given by

-\frac{b}{a}

.

Key Concept

Slope of a Line

Estimated Time:2m 0s

Question 154Question →

In the standard $(x, y)$ coordinate plane, one diagonal of a square lies along the line with equation $y = 3x - 4$ . If one of the vertices of the square that does not lie on this diagonal is located at the point $(2, 8)$ , which of the following equations represents the line containing the other diagonal of the square?

3x - y = -2

3x + y = 14

x + 3y = 26

x - 3y = -22

x + 3y = 40

Show answer & explanation

Answer: $x + 3y = 26$

Answer

The equation of the line containing the other diagonal is

x + 3y = 26

.

The correct equation is

x + 3y = 26

. Since the diagonals of a square are perpendicular, their slopes must be negative reciprocals of each other. The given diagonal has a slope of

3

, meaning the other diagonal has a slope of

-\frac{1}{3}

. Since the vertex

(2, 8)

does not satisfy the equation of the first diagonal, it must lie on the second diagonal. Substituting this point into the slope-intercept or point-slope equation yields the standard form equation

x + 3y = 26

.

Step-by-Step Solution

1

Identify the slope of the given diagonal line.

The slope of the line

y = 3x - 4

is

3

.

The equation is in slope-intercept form

y = mx + b

, where

m

is the slope.

2

Determine the relationship between the two diagonals of a square.

The diagonals of a square are perpendicular to each other.

By geometric definition, the diagonals of any square intersect at right angles.

3

Calculate the slope of the perpendicular diagonal.

The slope of the perpendicular diagonal is

-\frac{1}{3}

.

Perpendicular lines have slopes that are negative reciprocals of each other, so the perpendicular slope is

-\frac{1}{3}

.

4

Identify the point that the perpendicular diagonal passes through.

The perpendicular diagonal passes through the vertex

(2, 8)

.

The vertex

(2, 8)

does not lie on the first diagonal because

8 \neq 3(2) - 4

. Since a square only has two diagonals, any vertex not on the first diagonal must lie on the second diagonal.

5

Find the equation of the perpendicular diagonal using point-slope form.

The equation is

x + 3y = 26

.

Using the point-slope form

y - y_1 = m(x - x_1)

with

m = -\frac{1}{3}

and point

(2, 8)

yields

y - 8 = -\frac{1}{3}(x - 2)

, which simplifies to

3(y - 8) = -(x - 2) \Rightarrow 3y - 24 = -x + 2 \Rightarrow x + 3y = 26

.

Key Concept

Diagonals of a square are perpendicular, and perpendicular lines have slopes that are negative reciprocals.

Question 155Question →

A line segment in the standard $(x, y)$ coordinate plane has endpoints at $A(-1, 10)$ and $B(7, -2)$ . If point $M$ is the midpoint of this segment, what is the distance from $M$ to the origin $(0, 0)$ ?

Show answer & explanation

Answer: 5

Answer

The distance from the midpoint

M

to the origin is 5.

To find the distance from the midpoint

M

to the origin

(0,0)

, we first determine the coordinates of

M

by taking the average of the coordinates of

A(-1, 10)

and

B(7, -2)

. This results in

M\left(\frac{-1+7}{2}, \frac{10+(-2)}{2}\right) = (3, 4)

. Next, we apply the distance formula between

M(3,4)

and the origin

(0,0)

to get

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

.

Step-by-Step Solution

1

Calculate the coordinates of the midpoint

M

of the segment

AB

.

The midpoint is

M(3, 4)

.

The midpoint formula is

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

. Substituting the coordinates of

A(-1, 10)

and

B(7, -2)

gives the x-coordinate as

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

and the y-coordinate as

\frac{10 + (-2)}{2} = 4

.

2

Calculate the distance from the midpoint

M(3, 4)

to the origin

(0, 0)

.

The distance is 5.

The distance formula is

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

. Substituting the coordinates of

M(3, 4)

and the origin

(0, 0)

gives

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

.

Key Concept

Using the midpoint formula to find the center point of a line segment, and then using the distance formula to find the length between that point and another specified coordinate.

Question 156Question →

In the standard $(x, y)$ coordinate plane, a line passes through the points $(\frac{1}{2}, \frac{2}{3})$ and $(\frac{3}{4}, \frac{5}{6})$ . What is the $y$ -intercept of this line?

\frac{1}{3}

\frac{1}{6}

-1

-\frac{1}{12}

-\frac{1}{7}

Show answer & explanation

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

Answer

The

y

-intercept of the line is

\frac{1}{3}

.

The correct answer is

\frac{1}{3}

. By calculating the slope of the line as

m = \frac{2}{3}

, and using the point-slope formula

y - y_1 = m(x - x_1)

with either given point, we can isolate

y

to get the equation

y = \frac{2}{3}x + \frac{1}{3}

, where the constant term represents the

y

-intercept.

Step-by-Step Solution

1

Calculate the slope (

m

) of the line passing through the points

(\frac{1}{2}, \frac{2}{3})

and

(\frac{3}{4}, \frac{5}{6})

using the slope formula

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

.

m = \frac{\frac{5}{6} - \frac{2}{3}}{\frac{3}{4} - \frac{1}{2}} = \frac{\frac{1}{6}}{\frac{1}{4}} = \frac{2}{3}

To find the equation of a line, we first need to determine its slope.

2

Use the point-slope form

y - y_1 = m(x - x_1)

with the point

(\frac{1}{2}, \frac{2}{3})

and slope

m = \frac{2}{3}

to write the equation of the line.

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

This allows us to write the linear equation in a form where we can isolate the

y

-intercept.

3

Isolate

y

to convert the equation into slope-intercept form

y = mx + b

, where

b

represents the

y

-intercept.

y = \frac{2}{3}x + \frac{1}{3}

The constant term in slope-intercept form represents the

y

-intercept.

Key Concept

Linear Equations and Graphing

Alternative Method

Instead of using the point-slope form, you can substitute one of the points directly into the slope-intercept form

y = mx + b

. Substituting

(\frac{1}{2}, \frac{2}{3})

and

m = \frac{2}{3}

yields

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

, which simplifies to

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

. Solving for

b

gives

b = \frac{1}{3}

.

Estimated Time:1m 30s

Question 157Question →

On a coordinate grid, a circle is drawn such that its center is at the point $(4, -2)$ . If the circle is tangent to the $y$ -axis, what is the equation of this circle?

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

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

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

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

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

Show answer & explanation

Answer: $(x - 4)^2 + (y + 2)^2 = 16$

Answer

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

The correct equation is

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

. A circle with center

(h, k)

and radius

r

has the standard equation

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

. Substituting

(4, -2)

for the center gives

(x - 4)^2 + (y + 2)^2 = r^2

. Since the circle is tangent to the

y

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

x = 0

, which is

4

units. Squaring the radius gives

r^2 = 16

.

Step-by-Step Solution

1

Recall the standard form of the equation of a circle.

The standard equation of a circle is

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

, where

(h, k)

is the center and

r

is the radius.

This sets up the formula needed to write the circle's equation.

2

Substitute the center coordinates

(h, k) = (4, -2)

into the formula.

(x - 4)^2 + (y - (-2))^2 = r^2

, which simplifies to

(x - 4)^2 + (y + 2)^2 = r^2

.

To represent the translation of the circle's center from the origin.

3

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

Since the circle is tangent to the

y

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

(4, -2)

to the

y

-axis (the line

x = 0

). This distance is

|4 - 0| = 4

units, so

r = 4

.

To find the radius from the geometric properties of the circle.

4

Square the radius and write the complete equation.

r^2 = 4^2 = 16

. The completed equation is

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

.

To complete the standard equation form.

Key Concept

Standard form of a circle's equation and determining its radius from a tangency condition

Question 158Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices at $A(k, 4)$ , $B(-1, 2)$ , and $C(3, -6)$ . The altitude from vertex $A$ to side $BC$ intersects the $y$ -axis at $(0, -2)$ . What is the value of $k$ ?

-12

3

4

12

24

Show answer & explanation

Answer: 12

Answer

12

The slope of side

BC

is calculated as

m_{BC} = \frac{-6 - 2}{3 - (-1)} = -2

. Since the altitude from vertex

A

is perpendicular to side

BC

, its slope must be the negative reciprocal of

-2

, which is

\frac{1}{2}

. The equation of the line containing this altitude, with a given

y

-intercept of

(0, -2)

, is

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

. Substituting the coordinates of vertex

A(k, 4)

into the equation gives

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

. Solving for

k

yields

k = 12

.

Step-by-Step Solution

1

Calculate the slope of side

BC

using the slope formula

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

.

m_{BC} = \frac{-6 - 2}{3 - (-1)} = \frac{-8}{4} = -2

The altitude is perpendicular to the side

BC

, so we first need the slope of

BC

.

2

Find the slope of the altitude by taking the negative reciprocal of the slope of

BC

.

m_{\text{alt}} = -\frac{1}{m_{BC}} = -\frac{1}{-2} = \frac{1}{2}

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

3

Write the equation of the line containing the altitude using the slope-intercept form

y = mx + b

with the given

y

-intercept of

(0, -2)

.

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

The line has a slope of

\frac{1}{2}

and crosses the

y

-axis at

-2

.

4

Substitute the coordinates of vertex

A(k, 4)

into the equation and solve for

k

.

4 = \frac{1}{2}k - 2 \Rightarrow 6 = \frac{1}{2}k \Rightarrow k = 12

Vertex

A

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

Key Concept

The slope of a line perpendicular to a given line is the negative reciprocal of the given line's slope.

Alternative Method

Alternatively, we can use the vector dot product. The vector representing side

BC

is

\vec{BC} = (3 - (-1), -6 - 2) = (4, -8)

. The vector from the

y

-intercept to vertex

A

is

\vec{v} = (k - 0, 4 - (-2)) = (k, 6)

. Since the altitude is perpendicular to

BC

, the dot product of these two vectors must equal zero:

(4)(k) + (-8)(6) = 0 \Rightarrow 4k - 48 = 0 \Rightarrow 4k = 48 \Rightarrow k = 12

.

Estimated Time:2m 0s

Question 159Question →

A circle in the standard $(x, y)$ coordinate plane passes through the points $(0, 0)$ , $(8, 0)$ , and $(2, 6)$ . What are the coordinates of the center of this circle?

(3, 3)

(4, 2)

(4, 3)

(5, 3)

(2, 4)

Show answer & explanation

Answer: $(4, 2)$

Answer

The center of the circle is

(4, 2)

The correct answer is

(4, 2)

. A circle's center

(h, k)

is equidistant from all points on the circle. By setting up the distance formula from the center

(h, k)

to the points

(0,0)

,

(8,0)

, and

(2,6)

, we obtain the equations

h^2 + k^2 = (h-8)^2 + k^2

and

h^2 + k^2 = (h-2)^2 + (k-6)^2

. Solving the first equation yields

16h = 64

, which simplifies to

h = 4

. Substituting

h = 4

into the second equation yields

16 + k^2 = 4 + (k-6)^2

. Expanding and simplifying gives

16 = 40 - 12k

, which simplifies to

12k = 24

, or

k = 2

. Therefore, the center of the circle is

(4, 2)

.

Step-by-Step Solution

1

Set up the distance formula equations for the center of the circle,

(h, k)

, which must be equidistant from the three points on the circle:

(0, 0)

,

(8, 0)

, and

(2, 6)

.

The distances from

(h, k)

to the points are

\sqrt{h^2 + k^2}

,

\sqrt{(h-8)^2 + k^2}

, and

\sqrt{(h-2)^2 + (k-6)^2}

. Since they are radii, we equate them:

h^2 + k^2 = (h-8)^2 + k^2 = (h-2)^2 + (k-6)^2

.

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

2

Equate the distance to

(0,0)

and the distance to

(8,0)

, then solve for

h

.

h^2 + k^2 = (h-8)^2 + k^2 \implies h^2 = h^2 - 16h + 64 \implies 16h = 64 \implies h = 4

.

Since both points lie on the x-axis, the x-coordinate of the center must lie on the perpendicular bisector of the segment connecting them, which is

x = 4

.

3

Substitute

h = 4

into the equation equating the distance to

(0,0)

and

(2,6)

to solve for

k

.

4^2 + k^2 = (4-2)^2 + (k-6)^2 \implies 16 + k^2 = 4 + k^2 - 12k + 36 \implies 16 = 40 - 12k \implies 12k = 24 \implies k = 2

.

Substituting the known coordinate allows us to solve the remaining single-variable equation for the

y

-coordinate of the center.

4

Combine the coordinates to state the final center of the circle.

The center is

(4, 2)

.

Both coordinates

(h, k)

have been uniquely determined.

Key Concept

Circumcenter of a triangle / Center of a circle passing through three points on a coordinate plane

Estimated Time:2m 0s

Question 160Question →

In the standard $(x, y)$ coordinate plane, if the three distinct points $(k, 4)$ , $(4, k)$ , and $(5, 6)$ are collinear, then there are exactly two possible values for the slope of the line passing through these points.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false because the requirement that the three points must be distinct eliminates one of the algebraic solutions for

k

, resulting in only one possible slope for the line.

The statement is false because equating the slopes between the points leads to the quadratic equation

k^2 - 11k + 28 = 0

, which has the solutions

k = 4

and

k = 7

. However, substituting

k = 4

back into the coordinates results in the points

(4, 4)

and

(4, 4)

, which are not distinct. Thus, only

k = 7

is a valid solution, which yields exactly one line with a slope of

-1

.

Step-by-Step Solution

1

Set up the collinearity condition using the slope formula.

The slope of the segment connecting

(4, k)

and

(5, 6)

is

m_1 = \frac{6 - k}{5 - 4} = 6 - k

. The slope of the segment connecting

(k, 4)

and

(5, 6)

is

m_2 = \frac{6 - 4}{5 - k} = \frac{2}{5 - k}

. Equating the two slopes gives

6 - k = \frac{2}{5 - k}

.

For three points to be collinear, the slope between any two pairs of points must be equal.

2

Solve the rational equation for

k

.

Multiplying both sides by

5 - k

yields

(6 - k)(5 - k) = 2

, which expands to

30 - 11k + k^2 = 2

. Rearranging into standard quadratic form gives

k^2 - 11k + 28 = 0

. Factoring the quadratic yields

(k - 4)(k - 7) = 0

, giving the solutions

k = 4

or

k = 7

.

Finding the roots of the quadratic equation identifies the algebraic candidates for

k

.

3

Verify if the points are distinct for each candidate value of

k

.

If

k = 4

, the first two points are

(4, 4)

and

(4, 4)

, which are not distinct. If

k = 7

, the three points are

(7, 4)

,

(4, 7)

, and

(5, 6)

, which are distinct.

The problem specifies that the three points must be distinct, so any solution violating this condition must be discarded.

4

Determine the number of possible slopes.

For the only valid value

k = 7

, the slope of the line is

m = 6 - 7 = -1

. There is only one possible slope.

Comparing the number of valid slopes to the statement shows that the statement is false.

Key Concept

Collinearity of points in the coordinate plane and constraints on slope calculation.

Coordinate Geometry

Answer

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