Coordinate Geometry Practice Questions — ACT — Page 2

Question 21Question →

In the standard $(x,y)$ coordinate plane, a line with a negative slope passes through the point $(3, 4)$ . If the sum of the line's $x$ -intercept and $y$ -intercept is $14$ , which of the following could be the slope of this line?

-\frac{3}{4}

-\frac{4}{3}

-\frac{2}{3}

-4

\frac{4}{3}

Show answer & explanation

Answer: $-\frac{4}{3}$

Answer

-\frac{4}{3}

The correct answer is

-\frac{4}{3}

. By representing the line in point-slope form as

y - 4 = m(x - 3)

, we determine that the

y

-intercept is at

(0, -3m + 4)

and the

x

-intercept is at

\left(3 - \frac{4}{m}, 0\right)

. Adding these intercepts together and setting the sum equal to

14

gives the equation

-3m + 7 - \frac{4}{m} = 14

. Simplifying this equation leads to the quadratic expression

3m^2 + 7m + 4 = 0

, which factors as

(3m + 4)(m + 1) = 0

. This yields two possible negative slopes:

-\frac{4}{3}

and

-1

. Among the choices,

-\frac{4}{3}

is the only matching option.

Step-by-Step Solution

1

Write the general equation of a line passing through

(3, 4)

with slope

m

.

y - 4 = m(x - 3)

Using the point-slope form

y - y_1 = m(x - x_1)

is a direct way to parameterize the line in terms of its slope.

2

Find the

y

-intercept of the line by setting

x = 0

.

y = -3m + 4

The

y

-intercept is the point where the line crosses the

y

-axis, which occurs when

x = 0

.

3

Find the

x

-intercept of the line by setting

y = 0

and solving for

x

.

x = 3 - \frac{4}{m}

The

x

-intercept is the point where the line crosses the

x

-axis, which occurs when

y = 0

.

4

Set the sum of the

x

-intercept and

y

-intercept equal to

14

.

(-3m + 4) + \left(3 - \frac{4}{m}\right) = 14

This translates the given condition that the sum of the intercepts is

14

.

5

Simplify the equation and clear the fraction by multiplying by

-m

.

3m^2 + 7m + 4 = 0

Grouping like terms yields

-3m - \frac{4}{m} - 7 = 0

. Multiplying by

-m

transforms it into a standard quadratic equation.

6

Factor the quadratic equation to solve for

m

.

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

or

m = -1

Factoring helps find the values of

m

that satisfy the equation. Both values represent lines with negative slopes.

Key Concept

Using the point-slope form of a linear equation to find intercepts and solving the resulting quadratic equation to determine the slope.

Alternative Method

An alternative approach is using the intercept form of a line,

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

, where

a

and

b

are the

x

- and

y

-intercepts. Since the point

(3, 4)

lies on the line, we have

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

. We are also given

a + b = 14 \implies b = 14 - a

. Substituting

b

into the equation gives

\frac{3}{a} + \frac{4}{14-a} = 1

. Solving this equation by finding a common denominator results in

3(14-a) + 4a = a(14-a) \implies 42 + a = 14a - a^2 \implies a^2 - 13a + 42 = 0

. Factoring gives

(a-6)(a-7) = 0

, so

a = 6

or

a = 7

. If

a = 6

, then

b = 8

, and the slope

m = -\frac{b}{a} = -\frac{8}{6} = -\frac{4}{3}

. If

a = 7

, then

b = 7

, and the slope

m = -\frac{7}{7} = -1

. This confirms the possible slopes.

Estimated Time:2m 30s

Question 22Question →

In the standard $(x, y)$ coordinate plane, a hyperbola is defined by the equation $9x^2 - 16y^2 - 54x - 64y - 127 = 0$ . What is the shortest distance from the focus of the hyperbola with the larger $x$ -coordinate to the asymptote with the positive slope?

\frac{15}{7}

3

4

5

\frac{17}{5}

Show answer & explanation

Answer: 3

Answer

3

The correct answer is 3. Completing the square for

9x^2 - 16y^2 - 54x - 64y - 127 = 0

yields the standard form equation

\frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1

. The focus with the larger

x

-coordinate is at

(8, -2)

and the asymptote with the positive slope is

3x - 4y - 17 = 0

. Applying the point-to-line distance formula yields a distance of 3.

Step-by-Step Solution

1

Group the

x

-terms and

y

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

9(x^2 - 6x) - 16(y^2 + 4y) = 127

Grouping the terms allows us to complete the square for the

x

and

y

variables separately.

2

Complete the square for both the

x

and

y

expressions, adjusting the right side of the equation by adding the weighted constants.

9(x^2 - 6x + 9) - 16(y^2 + 4y + 4) = 127 + 9(9) - 16(4) \implies 9(x-3)^2 - 16(y+2)^2 = 144

Completing the square allows us to write the quadratic expressions as perfect squares to put the equation in standard form.

3

Divide both sides of the equation by 144 to obtain the standard form of the hyperbola.

\frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1

The standard form of a horizontal hyperbola is

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

, which reveals the center

(h, k)

and the semi-axes

a

and

b

.

4

Identify the key parameters of the hyperbola: center,

a

,

b

, and calculate the focal distance

c

.

Center is

(3, -2)

,

a = 4

,

b = 3

, and

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

.

These parameters are required to find the coordinates of the focus and the equation of the asymptote.

5

Find the coordinates of the focus with the larger

x

-coordinate.

Focus is

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

.

For a horizontal hyperbola, the foci are located at

(h \pm c, k)

. The focus with the larger

x

-coordinate is at

(h+c, k)

.

6

Determine the equation of the asymptote with the positive slope.

The asymptote equation is

y + 2 = \frac{3}{4}(x - 3) \implies 3x - 4y - 17 = 0

.

The asymptotes of a horizontal hyperbola are given by

y - k = \pm \frac{b}{a}(x - h)

. The one with the positive slope uses

+\frac{b}{a}

.

7

Use the point-to-line distance formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

to find the distance from the focus

(8, -2)

to the asymptote line

3x - 4y - 17 = 0

.

d = \frac{|3(8) - 4(-2) - 17|}{\sqrt{3^2 + (-4)^2}} = \frac{|24 + 8 - 17|}{\sqrt{25}} = \frac{15}{5} = 3

.

Calculating this gives the shortest distance from the focus to the asymptote.

Key Concept

Rewriting the general equation of a hyperbola into standard form by completing the square, identifying its center, foci, and asymptotes, and applying the distance formula from a point to a line.

Question 23Question →

A circle in the standard $(x,y)$ coordinate plane has its center at $(2, 3)$ and passes through the point $(8, 11)$ . What is the length of the radius of this circle?

10

14

28

70

100

Show answer & explanation

Answer: 10

Answer

10

The correct answer is 10. The radius of the circle is the distance from the center

(2,3)

to the point

(8,11)

on the circle. Substituting these coordinates into the distance formula yields

\sqrt{(8-2)^2 + (11-3)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10

.

Step-by-Step Solution

1

Identify the formula for the radius of a circle given its center and a point on the circle.

The radius

r

is the distance between the center

(x_1, y_1) = (2, 3)

and the point

(x_2, y_2) = (8, 11)

.

By definition, the radius of a circle is the straight-line distance from the center to any point on its outer boundary.

2

Substitute the coordinates into the distance formula

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

.

r = \sqrt{(8 - 2)^2 + (11 - 3)^2}

This sets up the calculation for the horizontal and vertical changes between the center and the point on the circle.

3

Simplify the terms inside the parentheses and evaluate the exponents.

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

Following the order of operations, we first perform the subtraction within the parentheses, and then evaluate the squares.

4

Add the values inside the radical and take the square root.

r = \sqrt{100} = 10

We must sum the terms under the square root before taking the square root of the total sum.

Key Concept

Using the distance formula to find the radius of a circle given its center and a point on the circle.

Estimated Time:45s

Question 24Question →

In the standard $(x,y)$ coordinate plane, a line has an $x$ -intercept of $(a, 0)$ , where $a \neq 0$ , and a $y$ -intercept of $(0, 2a)$ . If the line passes through the point $(4, -3)$ , what is the value of $a$ ?

Show answer & explanation

Answer: 2.5

Answer

The value of

a

is

2.5

.

Representing the intercepts as

(a, 0)

and

(0, 2a)

lets us find the slope of the line, which is

m = \frac{2a - 0}{0 - a} = -2

. The line can then be written as

y = -2x + 2a

. Substituting the point

(4, -3)

into the equation yields

-3 = -2(4) + 2a

, which simplifies to

2a = 5

, and therefore

a = 2.5

.

Step-by-Step Solution

1

Find the slope of the line in terms of the variable

a

.

The slope is

m = -2

.

Applying the slope formula to the points

(a, 0)

and

(0, 2a)

yields

m = \frac{2a - 0}{0 - a} = -2

.

2

Write the general equation of the line.

The equation is

y = -2x + 2a

.

The slope is

-2

and the

y

-intercept is

2a

, so the equation in slope-intercept form is

y = mx + b

.

3

Substitute the coordinates of the point

(4, -3)

to find

a

.

a = 2.5

.

Substituting

x = 4

and

y = -3

into

y = -2x + 2a

gives

-3 = -8 + 2a

, which simplifies to

2a = 5

and

a = 2.5

.

Key Concept

Linear Equations and Graphing

Question 25Question →

In the standard $(x,y)$ coordinate plane, the midpoint of a line segment is $(2, 5)$ . If one of the endpoints of the segment is $(-2, 1)$ , what is the $x$ -coordinate of the other endpoint?

Show answer & explanation

Answer: 6

Answer

The

x

-coordinate of the other endpoint is 6.

According to the midpoint formula, the

x

-coordinate of the midpoint is the average of the

x

-coordinates of the two endpoints. Substituting the given values yields the equation

2 = \frac{-2 + x_2}{2}

. Multiplying both sides by 2 gives

4 = -2 + x_2

, and adding 2 to both sides results in

x_2 = 6

.

Step-by-Step Solution

1

Set up the midpoint equation for the

x

-coordinate using the midpoint formula

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

.

2 = \frac{-2 + x_2}{2}

The

x

-coordinate of the midpoint is the average of the

x

-coordinates of the endpoints.

2

Multiply both sides of the equation by 2 to solve for the numerator.

4 = -2 + x_2

Multiplying by 2 eliminates the denominator on the right side.

3

Add 2 to both sides of the equation to isolate

x_2

.

x_2 = 6

Adding 2 to both sides isolates the variable

x_2

.

Key Concept

Midpoint Formula

Question 26Question →

In the standard $(x, y)$ coordinate plane, a line $L_1$ with a positive slope $m$ and a negative $y$ -intercept $b$ passes through the point $(4, 3)$ . The region in the fourth quadrant bounded by the line $L_1$ , the $x$ -axis, and the $y$ -axis has an area of exactly $8$ square units. What is the $y$ -coordinate of the $y$ -intercept of line $L_1$ ?

Show answer & explanation

Answer: -6

Answer

The y-coordinate of the y-intercept of line

L_1

is

-6

.

The correct answer is

-6

. Substituting

(4, 3)

into the slope-intercept equation

y = mx + b

gives

3 = 4m + b

, or

m = \frac{3-b}{4}

. The area of the right triangle in the fourth quadrant is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (-\frac{b}{m})(-b) = \frac{b^2}{2m} = 8

. Substituting

m

yields

b^2 = 16\left(\frac{3-b}{4}\right)

, which simplifies to the quadratic equation

b^2 + 4b - 12 = 0

. Factoring gives

(b+6)(b-2) = 0

. Since the y-intercept

b

must be negative, we have

b = -6

.

Step-by-Step Solution

1

Substitute the given point into the slope-intercept equation

m = \frac{3 - b}{4}

Since the line passes through

(4, 3)

, substituting these coordinates into

y = mx + b

allows us to express the slope

m

in terms of the y-intercept

b

.

2

Determine the intercepts and the dimensions of the bounded region

Base

= -\frac{b}{m}

and Height

= -b

The boundary of the region in the fourth quadrant is a right triangle formed by the origin, the x-intercept

(-\frac{b}{m}, 0)

, and the y-intercept

(0, b)

.

3

Set up the area of the triangle and equate it to 8

b^2 = 16m

The area of a right triangle is

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

, so

\frac{1}{2} \left(-\frac{b}{m}\right)(-b) = 8

simplifies to

b^2 = 16m

.

4

Substitute

m

into the area equation and solve the resulting quadratic equation

b = -6

(discarding

b = 2

)

Substituting

m = \frac{3 - b}{4}

yields

b^2 + 4b - 12 = 0

, which factors into

(b + 6)(b - 2) = 0

. Since the region is in the fourth quadrant, the y-intercept must be negative (

b < 0

).

Key Concept

Using linear equation intercepts to calculate bounded areas on the coordinate plane and relating variables using point substitution.

Question 27Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices $A(1, 2)$ , $B(7, 2)$ , and $C$ . The midpoint of side $AC$ lies on the line $y = 3x - 3$ , and the midpoint of side $BC$ lies on the line $y = -2x + 13$ . What is the distance between point $C$ and the midpoint of side $AB$ ?

3

4

\sqrt{5}

2\sqrt{2}

2\sqrt{5}

Show answer & explanation

Answer: $\sqrt{5}$

Answer

The distance between point

C

and the midpoint of side

AB

is

\sqrt{5}

.

The coordinates of point

C(3, 4)

are determined by setting up the midpoint coordinates for sides

AC

and

BC

and substituting them into their respective line equations. The midpoint of

AB

is calculated to be

(4, 2)

. Using the distance formula between

C(3, 4)

and

(4, 2)

yields

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

.

Step-by-Step Solution

1

Express the midpoint of side

AC

in terms of the unknown coordinates of point

C(x, y)

and substitute it into the given line equation.

The midpoint of

AC

is

M_{AC} = \left(\frac{x+1}{2}, \frac{y+2}{2}\right)

. Substituting this into

y = 3x - 3

yields:

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

, which simplifies to

y = 3x - 5

.

Since the midpoint of

AC

lies on the line

y = 3x - 3

, its coordinates must satisfy the equation of the line.

2

Express the midpoint of side

BC

in terms of the unknown coordinates of point

C(x, y)

and substitute it into the second given line equation.

The midpoint of

BC

is

M_{BC} = \left(\frac{x+7}{2}, \frac{y+2}{2}\right)

. Substituting this into

y = -2x + 13

yields:

\frac{y+2}{2} = -2\left(\frac{x+7}{2}\right) + 13

, which simplifies to

y = -2x + 10

.

Since the midpoint of

BC

lies on the line

y = -2x + 13

, its coordinates must satisfy this equation.

3

Solve the system of two linear equations to find the coordinates of point

C(x, y)

.

Equating the two expressions for

y

:

3x - 5 = -2x + 10 \Rightarrow 5x = 15 \Rightarrow x = 3

. Substituting

x = 3

back into the first equation:

y = 3(3) - 5 = 4

. Thus,

C = (3, 4)

.

Point

C

must simultaneously satisfy the midpoint constraints on both sides

AC

and

BC

.

4

Find the coordinates of the midpoint of side

AB

.

The midpoint of segment

AB

with endpoints

A(1, 2)

and

B(7, 2)

is

M_{AB} = \left(\frac{1+7}{2}, \frac{2+2}{2}\right) = (4, 2)

.

The question asks for the distance between point

C

and the midpoint of

AB

, so we need to determine the coordinates of this midpoint first.

5

Calculate the distance between point

C(3, 4)

and the midpoint

M_{AB}(4, 2)

using the distance formula.

The distance

d

is:

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

.

The distance formula

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

is used to find the straight-line distance between two points in a coordinate plane.

Key Concept

Applying the midpoint and distance formulas within coordinate geometry constraint systems.

Alternative Method

Instead of algebraically solving for the lines of midpoints, one can translate the lines using vectors. The set of possible points

C

when the midpoint of

AC

lies on line

L_1

is a line

L'_1

obtained by dilating

L_1

by a factor of 2 with respect to center

A

. Dilating

y = 3x - 3

from

A(1, 2)

gives the line

y = 3x - 5

. Similarly, dilating

y = -2x + 13

from

B(7, 2)

by a factor of 2 gives

y = -2x + 10

. The intersection of these two dilated lines is point

C(3, 4)

.

Estimated Time:2m 30s

Question 28Question →

In the standard $(x,y)$ coordinate plane, a line with a negative slope passes through the point $(3, 2)$ and has positive integer intercepts $(a, 0)$ and $(0, b)$ . What is the sum of all possible values of $a$ ?

10

15

24

20

27

Show answer & explanation

Answer: 24

Answer

The sum of all possible values of

a

is 24.

The correct answer is the sum 24. Writing the line's equation in intercept form gives

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

. Substituting the point

(3, 2)

yields

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

. Solving for

b

results in

b = 2 + \frac{6}{a-3}

. Because both

a

and

b

must be positive integers, the expression

a-3

must be a positive divisor of

6

. The positive divisors of

6

are

1, 2, 3,

and

6

, which correspond to the

a

values of

4, 5, 6,

and

9

. Summing these values gives

4 + 5 + 6 + 9 = 24

.

Step-by-Step Solution

1

Set up the intercept form of the linear equation.

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

A line with

x

-intercept

(a,0)

and

y

-intercept

(0,b)

can be written in intercept form.

2

Substitute the given point

(3, 2)

into the equation.

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

Since the line passes through

(3,2)

, these coordinates must satisfy the equation.

3

Solve the equation for

b

in terms of

a

.

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

Expressing

b

in terms of

a

helps analyze the integer constraints.

4

Rewrite the expression for

b

to isolate the fractional part.

b = \frac{2(a-3) + 6}{a-3} = 2 + \frac{6}{a-3}

This form allows us to see when

b

will be an integer based on the divisors of the numerator.

5

Determine the positive integer solutions for

a

and

b

.

a-3

must be a positive divisor of

6

. The positive divisors of

6

are

1, 2, 3,

and

6

. This yields:
- If

a-3=1 \implies a=4, b=8

- If

a-3=2 \implies a=5, b=5

- If

a-3=3 \implies a=6, b=4

- If

a-3=6 \implies a=9, b=3

Since

a

and

b

must be positive integers,

a-3

must be positive and divide 6 evenly. Negative divisors (like

-1

or

-2

) would make

b

negative or zero.

6

Sum the possible values of

a

.

4 + 5 + 6 + 9 = 24

To find the final answer, we sum all the valid

x

-intercept values.

Key Concept

Using the intercept form of a linear equation and applying integer constraints to find coordinates.

Question 29Question →

A parabola in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 - 6x - 8y + 25 = 0$ . What is the distance, in coordinate units, between the focus and the directrix of this parabola?

Show answer & explanation

Answer: 4

Answer

The distance between the focus and the directrix of the parabola is 4.

By completing the square on the equation

x^2 - 6x - 8y + 25 = 0

, we get

(x-3)^2 = 8(y-2)

. Since the coefficient of the linear factor is

8

, we set

4p = 8

, which yields

p = 2

. The distance from the focus to the directrix is

2p = 2(2) = 4

.

Step-by-Step Solution

1

Isolate the terms containing

x

on one side of the equation.

x^2 - 6x = 8y - 25

To set up the equation for completing the square on the

x

terms.

2

Complete the square for the quadratic expression in

x

by adding

9

to both sides.

x^2 - 6x + 9 = 8y - 16 \implies (x-3)^2 = 8y - 16

Adding

( -6/2 )^2 = 9

creates a perfect square trinomial on the left side.

3

Factor out the coefficient of

y

on the right side to write the equation in standard form.

(x-3)^2 = 8(y-2)

This matches the standard form equation

(x-h)^2 = 4p(y-k)

for a vertical parabola.

4

Determine the value of the focal parameter

p

from the standard form.

4p = 8 \implies p = 2

Comparing the standard form coefficient

4p

with the value

8

gives

p = 2

.

5

Calculate the total distance between the focus and the directrix.

2p = 2(2) = 4

The vertex is situated halfway between the focus and the directrix, making the distance between them

2p

.

Key Concept

Finding the geometric properties of a parabola by completing the square to convert its general equation to standard form.

Question 30Question →

In the standard $(x,y)$ coordinate plane, a line is defined by the equation $3x - 2y = 12$ . What is the sum of the $x$ -intercept and the $y$ -intercept of this line?

-2

10

2

16

1

Show answer & explanation

Answer: -2

Answer

The sum of the

x

-intercept and the

y

-intercept of the line is

-2

.

To find the

x

-intercept, set

y = 0

in the equation

3x - 2y = 12

, which gives

3x = 12

, so

x = 4

. To find the

y

-intercept, set

x = 0

, which gives

-2y = 12

, so

y = -6

. Adding these two values together gives

4 + (-6) = -2

.

Step-by-Step Solution

1

Find the

x

-intercept by setting

y = 0

in the equation.

Substitute

y = 0

into

3x - 2y = 12

to get

3x - 2(0) = 12

, which simplifies to

3x = 12

, yielding

x = 4

.

The

x

-intercept of a line is the point where the line crosses the

x

-axis, which occurs when the

y

-coordinate is

0

.

2

Find the

y

-intercept by setting

x = 0

in the equation.

Substitute

x = 0

into

3x - 2y = 12

to get

3(0) - 2y = 12

, which simplifies to

-2y = 12

, yielding

y = -6

.

The

y

-intercept of a line is the point where the line crosses the

y

-axis, which occurs when the

x

-coordinate is

0

.

3

Add the

x

-intercept and

y

-intercept together.

Calculate

4 + (-6) = -2

.

The question asks for the sum of the

x

-intercept and the

y

-intercept.

Key Concept

Finding the

x

- and

y

-intercepts of a linear equation in standard form.

Estimated Time:1m 0s

Question 31Question →

An ellipse in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 + 4y^2 + 6x - 8y + 9 = 0$ . What is the length of the major axis of this ellipse?

1

2

4

8

16

Show answer & explanation

Answer: 4

Answer

4

To find the length of the major axis, rewrite the general equation of the ellipse in standard form by completing the square. Grouping the terms yields

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

. Completing the square for both variables gives

(x+3)^2 - 9 + 4[(y-1)^2 - 1] = -9

, which simplifies to

(x+3)^2 + 4(y-1)^2 = 4

. Dividing both sides by 4 gives the standard form

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

. In this form, the horizontal axis is the major axis because the denominator under the

x

-term (

a^2 = 4

) is larger than the denominator under the

y

-term (

b^2 = 1

). Since

a^2 = 4

, the semi-major axis is

a = 2

. Therefore, the total length of the major axis is

2a = 2(2) = 4

.

Step-by-Step Solution

1

Group the

x

-terms and

y

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

(x^2 + 6x) + (4y^2 - 8y) = -9

Grouping like terms allows completing the square for each variable independently.

2

Factor out the coefficient of

y^2

from the

y

-terms.

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

Before completing the square, the leading coefficient of the squared terms inside the parentheses must be 1.

3

Complete the square for both the

x

and

y

expressions by adding and subtracting the square of half of the linear coefficients.

((x+3)^2 - 9) + 4((y-1)^2 - 1) = -9

This rewrites the quadratic expressions into perfect square trinomial form.

4

Distribute the coefficients and simplify the constant terms.

(x+3)^2 - 9 + 4(y-1)^2 - 4 = -9 \implies (x+3)^2 + 4(y-1)^2 - 13 = -9 \implies (x+3)^2 + 4(y-1)^2 = 4

Isolating the squared terms on one side helps convert the equation to the standard form of an ellipse.

5

Divide both sides of the equation by 4 to set the right side equal to 1.

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

The standard form of a horizontal ellipse is

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

.

6

Identify the values of

a^2

and

b^2

and calculate the length of the major axis.

a^2 = 4 \implies a = 2

. The major axis length is

2a = 2(2) = 4

.

The length of the major axis is twice the length of the semi-major axis (

a

).

Key Concept

Rewriting the general equation of an ellipse into standard form by completing the square to find its key features, such as the length of the major axis.

Alternative Method

Another way to find the length of the major axis is to find the vertices of the ellipse by finding the maximum and minimum x-values where the equation has real solutions for y, though completing the square is the standard and most direct method.

Estimated Time:1m 30s

Question 32Question →

In the standard $(x, y)$ coordinate plane, a line $L$ passes through the point $(2, -3)$ and has a $y$ -intercept of $(0, b)$ , where $b > 0$ . If the area of the triangular region bounded by the line $L$ , the $x$ -axis, and the $y$ -axis is $4$ square units, what is the value of $b$ ?

2

4

6

8

12

Show answer & explanation

Answer: 6

Answer

6

The correct answer is the option containing 6. Using the two points

(2, -3)

and

(0, b)

, we find the slope of the line is

m = -\frac{b+3}{2}

, which gives the equation

y = -\frac{b+3}{2}x + b

. Setting

y=0

shows that the

x

-intercept is at

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

. The area of the right triangle formed by the intercepts and the origin is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{2b}{b+3} \times b = \frac{b^2}{b+3}

. Setting this area equal to

4

yields

b^2 - 4b - 12 = 0

. Factoring gives

(b - 6)(b + 2) = 0

. Since we are given

b > 0

,

b

must be

6

.

Step-by-Step Solution

1

Find the slope of the line

L

in terms of

b

.

The slope

m

is given by

\frac{-3 - b}{2 - 0} = -\frac{b+3}{2}

.

The line passes through

(2, -3)

and its

y

-intercept

(0, b)

.

2

Write the equation of the line

L

and determine its

x

-intercept.

The equation is

y = -\frac{b+3}{2}x + b

. Setting

y = 0

gives the

x

-intercept

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

.

The

x

-intercept is the point where the line crosses the

x

-axis (

y = 0

).

3

Set up the area equation for the triangle formed by the axes and the line.

The area is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{2b}{b+3} \times b = \frac{b^2}{b+3} = 4

.

The base of the triangle is the

x

-intercept, and the height is the

y

-intercept

b

(since both are positive for

b > 0

).

4

Solve the quadratic equation for

b

.

b^2 = 4(b + 3) \Rightarrow b^2 - 4b - 12 = 0 \Rightarrow (b - 6)(b + 2) = 0

. Since

b > 0

,

b = 6

.

We must solve the equation and select the positive solution because the problem specifies

b > 0

.

Key Concept

Using the coordinates of a point and intercepts to write a linear equation, finding intercepts, and calculating the area of a coordinate triangle.

Alternative Method

Instead of setting up the area algebraically first, you can test the answer choices. For example, testing the correct value 6: the y-intercept is

(0, 6)

. The slope of the line passing through

(0, 6)

and

(2, -3)

is

m = \frac{-3 - 6}{2 - 0} = -4.5

. The equation of the line is

y = -4.5x + 6

. The x-intercept is found by setting

y = 0

, giving

x = \frac{6}{4.5} = \frac{4}{3}

. The area of the triangle is

\frac{1}{2} \times \frac{4}{3} \times 6 = 4

, which matches the given area of 4 square units.

Estimated Time:1m 30s

Question 33Question →

The equation of a parabola is given by $(x - 4)^2 = 12(y + 1)$ . What is the $y$ -coordinate of the focus of this parabola?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

The equation

(x - 4)^2 = 12(y + 1)

is a parabola with a vertical axis of symmetry, vertex at

(4, -1)

, and focal length

p = 3

. The focus is located

p

units above the vertex, yielding a

y

-coordinate of

-1 + 3 = 2

.

Step-by-Step Solution

1

Identify the standard form of the parabola's equation.

The equation

(x - 4)^2 = 12(y + 1)

matches

(x - h)^2 = 4p(y - k)

.

This form allows us to find the vertex and the focal distance

p

directly.

2

Determine the vertex and focal distance

p

.

The vertex is

(4, -1)

and

p = 3

since

4p = 12

.

Matching the given equation terms to the standard form reveals these properties.

3

Find the coordinates of the focus.

The focus is at

(4, 2)

.

The focus is located

p

units vertically above the vertex for a parabola opening upward.

Key Concept

Focus of a Parabola

Question 34Question →

In the standard $(x, y)$ coordinate plane, a line passes through the point $(3, 2)$ and has a $y$ -intercept of $-4$ . If the point $(k, 5k + 2)$ also lies on this line, what is the value of $k$ ?

Show answer & explanation

Answer: -2

Answer

The value of

k

is

-2

.

The line has a slope of

2

and a y-intercept of

-4

, giving the equation

y = 2x - 4

. Substituting the coordinates of

(k, 5k + 2)

results in

5k + 2 = 2k - 4

, which simplifies to

3k = -6

, yielding

k = -2

.

Step-by-Step Solution

1

Identify the coordinates of the y-intercept.

The y-intercept of

-4

corresponds to the point

(0, -4)

.

The y-intercept is the point where the line crosses the y-axis, meaning the x-coordinate is 0.

2

Calculate the slope of the line.

The slope

m

is

2

.

Using the slope formula

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

with the points

(3, 2)

and

(0, -4)

gives

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

.

3

Write the equation of the line.

The equation of the line is

y = 2x - 4

.

Using the slope-intercept form

y = mx + b

, where the slope

m = 2

and the y-intercept

b = -4

.

4

Substitute the point

(k, 5k + 2)

into the line's equation.

The equation becomes

5k + 2 = 2k - 4

.

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

5

Solve the linear equation for

k

.

k = -2

.

Subtracting

2k

from both sides gives

3k + 2 = -4

. Subtracting

2

from both sides gives

3k = -6

. Dividing by

3

gives

k = -2

.

Key Concept

Finding the equation of a line from a point and an intercept, and solving for parameters of points on that line.

Question 35Question →

In the standard $(x, y)$ coordinate plane, an ellipse is defined by the equation $7x^2 + 16y^2 - 42x - 32y - 33 = 0$ . A parabola has its vertex at the focus of the ellipse with the smaller $x$ -coordinate, and its focus at the focus of the ellipse with the larger $x$ -coordinate. What is the larger of the two $y$ -coordinates of the points on the parabola that have an $x$ -coordinate of 6?

Show answer & explanation

Answer: 13

Answer

The larger of the two

y

-coordinates of the points on the parabola is 13.

By completing the square on the general ellipse equation, we get

\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1

. The center is

(3, 1)

and the focal distance is

c = \sqrt{16-7} = 3

, meaning the foci are at

(0, 1)

and

(6, 1)

. The parabola has its vertex at

(0, 1)

and focus at

(6, 1)

, which means it opens to the right with

p = 6

. Its equation is

(y - 1)^2 = 24x

. Substituting

x = 6

yields

(y - 1)^2 = 144

, so

y - 1 = \pm 12

. The two possible

y

-coordinates are

13

and

-11

, of which

13

is the larger value.

Step-by-Step Solution

1

Complete the square for the given ellipse equation to rewrite it in standard form.

\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1

Converting the equation to standard form is necessary to determine the center and semi-axis lengths of the ellipse.

2

Find the focal distance

c

and calculate the coordinates of the foci.

Focal distance

c = 3

; Foci at

(0, 1)

and

(6, 1)

For an ellipse, the distance

c

from the center

(h, k)

to the foci is

\sqrt{a^2 - b^2}

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

(h \pm c, k)

.

3

Use the foci coordinates to identify the vertex and focus of the parabola.

Vertex:

(0, 1)

; Focus:

(6, 1)

The problem defines the parabola's vertex as the ellipse focus with the smaller

x

-coordinate, and the parabola's focus as the ellipse focus with the larger

x

-coordinate.

4

Determine the equation of the parabola using its vertex and focus.

(y - 1)^2 = 24x

The parabola is horizontal and opens to the right with focal distance

p = 6

. The standard form is

(y - k)^2 = 4p(x - h)

.

5

Substitute

x = 6

into the parabola equation and solve for the larger

y

-value.

y = 13

Substituting

x = 6

yields

(y - 1)^2 = 144

, which gives

y = 1 + 12 = 13

or

y = 1 - 12 = -11

. The larger value is 13.

Key Concept

Determining the equations and key features (foci, vertices, focal parameters) of ellipses and parabolas by rewriting equations into standard forms.

Alternative Method

Once the equation

(y - 1)^2 = 24x

is established, recognize that at

x = 6

(which is the

x

-coordinate of the focus), the points on the parabola form the endpoints of the latus rectum. The length of the latus rectum is

4p = 24

, so the points lie at distance

2p = 12

vertically above and below the focus

(6, 1)

. Thus, the

y

-coordinates are

1 \pm 12

, immediately yielding the larger coordinate as 13.

Estimated Time:3m 0s

Question 36Question →

A line graphed in the standard $(x,y)$ coordinate plane has an $x$ -intercept of $6$ and a $y$ -intercept of $-3$ . What is the slope of this line?

Show answer & explanation

Answer: 0.5

Answer

The slope of the line is

0.5

(or

\frac{1}{2}

).

The correct slope is

0.5

. By identifying the

x

-intercept as

(6, 0)

and the

y

-intercept as

(0, -3)

, we can apply the slope formula

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

to find

m = \frac{-3 - 0}{0 - 6} = \frac{-3}{-6} = 0.5

.

Step-by-Step Solution

1

Identify the coordinates of the intercepts on the coordinate plane.

The points are

(6, 0)

and

(0, -3)

.

An

x

-intercept of

6

means the line crosses the

x

-axis at

(6, 0)

. A

y

-intercept of

-3

means the line crosses the

y

-axis at

(0, -3)

.

2

Apply the slope formula with the identified coordinates.

m = \frac{-3 - 0}{0 - 6} = \frac{-3}{-6} = 0.5

The slope formula is defined as

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

for any two points on a line.

Key Concept

Calculating the slope of a line from its intercepts

Question 37Question →

In the standard $(x, y)$ coordinate plane, the points $A(1, 2)$ and $B(9, 8)$ are the endpoints of a diameter of a circle $C$ . A line $L$ passes through the center of $C$ and is perpendicular to segment $AB$ . A point $P(x, y)$ lies on line $L$ such that the distance from $P$ to the center of $C$ is equal to the radius of $C$ . If the $x$ -coordinate of $P$ is greater than the $x$ -coordinate of the center of $C$ , what is the $y$ -coordinate of $P$ ?

Show answer & explanation

Answer: 1

Answer

The y-coordinate of the point P is 1.

The correct answer is 1. The center of circle C is the midpoint of the diameter AB, which is calculated as M(5, 5). The radius is half the length of AB, which is 5. The line L passing through M perpendicular to AB has a slope of -4/3. Points on this line at a distance of 5 from M are found by changing the coordinates by (+3, -4) or (-3, +4), yielding (8, 1) and (2, 9). Since the x-coordinate must be greater than the center's x-coordinate of 5, the correct point is (8, 1), which has a y-coordinate of 1.

Step-by-Step Solution

1

Calculate the center of the circle C by finding the midpoint of the diameter AB.

The center is M(5, 5).

The center of a circle is the midpoint of any of its diameters.

2

Calculate the radius of circle C by finding half the distance between A(1, 2) and B(9, 8).

The radius is 5.

The distance formula gives the diameter length as 10, and the radius is half the diameter.

3

Find the slope of line L perpendicular to AB.

The slope of L is -4/3.

The slope of AB is 3/4, and perpendicular lines have slopes that are negative reciprocals of each other.

4

Determine the coordinates of point P using the distance from the center and the slope of line L.

The possible points are (8, 1) and (2, 9).

Moving a distance of 5 along a line with slope -4/3 from (5, 5) results in a change of +/-3 in the x-coordinate and -/+4 in the y-coordinate.

5

Apply the constraint that the x-coordinate of P must be greater than the x-coordinate of the center (5).

P is (8, 1), so the y-coordinate is 1.

Comparing the two candidate points, only (8, 1) has an x-coordinate greater than 5.

Key Concept

Applying midpoint, distance, and perpendicular slope relationships in coordinate geometry to locate points.

Estimated Time:2m 30s

Question 38Question →

In the standard $(x, y)$ coordinate plane, the lines with equations $y = \frac{1}{2}x + 1$ , $y = -x + 7$ , and $y = k$ enclose a triangular region with an area of $6$ square units. What is a possible value of $k$ ?

2

3

4

5

7

Show answer & explanation

Answer: 5

Answer

The correct value of k is 5.

The correct answer is 5. Finding the intersection of the two boundary lines yields the vertex

P(4, 3)

. Calculating the intersection points of the horizontal line

y = k

with the boundary lines gives the

x

-coordinates

2k - 2

and

7 - k

. The distance between these coordinates represents the base of the triangle,

3|k - 3|

, while the height is the vertical distance

|k - 3|

. Substituting these into the triangle area formula

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

results in

6 = \frac{3}{2}(k - 3)^2

. Solving this quadratic equation gives

(k - 3)^2 = 4

, which leads to

k = 5

or

k = 1

. Therefore, 5 is the correct possible value.

Step-by-Step Solution

1

Find the intersection point of the two non-horizontal lines by setting their equations equal to each other.

\frac{1}{2}x + 1 = -x + 7 \Rightarrow \frac{3}{2}x = 6 \Rightarrow x = 4

. Substituting

x = 4

back into either equation gives

y = 3

. The intersection point is

P(4, 3)

.

The intersection point serves as the third vertex of the triangle, and its

y

-coordinate helps determine the height of the triangle relative to the horizontal boundary line

y = k

.

2

Find the

x

-coordinates of the intersection points between the horizontal line

y = k

and the other two lines.

For

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

, setting

y = k

gives

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

. For

y = -x + 7

, setting

y = k

gives

k = -x + 7 \Rightarrow x = 7 - k

.

These two points define the base of the triangle along the line

y = k

.

3

Express the base length and height of the triangle in terms of

k

, then set up the area equation.

The base is the distance between the two

x

-coordinates:

|(2k - 2) - (7 - k)| = |3k - 9| = 3|k - 3|

. The height is the vertical distance from the line

y = k

to the point

P(4, 3)

:

|k - 3|

. The area is

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{3}{2}(k - 3)^2

.

This sets up a solvable algebraic equation using the given area of

6

square units.

4

Solve the area equation for

k

.

\frac{3}{2}(k - 3)^2 = 6 \Rightarrow (k - 3)^2 = 4 \Rightarrow k - 3 = \pm 2

. Thus,

k = 5

or

k = 1

. Since

5

is among the options, it is the correct choice.

This yields the possible values of

k

that satisfy the geometric conditions.

Key Concept

Linear Equations and Graphing

Alternative Method

Instead of solving the algebraic quadratic equation, you can test the given choices for

k

. For example, if you test the value

5

, the line is

y = 5

. The intersection of

y = 5

and

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

is

(8, 5)

, and the intersection of

y = 5

and

y = -x + 7

is

(2, 5)

. The base of the triangle is the horizontal distance from

x = 2

to

x = 8

, which is

6

. The height of the triangle is the vertical distance from

y = 3

(the intersection vertex) to

y = 5

, which is

2

. The area is

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

. This matches the problem statement, confirming that

5

is the correct answer.

Estimated Time:2m 0s

Question 39Question →

In the standard $(x,y)$ coordinate plane, a region in the first quadrant is bounded by the $x$ -axis, the $y$ -axis, and the line with equation $ax + by = c$ , where $a$ , $b$ , and $c$ are positive constants. The line passes through the point $(8, 18)$ . If the area of this region is minimized when $a = 3$ , what is the value of $c$ ?

Show answer & explanation

Answer: 48

Answer

48

Substituting the given point and

a = 3

into the equation yields

c = 24 + 18b

. The area of the triangle formed by the intercepts is

A = \frac{c^2}{6b}

. Substituting

c

gives

A = 6(9b + 24 + \frac{16}{b})

. Using AM-GM, the minimum occurs when

9b = \frac{16}{b}

, resulting in

b = \frac{4}{3}

. Using this value, we find

c = 48

.

Step-by-Step Solution

1

Substitute the point

(8, 18)

and

a = 3

into the equation

ax + by = c

.

24 + 18b = c

This establishes a relationship between the constants

b

and

c

based on the given point that lies on the line.

2

Calculate the

x

-intercept and

y

-intercept of the line.

x

-intercept is at

x = \frac{c}{3}

, and

y

-intercept is at

y = \frac{c}{b}

.

The boundary of the region in the first quadrant is defined by these coordinate intercepts.

3

Formulate the area

A

of the right triangle bounded by the axes and the line.

A = \frac{c^2}{6b}

The area of a right triangle with vertices at the origin and the intercepts is

\frac{1}{2} \cdot \text{base} \cdot \text{height}

.

4

Substitute

c = 24 + 18b

into the area formula and simplify.

A = 6\left(9b + 24 + \frac{16}{b}\right)

Expressing the area as a single-variable function of

b

allows us to find its minimum value.

5

Apply the AM-GM inequality to minimize the variable term

9b + \frac{16}{b}

.

b = \frac{4}{3}

minimizes the expression.

The sum of two positive terms is minimized when the terms are equal, so

9b = \frac{16}{b} \implies b^2 = \frac{16}{9} \implies b = \frac{4}{3}

.

6

Calculate the value of

c

using the minimizing value of

b

.

c = 48

Substituting

b = \frac{4}{3}

into the relation

c = 24 + 18b

yields the constant value

c

for the minimum area.

Key Concept

Minimizing the area bounded by a line and the coordinate axes using linear equation forms and algebraic minimization.

Alternative Method

Instead of using the AM-GM inequality, you can find the minimum by taking the derivative of the area function

A(b) = 54b + 144 + \frac{96}{b}

with respect to

b

. Setting the derivative

A'(b) = 54 - \frac{96}{b^2} = 0

yields

b^2 = \frac{96}{54} = \frac{16}{9}

, which gives

b = \frac{4}{3}

for

b > 0

.

Estimated Time:3m 0s

Question 40Question →

In the standard $(x, y)$ coordinate plane, an ellipse is centered at the origin $(0, 0)$ and has vertices at $(-5, 0)$ and $(5, 0)$ . If the distance between the two foci of the ellipse is $8$ , what is the length of the minor axis of the ellipse?

Show answer & explanation

Answer: 6

Answer

The correct answer is 6.

The correct answer is 6 because the ellipse has a horizontal major axis with

a = 5

and focal distance

c = 4

. Using the relationship

c^2 = a^2 - b^2

, we solve for the semi-minor axis

b

to get

b = 3

. The total length of the minor axis is

2b = 6

.

Step-by-Step Solution

1

Determine the semi-major axis length

a

.

a = 5

The vertices are at

(\pm 5, 0)

, which are

5

units from the center

(0, 0)

along the major axis.

2

Determine the distance from the center to each focus

c

.

c = 4

The distance between the two foci is

2c = 8

, so the distance from the center to a focus is

c = 4

.

3

Find the semi-minor axis length

b

.

b = 3

Using the relation

c^2 = a^2 - b^2

for ellipses, we get

4^2 = 5^2 - b^2 \implies b^2 = 9 \implies b = 3

.

4

Calculate the full length of the minor axis.

6

The length of the minor axis is

2b = 2(3) = 6

.

Key Concept

The relationship between the semi-major axis, semi-minor axis, and focal distance of an ellipse.

Estimated Time:1m 30s

Coordinate Geometry

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer