Coordinate Geometry Practice Questions — ACT — Page 5

Question 81Question →

In the standard $(x, y)$ coordinate plane, a line segment has midpoint $M(4, 1)$ and one endpoint $P(x, 5)$ . If the total length of the line segment is $10$ units, and $x < 4$ , what is the value of $x$ ?

Show answer & explanation

Answer: 1

Answer

1

The midpoint

M

divides the line segment into two equal segments, meaning the distance from endpoint

P

to midpoint

M

is half of the total length of the segment,

\frac{10}{2} = 5

units. Applying the distance formula between

P(x, 5)

and

M(4, 1)

gives

\sqrt{(x - 4)^2 + (5 - 1)^2} = 5

. Squaring both sides yields

(x - 4)^2 + 16 = 25

, which simplifies to

(x - 4)^2 = 9

. Solving for

x

gives

x - 4 = 3

or

x - 4 = -3

, meaning

x = 7

or

x = 1

. The problem specifies that

x < 4

, so the only valid value is

1

.

Step-by-Step Solution

1

Find the distance from the endpoint to the midpoint.

The distance

PM

is

5

units.

The midpoint divides the line segment into two equal parts, so the distance from any endpoint to the midpoint is half of the total length of the segment:

\frac{10}{2} = 5

.

2

Set up the distance formula equation for the segment

PM

.

\sqrt{(x - 4)^2 + (5 - 1)^2} = 5

Using the coordinates of

P(x, 5)

and

M(4, 1)

with the distance formula

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

.

3

Simplify the equation and square both sides.

(x - 4)^2 + 16 = 25

Calculating

(5 - 1)^2 = 16

and squaring both sides of the equation to eliminate the radical.

4

Isolate the squared term and solve for

x

.

x = 7

or

x = 1

Subtracting

16

from both sides gives

(x - 4)^2 = 9

. Taking the square root gives

x - 4 = 3

or

x - 4 = -3

, yielding

x = 7

or

x = 1

.

5

Apply the constraint

x < 4

.

x = 1

Since the question states that

x < 4

, we choose

x = 1

instead of

x = 7

.

Key Concept

The midpoint divides a segment into two segments of equal length, and the distance between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

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

.

Question 82Question →

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

Show answer & explanation

Answer: 4.5

Answer

The value of

p

is

4.5

.

The line equation is

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

. By substituting

y = 0

for the

x

-intercept, the equation becomes

-1 = -\frac{2}{3}(p - 3)

. Multiplying both sides by

-3

gives

3 = 2(p - 3)

, which simplifies to

2p = 9

, resulting in

p = 4.5

.

Step-by-Step Solution

1

Write the point-slope equation of the line.

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

We use the point-slope form

y - y_1 = m(x - x_1)

with the point

(3, 1)

and slope

-\frac{2}{3}

.

2

Substitute the point

(p, 0)

into the line's equation.

-1 = -\frac{2}{3}(p - 3)

The

x

-intercept

(p, 0)

lies on the line, so its coordinates must satisfy the equation of the line.

3

Solve for the variable

p

.

p = 4.5

Multiply by

-3

to get

3 = 2(p - 3)

, add

6

to both sides to get

2p = 9

, and divide by

2

.

Key Concept

Linear equations and graphing, specifically point-slope form and x-intercept calculation.

Question 83Question →

An ellipse in the standard $(x, y)$ coordinate plane is defined by the equation $\frac{(x-4)^2}{169} + \frac{(y+3)^2}{144} = 1$ . What is the distance between the two foci of this ellipse?

Show answer & explanation

Answer: 10

Answer

The distance between the two foci of the ellipse is 10.

By comparing the given equation to the standard form of an ellipse, we find

a^2 = 169

and

b^2 = 144

. The distance from the center to each focus,

c

, is given by

c = \sqrt{a^2 - b^2} = \sqrt{169 - 144} = \sqrt{25} = 5

. The total distance between the two foci is

2c = 2(5) = 10

.

Step-by-Step Solution

1

Identify

a^2

and

b^2

from the given equation of the ellipse.

a^2 = 169

and

b^2 = 144

The standard equation of a horizontal ellipse is

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

, where

a^2

is the larger denominator.

2

Calculate the value of

c

, the distance from the center to a focus.

c = 5

For an ellipse, the focal distance

c

is related to the semi-major axis

a

and semi-minor axis

b

by the equation

c^2 = a^2 - b^2

.

3

Calculate the distance between the two foci, which is

2c

.

10

The distance between the two foci of an ellipse is twice the distance from the center to each focus (

2c

).

Key Concept

Focal distance of an ellipse

Estimated Time:1m 15s

Question 84Question →

In the standard $(x, y)$ coordinate plane, a line with a non-zero slope $m$ passes through the point $(8, 2)$ . If the product of the line's $x$ -intercept and its $y$ -intercept is $64$ , what is the value of $m$ ?

Show answer & explanation

Answer: -0.25

Answer

The value of the slope

m

is

-0.25

(or

-\frac{1}{4}

).

By writing the line as

y = mx + b

, the given point

(8, 2)

establishes that

b = 2 - 8m

. Since the

x

-intercept is

-\frac{b}{m}

, the product of the intercepts is

-\frac{b^2}{m}

. Setting this product to

64

gives

-\frac{(2-8m)^2}{m} = 64

, which simplifies to the quadratic equation

16m^2 + 8m + 1 = 0

. Factoring this perfect square trinomial gives

(4m + 1)^2 = 0

, which yields the unique solution

m = -0.25

.

Step-by-Step Solution

1

Express the

y

-intercept in terms of

m

.

b = 2 - 8m

The line equation is

y = mx + b

and it passes through

(8, 2)

, so

2 = 8(m) + b

.

2

Express the

x

-intercept in terms of

m

.

x = -\frac{b}{m}

The

x

-intercept is the value of

x

when

y = 0

in the line equation

y = mx + b

.

3

Write the product of the intercepts and set it equal to

64

.

b^2 = -64m

The product of the intercepts is

b \cdot \left(-\frac{b}{m}\right) = -\frac{b^2}{m} = 64

.

4

Substitute

b = 2 - 8m

into the product equation and simplify to a quadratic equation.

64m^2 + 32m + 4 = 0

Substituting

b

gives

(2 - 8m)^2 = -64m

. Expanding and combining like terms yields the quadratic equation.

5

Solve the quadratic equation for

m

.

m = -0.25

Dividing by

4

gives

16m^2 + 8m + 1 = (4m + 1)^2 = 0

, which has the single real solution

m = -0.25

.

Key Concept

Using linear equation forms, point-slope relationships, and intercept properties to solve coordinate geometry problems.

Question 85Question →

In the standard $(x, y)$ coordinate plane, what is the distance between the points $(1, \frac{1}{2})$ and $(5, \frac{7}{2})$ ?

7

2\sqrt{5}

5

25

2\sqrt{13}

Show answer & explanation

Answer: $5$

Answer

The distance between the two points is

5

.

The distance

d

between

(x_1, y_1)

and

(x_2, y_2)

is given by the formula

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

. Substituting

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

and

(5, \frac{7}{2})

gives

d = \sqrt{(5-1)^2 + (\frac{7}{2} - \frac{1}{2})^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Calculate the horizontal change (difference in

x

-coordinates)

x_2 - x_1 = 5 - 1 = 4

This determines the horizontal leg of the right triangle formed between the two points.

2

Calculate the vertical change (difference in

y

-coordinates)

y_2 - y_1 = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3

This determines the vertical leg of the right triangle formed between the two points.

3

Apply the Pythagorean theorem (distance formula) to find the distance

d

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

The straight-line distance is the hypotenuse of the right triangle with legs of lengths 4 and 3.

Key Concept

The Distance Formula

Estimated Time:45s

Question 86Question →

Line $P$ is perpendicular to a line with a slope of $3$ . If line $P$ passes through the point $(2, 4)$ , what is the $y$ -intercept of line $P$ ?

-2

-\frac{2}{3}

\frac{3}{2}

\frac{14}{3}

10

Show answer & explanation

Answer: $\frac{14}{3}$

Answer

The

y

-intercept of line

P

is

\frac{14}{3}

.

The slope of a line perpendicular to a line with a slope of

3

is its negative reciprocal,

-\frac{1}{3}

. Using the point-slope form with the point

(2, 4)

gives the equation

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

. Setting

x = 0

to find the

y

-intercept yields

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

, which simplifies to

y = \frac{14}{3}

.

Step-by-Step Solution

1

Determine the slope of line

P

.

The slope of line

P

is

-\frac{1}{3}

.

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

3

is

-\frac{1}{3}

.

2

Write the equation of line

P

in point-slope form using the point

(2, 4)

.

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

The point-slope form is

y - y_1 = m(x - x_1)

, where

m

is the slope and

(x_1, y_1)

is the given point.

3

Solve for the

y

-intercept by setting

x = 0

.

y = \frac{14}{3}

The

y

-intercept occurs where

x = 0

. Substituting

x = 0

gives

y - 4 = -\frac{1}{3}(0 - 2) \Rightarrow y - 4 = \frac{2}{3} \Rightarrow y = \frac{2}{3} + 4 = \frac{14}{3}

.

Key Concept

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

Question 87Question →

In the standard $(x,y)$ coordinate plane, a right isosceles triangle $PQR$ has its right angle at vertex $Q(2, -1)$ and another vertex at $P(-1, 3)$ . If the third vertex $R(x, y)$ is located in the first quadrant, what is the value of $3x - y$ ?

16

12

-2

8

21

Show answer & explanation

Answer: 16

Answer

16

The correct answer is 16. To find the coordinates of vertex

R(x, y)

, we find that the vector

\vec{QP} = (-3, 4)

has a length of 5 and a slope of

-4/3

. Since

\triangle PQR

is a right isosceles triangle with the right angle at

Q

, the segment

QR

must be perpendicular to

QP

(slope of

3/4

) and equal in length (

QR = 5

). This yields two possible locations for

R

:

(6, 2)

and

(-2, -4)

. Because

R

must lie in the first quadrant, its coordinates are

(6, 2)

. Evaluating

3x - y

gives

3(6) - 2 = 16

.

Step-by-Step Solution

1

Calculate the vector

\vec{QP}

and the length of segment

QP

.

\vec{QP} = P - Q = (-1 - 2, 3 - (-1)) = (-3, 4)

. The distance is

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

.

Since the triangle is right isosceles with the right angle at

Q

, the leg

QR

must be perpendicular to

QP

and have the same length of 5.

2

Find the slope of

QP

and determine the slope of the perpendicular line containing

QR

.

Slope of

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

. The perpendicular slope of

QR

is the negative reciprocal, which is

\frac{3}{4}

.

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

3

Determine the possible coordinates of

R

by scaling the unit perpendicular vector.

The vector

\vec{QR}

must be of the form

(4k, 3k)

for some scalar

k

. Since the length is 5,

(4k)^2 + (3k)^2 = 25 \implies 25k^2 = 25 \implies k = \pm 1

. This yields two possible coordinates for

R

:

R = (2+4, -1+3) = (6, 2)

or

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

.

Using the slope components and the distance constraint ensures

QR

is perpendicular and equal in length to

QP

.

4

Apply the quadrant constraint and evaluate the final expression

3x - y

.

Since

R

is in the first quadrant,

R = (6, 2)

with

x=6, y=2

. Thus,

3x - y = 3(6) - 2 = 16

.

The coordinates of a point in the first quadrant must both be positive.

Key Concept

Using slopes, vectors, and distance formulas to determine the vertices of geometric figures on the coordinate plane.

Question 88Question →

A drone's position on a coordinate grid is represented by the point $(3, -5)$ . If the drone undergoes a translation of $4$ units to the left and $2$ units up, what are the coordinates of its new position on the grid?

(7, -7)

(-1, -7)

(7, -3)

(1, -1)

(-1, -3)

Show answer & explanation

Answer: $(-1, -3)$

Answer

(-1, -3)

The correct answer is

(-1, -3)

. A horizontal translation of

4

units left subtracts

4

from the original

x

-coordinate of

3

, resulting in a new

x

-coordinate of

3 - 4 = -1

. A vertical translation of

2

units up adds

2

to the original

y

-coordinate of

-5

, resulting in a new

y

-coordinate of

-5 + 2 = -3

. This places the drone at

(-1, -3)

.

Step-by-Step Solution

1

Calculate the new x-coordinate

x' = 3 - 4 = -1

Translating

4

units to the left means subtracting

4

from the initial

x

-coordinate of

3

.

2

Calculate the new y-coordinate

y' = -5 + 2 = -3

Translating

2

units up means adding

2

to the initial

y

-coordinate of

-5

.

3

Combine the new coordinates to write the final position

(-1, -3)

Combining the calculated

x

-coordinate and

y

-coordinate yields the final position of the drone.

Key Concept

Performing translations in the coordinate plane

Estimated Time:45s

Question 89Question →

In the standard $(x, y)$ coordinate plane, triangle $XYZ$ has vertices $X(1, 2)$ , $Y(4, 2)$ , and $Z(1, 6)$ . The triangle is reflected across the line $y = x$ , and then the resulting image is dilated by a scale factor of $3$ with the center of dilation at $(2, 1)$ to form triangle $X''Y''Z''$ . What are the coordinates of the vertex $Z''$ ?

(12, 0)

(14, 1)

(16, -1)

(18, 3)

(-22, -5)

Show answer & explanation

Answer: $(14, 1)$

Answer

The correct coordinates of the vertex

Z''

are

(14, 1)

.

The correct coordinate pair is

(14, 1)

. Swapping the coordinates of

Z(1, 6)

across the reflection line

y = x

gives

Z'(6, 1)

. The dilation of

Z'(6, 1)

by a scale factor of 3 relative to the center

(2, 1)

involves scaling the displacement vector

(4, 0)

to

(12, 0)

and adding the center back, resulting in

(14, 1)

.

Step-by-Step Solution

1

Reflect the point

Z(1, 6)

across the line

y = x

.

The intermediate image is

Z'(6, 1)

.

Reflecting a point across the line

y = x

swaps its

x

- and

y

-coordinates.

2

Calculate the displacement of

Z'(6, 1)

relative to the center of dilation

(2, 1)

.

(6 - 2, 1 - 1) = (4, 0)

.

To perform a dilation centered at a point other than the origin, the point's coordinate must first be measured relative to the center.

3

Multiply the displacement by the scale factor of

3

.

3 \times (4, 0) = (12, 0)

.

Dilation scales the distance from the center of dilation by the given factor.

4

Add the scaled displacement back to the center of dilation

(2, 1)

to find the final absolute coordinates.

(2 + 12, 1 + 0) = (14, 1)

.

This translates the relative coordinates back into the standard coordinate plane.

Key Concept

Applying composite transformations in the coordinate plane, specifically combining a line reflection and a dilation about a non-origin center.

Estimated Time:2m 0s

Question 90Question →

In the standard $(x, y)$ coordinate plane, rectangle $ABCD$ has vertices $A(-3, 1)$ , $B(-1, 1)$ , $C(-1, 4)$ , and $D(-3, 4)$ . The rectangle is first reflected across the $y$ -axis and then translated $3$ units to the left and $2$ units down. What are the coordinates of the final image of vertex $C$ ?

(-2, 2)

(4, 2)

(-2, 6)

(-4, -6)

(-4, 2)

Show answer & explanation

Answer: $(-2, 2)$

Answer

(-2, 2)

To find the final image of vertex

C(-1, 4)

, we first apply the reflection across the

y

-axis. The rule for reflecting a point across the

y

-axis is

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

. Applying this to

C(-1, 4)

gives

C'(1, 4)

. Next, we apply the translation of

3

units to the left and

2

units down. The rule for this translation is

(x, y) \rightarrow (x - 3, y - 2)

. Applying this to

C'(1, 4)

gives

C''(1 - 3, 4 - 2) = (-2, 2)

.

Step-by-Step Solution

1

Identify the initial coordinates of vertex

C

.

C(-1, 4)

We need to find the final image of vertex

C

, so we start with its given coordinates.

2

Apply the reflection across the

y

-axis to vertex

C

.

C'(1, 4)

Reflecting a point

(x, y)

across the

y

-axis negates the x-coordinate, changing

(x, y)

to

(-x, y)

.

3

Apply the translation of

3

units to the left and

2

units down to the reflected point

C'

.

C''(-2, 2)

Translating a point

(x, y)

by

3

units to the left subtracts

3

from the x-coordinate (

x - 3

), and translating

2

units down subtracts

2

from the y-coordinate (

y - 2

).

Key Concept

Composite transformations in the coordinate plane involve applying multiple geometric transformations in a specific sequence.

Question 91Question →

In the standard $(x, y)$ coordinate plane, line $L_1$ has a negative slope and passes through the point $(1, 4)$ . Line $L_2$ has a positive slope and passes through the point $(7, 3)$ . Both lines intersect the $x$ -axis at the same point $P$ . If the product of the slopes of $L_1$ and $L_2$ is $-\frac{4}{3}$ , and line $L_3$ is perpendicular to $L_1$ and passes through the point $(2, -2)$ , what is the $y$ -intercept of $L_3$ ?

Show answer & explanation

Answer: -3.5

Answer

The

y

-intercept of line

L_3

is

-3.5

.

By writing the slope of the first line as

\frac{-4}{x_p - 1}

and the slope of the second line as

\frac{-3}{x_p - 7}

, their product is set to

-\frac{4}{3}

. Solving the resulting quadratic equation

(x_p - 4)^2 = 0

yields

x_p = 4

. Substituting this back gives a slope of

-\frac{4}{3}

for the first line. The perpendicular line must have a slope of

\frac{3}{4}

. Using the point-slope formula with point

(2, -2)

and slope

\frac{3}{4}

gives the line

y = \frac{3}{4}x - 3.5

, which crosses the

y

-axis at

-3.5

.

Step-by-Step Solution

1

Express the slopes of

L_1

and

L_2

in terms of the unknown

x

-coordinate of point

P

.

m_1 = \frac{0 - 4}{x_p - 1} = \frac{-4}{x_p - 1}

and

m_2 = \frac{0 - 3}{x_p - 7} = \frac{-3}{x_p - 7}

Since

P

lies on the

x

-axis, its coordinates are

(x_p, 0)

. We apply the slope formula

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

to the points on each line.

2

Set up the equation for the product of the slopes.

\left(\frac{-4}{x_p - 1}\right)\left(\frac{-3}{x_p - 7}\right) = -\frac{4}{3} \implies \frac{12}{(x_p - 1)(x_p - 7)} = -\frac{4}{3}

The problem states that the product of the slopes of

L_1

and

L_2

is

-\frac{4}{3}

.

3

Solve the equation for

x_p

.

x_p = 4

Cross-multiplying gives

36 = -4(x_p^2 - 8x_p + 7)

, which simplifies to

-9 = x_p^2 - 8x_p + 7 \implies x_p^2 - 8x_p + 16 = 0 \implies (x_p - 4)^2 = 0

.

4

Find the slope of

L_1

.

m_1 = \frac{-4}{4 - 1} = -\frac{4}{3}

Substituting

x_p = 4

back into the expression for

m_1

yields the slope of

L_1

.

5

Find the slope of

L_3

.

m_3 = \frac{3}{4}

Since

L_3

is perpendicular to

L_1

, its slope is the negative reciprocal of

m_1

.

6

Write the linear equation for

L_3

and determine its

y

-intercept.

y = \frac{3}{4}x - 3.5

, so the

y

-intercept is

-3.5

.

Using the point-slope form with point

(2, -2)

and slope

m_3 = \frac{3}{4}

, we get

y - (-2) = \frac{3}{4}(x - 2) \implies y + 2 = \frac{3}{4}x - 1.5 \implies y = \frac{3}{4}x - 3.5

.

Key Concept

The slope of a line is defined as

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

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

m_1 \cdot m_2 = -1

. The

y

-intercept of a line is the value of

y

when

x = 0

.

Question 92Question →

In the standard $(x, y)$ coordinate plane, a line passes through the points $(1, 3)$ and $(4, 9)$ . Is the statement that this line has a slope of $2$ true or false?

True

False

Show answer & explanation

Answer: True

Answer

true

Applying the slope formula

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

to the points

(1, 3)

and

(4, 9)

yields

\frac{9 - 3}{4 - 1} = \frac{6}{3} = 2

, which confirms that the statement is true.

Step-by-Step Solution

1

Identify the coordinates of the two points on the line.

The points are

(x_1, y_1) = (1, 3)

and

(x_2, y_2) = (4, 9)

.

These coordinates are needed to apply the slope formula.

2

Apply the slope formula

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

using the identified points.

m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2

.

This calculates the rate of change of the line, which is its slope.

3

Compare the calculated slope to the value in the statement.

The calculated slope is

2

, which matches the statement.

To determine whether the statement is true or false.

Key Concept

The slope of a line is the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line.

Estimated Time:45s

Question 93Question →

In a coordinate plane, two perpendicular lines, $L_1$ and $L_2$ , intersect at a point on the positive $y$ -axis. The line $L_1$ passes through the point $(-4, 3)$ , and the line $L_2$ has an $x$ -intercept at $(2.5, 0)$ . What is the $y$ -coordinate of the intersection point of $L_1$ and $L_2$ ?

Show answer & explanation

Answer: 5

Answer

The y-coordinate of the intersection point is 5.

The intersection point on the positive

y

-axis has coordinates

(0, 5)

. The slope of the line passing through

(-4, 3)

and

(0, 5)

is

0.5

. The slope of the line passing through

(2.5, 0)

and

(0, 5)

is

-2

. Since the product of these slopes is

0.5 \times (-2) = -1

, the lines are perpendicular.

Step-by-Step Solution

1

Identify the coordinates of the intersection point.

The intersection point of

L_1

and

L_2

is

(0, b)

, where

b > 0

.

Since the intersection point lies on the positive

y

-axis, its

x

-coordinate is

0

and its

y

-coordinate

b

must be positive.

2

Calculate the slope of

L_1

.

The slope of

L_1

is

m_1 = \frac{b - 3}{0 - (-4)} = \frac{b - 3}{4}

.

The slope formula

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

is applied to the points

(-4, 3)

and

(0, b)

.

3

Calculate the slope of

L_2

.

The slope of

L_2

is

m_2 = \frac{b - 0}{0 - 2.5} = -\frac{b}{2.5}

.

The line

L_2

passes through the

y

-intercept

(0, b)

and its

x

-intercept

(2.5, 0)

.

4

Apply the perpendicular slope condition.

m_1 \cdot m_2 = -1 \implies \left(\frac{b - 3}{4}\right)\left(-\frac{b}{2.5}\right) = -1 \implies b(b - 3) = 10

.

Perpendicular lines have slopes that are negative reciprocals of each other, meaning their product is

-1

.

5

Solve the quadratic equation for

b

.

Expanding the equation gives

b^2 - 3b - 10 = 0

, which factors into

(b - 5)(b + 2) = 0

. Since

b > 0

, the only valid solution is

b = 5

.

Solving the factored equation yields

b = 5

or

b = -2

. The constraint that the intersection lies on the positive

y

-axis rules out the negative value.

Key Concept

Perpendicular line slopes and coordinate intercepts

Estimated Time:2m 30s

Question 94Question →

In the standard $(x, y)$ coordinate plane, the midpoint of the line segment with endpoints $A(2, a)$ and $B(b, -3)$ lies on the line $y = 3x - 4$ . If the distance between $A$ and $B$ is $5\sqrt{2}$ and $a > 0$ , what is the value of $a + b$ ?

13

-11

5

3

1

Show answer & explanation

Answer: 5

Answer

The value of

a + b

is

5

.

The midpoint of the segment with endpoints

(2, a)

and

(b, -3)

is

\left(\frac{2+b}{2}, \frac{a-3}{2}\right)

. Substituting these coordinates into the line

y = 3x - 4

gives

\frac{a-3}{2} = 3\left(\frac{2+b}{2}\right) - 4

, which simplifies to

a = 3b + 1

. The squared distance between the endpoints is

(b-2)^2 + (-3-a)^2 = (5\sqrt{2})^2 = 50

. Substituting

a = 3b + 1

into the distance equation yields

(b-2)^2 + (3b+4)^2 = 50

, which simplifies to the quadratic equation

b^2 + 2b - 3 = 0

. Factoring gives

(b+3)(b-1) = 0

, so

b = 1

or

b = -3

. The constraint

a > 0

means

a = 3(1) + 1 = 4

is the only valid solution. Therefore, the sum is

4 + 1 = 5

.

Step-by-Step Solution

1

Find the midpoint of the segment

AB

using the midpoint formula.

M = \left(\frac{2+b}{2}, \frac{a-3}{2}\right)

The midpoint is defined as the average of the x-coordinates and the average of the y-coordinates.

2

Substitute the midpoint coordinates into the equation of the line

y = 3x - 4

to relate

a

and

b

.

a = 3b + 1

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

3

Use the distance formula to set up an equation for the distance between

A

and

B

.

(b-2)^2 + (a+3)^2 = 50

The squared distance between

(2, a)

and

(b, -3)

is

(5\sqrt{2})^2 = 50

.

4

Substitute the relation

a = 3b + 1

into the distance equation and solve for

b

.

b = 1

or

b = -3

This yields the quadratic equation

b^2 + 2b - 3 = 0

, which factors as

(b+3)(b-1) = 0

.

5

Apply the constraint

a > 0

to find the correct values of

a

and

b

.

a = 4

and

b = 1

If

b = -3

, then

a = 3(-3) + 1 = -8

, which is not greater than

0

. If

b = 1

, then

a = 3(1) + 1 = 4 > 0

.

6

Calculate the sum

a + b

.

4 + 1 = 5

To find the final requested value.

Key Concept

Using the distance and midpoint formulas in combination with linear equations to solve for unknown coordinates.

Estimated Time:2m 30s

Question 95Question →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 + y^2 - 4x + 6y - 12 = 0$ . What is the radius of this circle?

25

2\sqrt{3}

5

8

\sqrt{17}

Show answer & explanation

Answer: $5$

Answer

The radius of the circle is 5.

The correct answer is 5. Grouping the

x

-terms and

y

-terms of the equation

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

gives

(x^2 - 4x) + (y^2 + 6y) = 12

. Completing the square requires adding

(\frac{-4}{2})^2 = 4

and

(\frac{6}{2})^2 = 9

to both sides, yielding

(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9

, which simplifies to

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

. In the standard circle equation form

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

,

r^2 = 25

, meaning the radius

r

is

\sqrt{25} = 5

.

Step-by-Step Solution

1

Group the

x

-terms and

y

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

(x^2 - 4x) + (y^2 + 6y) = 12

This prepares the quadratic expression to be written in standard circle form by completing the square.

2

Find the constant values needed to complete the square for both the

x

and

y

variables. Add these constants to both sides of the equation.

(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9

The constant for

x

is

(\frac{-4}{2})^2 = 4

, and the constant for

y

is

(\frac{6}{2})^2 = 9

. Adding them to both sides maintains equality.

3

Factor the perfect square trinomials on the left side and simplify the right side.

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

This puts the equation in the standard form

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

.

4

Identify

r^2

from the standard form and solve for the radius

r

.

r = \sqrt{25} = 5

Taking the square root of the constant on the right side yields the radius of the circle.

Key Concept

Rewriting a circle's equation from general form to standard form by completing the square to find its radius.

Alternative Method

For a circle given in the general form

x^2 + y^2 + Dx + Ey + F = 0

, the radius can be calculated directly using the formula

r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}

. Substituting the coefficients

D = -4

,

E = 6

, and

F = -12

gives:

r = \frac{1}{2}\sqrt{(-4)^2 + 6^2 - 4(-12)} = \frac{1}{2}\sqrt{16 + 36 + 48} = \frac{1}{2}\sqrt{100} = \frac{1}{2}(10) = 5

.

Estimated Time:1m 30s

Question 96Question →

In the standard $(x, y)$ coordinate plane, line $L_1$ passes through the points $(-1, 2)$ and $(3, 14)$ . If line $L_2$ is parallel to line $L_1$ , what is the slope of line $L_2$ ?

Show answer & explanation

Answer: 3

Answer

The slope of line

L_2

is

3

.

To find the slope of line

L_2

, we calculate the slope of line

L_1

using the formula

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

. Substituting the coordinates

(-1, 2)

and

(3, 14)

yields

m = \frac{14 - 2}{3 - (-1)} = \frac{12}{4} = 3

. Because parallel lines have the same slope, the slope of line

L_2

is also

3

.

Step-by-Step Solution

1

Calculate the slope of line

L_1

using the coordinates of the two given points,

(-1, 2)

and

(3, 14)

.

m = 3

The slope formula is

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

, which evaluates to

m = \frac{14 - 2}{3 - (-1)} = \frac{12}{4} = 3

.

2

Determine the slope of line

L_2

based on its relationship to line

L_1

.

The slope of line

L_2

is

3

.

Parallel lines always have identical slopes.

Key Concept

Parallel lines have the same slope.

Question 97Question →

In the standard $(x, y)$ coordinate plane, a line segment has endpoints $A(-3, 2)$ and $B(1, 4)$ . The segment is reflected across the line $y = -x$ to form segment $A'B'$ , which is then translated $5$ units to the right and $3$ units down to form segment $A''B''$ . What are the coordinates of the midpoint of segment $A''B''$ ?

(2, -2)

(-8, 4)

(8, -4)

(0, -4)

(6, 0)

Show answer & explanation

Answer: $(2, -2)$

Answer

(2, -2)

The correct answer is

(2, -2)

. The midpoint of the original segment

AB

is calculated as

M(-1, 3)

. Reflecting

M

across the line

y = -x

yields

M'(-3, 1)

. Translating

M'

by

5

units right and

3

units down results in the final midpoint

M''(2, -2)

.

Step-by-Step Solution

1

Find the coordinates of the midpoint of the original segment

AB

.

The midpoint

M

of

AB

is calculated using the midpoint formula:

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

.

Since translations, reflections, and dilations preserve midpoints, the midpoint of the final image segment

A''B''

is the same point obtained by applying the transformations to the midpoint of the original segment

AB

.

2

Apply the reflection across the line

y = -x

to the midpoint

M(-1, 3)

.

Using the reflection rule

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

, the reflected midpoint is

M' = (-3, 1)

.

A reflection across

y = -x

swaps the coordinates of a point and negates both values.

3

Apply the translation of

5

units right and

3

units down to the point

M'(-3, 1)

.

Using the translation rule

(x, y) \rightarrow (x + 5, y - 3)

, the final midpoint is

M'' = (-3 + 5, 1 - 3) = (2, -2)

.

Translating a point right increases the x-coordinate, and translating it down decreases the y-coordinate.

Key Concept

Applying composite transformations (reflection across

y = -x

followed by translation) to geometric objects and midpoints in the coordinate plane.

Estimated Time:2m 0s

Question 98Question →

In the standard $(x, y)$ coordinate plane, the line described by the equation $2x - 3y = 10$ is translated $2$ units to the left and $1$ unit up. What is the $y$ -intercept of the translated line?

-\frac{17}{3}

-3

-\frac{7}{3}

-\frac{5}{3}

-1

Show answer & explanation

Answer: $-1$

Answer

The correct answer is

-1

.

The correct answer is

-1

. Translating a line 2 units to the left shifts every point on the line by

-2

in the

x

-direction, which is represented algebraically by substituting

x + 2

for

x

. Translating it 1 unit up shifts every point by

+1

in the

y

-direction, represented by substituting

y - 1

for

y

. Substituting these into the original equation

2x - 3y = 10

yields

2(x + 2) - 3(y - 1) = 10

. Simplifying this equation gives

2x + 4 - 3y + 3 = 10

, which becomes

2x - 3y = 3

. To find the

y

-intercept of this new line, we set

x = 0

, giving

-3y = 3

, which solves to

y = -1

. Alternatively, we can find the original

y

-intercept of

(0, -10/3)

. Under the translation, this point shifts to

(0 - 2, -10/3 + 1) = (-2, -7/3)

. Since the slope of the line remains

2/3

, the new line equation is

y - (-7/3) = (2/3)(x - (-2))

, which simplifies to

y = (2/3)x - 1

, confirming the

y

-intercept is

-1

.

Step-by-Step Solution

1

Express the translation of the coordinates mathematically. Translating a line 2 units to the left replaces

x

with

x + 2

. Translating it 1 unit up replaces

y

with

y - 1

.

The substitution expressions are

x \to x + 2

and

y \to y - 1

.

This sets up the algebraic transformation of the original line's equation.

2

Substitute the expressions into the original equation

2x - 3y = 10

and simplify.

2(x + 2) - 3(y - 1) = 10 \implies 2x + 4 - 3y + 3 = 10 \implies 2x - 3y + 7 = 10 \implies 2x - 3y = 3

.

This derives the equation of the translated line.

3

Find the

y

-intercept of the translated line

2x - 3y = 3

by setting

x = 0

.

2(0) - 3y = 3 \implies -3y = 3 \implies y = -1

.

The

y

-intercept is the point on the line where

x = 0

.

Key Concept

Translating linear equations in the standard coordinate plane

Estimated Time:1m 30s

Question 99Question →

A circle in the standard $(x,y)$ coordinate plane is defined by the equation $3x^2 + 3y^2 - 24x + 12y + k = 0$ , where $k$ is a constant. If the circle is tangent to the line $y = 2$ , what is the value of $k$ ?

-48

0

12

48

60

Show answer & explanation

Answer: 12

Answer

12

The correct answer is 12. Dividing the given equation

3x^2 + 3y^2 - 24x + 12y + k = 0

by

3

gives

x^2 + y^2 - 8x + 4y + \frac{k}{3} = 0

. Completing the square for both variables yields

(x-4)^2 + (y+2)^2 = 20 - \frac{k}{3}

. This represents a circle centered at

(4, -2)

with a radius squared of

r^2 = 20 - \frac{k}{3}

. A circle centered at

(4, -2)

that is tangent to the line

y = 2

has a radius equal to the vertical distance between

y = -2

and

y = 2

, which is

|-2 - 2| = 4

. Thus,

r^2 = 16

. Setting

20 - \frac{k}{3} = 16

yields

\frac{k}{3} = 4

, which gives

k = 12

.

Step-by-Step Solution

1

Divide the entire equation by 3 to normalize the coefficients of the quadratic terms.

x^2 + y^2 - 8x + 4y + \frac{k}{3} = 0

The standard equation of a circle requires the coefficients of

x^2

and

y^2

to be 1.

2

Complete the square for the

x

and

y

terms.

(x - 4)^2 - 16 + (y + 2)^2 - 4 + \frac{k}{3} = 0 \Rightarrow (x - 4)^2 + (y + 2)^2 = 20 - \frac{k}{3}

This rewrites the equation in the standard form

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

to identify the center and radius.

3

Identify the center of the circle and set up the radius relationship.

Center is

(4, -2)

and the radius squared is

r^2 = 20 - \frac{k}{3}

.

Comparing

(x - 4)^2 + (y + 2)^2 = 20 - \frac{k}{3}

to the standard form gives the center

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

and

r^2

.

4

Determine the radius using the tangency condition with the line

y = 2

.

Radius

r = |y_{\text{center}} - 2| = |-2 - 2| = 4

Since the circle is tangent to the horizontal line

y = 2

, the radius is the vertical distance from the center's y-coordinate to the line.

5

Solve for the constant

k

using the radius squared value.

r^2 = 16 \Rightarrow 20 - \frac{k}{3} = 16 \Rightarrow 4 = \frac{k}{3} \Rightarrow k = 12

Equating the two expressions for

r^2

allows us to solve for

k

.

Key Concept

Converting the general equation of a circle to standard form by completing the square and using geometric tangency conditions to solve for unknown constants.

Estimated Time:2m 0s

Question 100Question →

In the standard $(x, y)$ coordinate plane, line segment $CD$ has endpoints $C(-2, 2)$ and $D(-2, 8)$ . If segment $CD$ is reflected across the $y$ -axis to form segment $C'D'$ , what are the coordinates of the midpoint of segment $C'D'$ ?

(2, 5)

(-2, 5)

(-2, -5)

(2, -5)

(5, 2)

Show answer & explanation

Answer: $(2, 5)$

Answer

The coordinate point

(2, 5)

The correct answer is

(2, 5)

because the midpoint of the original segment

CD

is located at

(-2, 5)

. When a point is reflected across the

y

-axis, its

x

-coordinate changes sign while its

y

-coordinate remains the same. Therefore, the image of the midpoint is

(2, 5)

.

Step-by-Step Solution

1

Find the midpoint of the original segment

CD

.

Midpoint

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

The midpoint formula is

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

.

2

Apply the reflection across the

y

-axis to the midpoint.

M' = (-(-2), 5) = (2, 5)

Reflecting a point

(x, y)

across the

y

-axis maps it to

(-x, y)

.

Key Concept

Reflection of points and segments across the axes in the coordinate plane

Alternative Method

Instead of finding the midpoint first, you can reflect the endpoints

C(-2, 2)

and

D(-2, 8)

across the

y

-axis to get

C'(2, 2)

and

D'(2, 8)

. Then, apply the midpoint formula to these new coordinates to get the midpoint:

\left(\frac{2 + 2}{2}, \frac{2 + 8}{2}\right) = (2, 5)

.

Estimated Time:45s

Coordinate Geometry

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