Coordinate Geometry Alıştırma Soruları — ACT

Soru 1Soru →

Line $p$ is defined by the equation $2x + 5y = 10$ . Line $q$ is perpendicular to line $p$ and has a $y$ -intercept of $-3$ . If line $q$ passes through the point $(a, 7)$ , what is the value of $a$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of

a

is 4.

First, the equation of line

p

,

2x + 5y = 10

, is converted to slope-intercept form to find its slope:

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

. This indicates the slope of line

p

is

-\frac{2}{5}

. Because line

q

is perpendicular to line

p

, its slope must be the negative reciprocal of

-\frac{2}{5}

, which is

\frac{5}{2}

. Using the given

y

-intercept of

-3

, the equation of line

q

is written as

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

. Substituting the coordinates of the point

(a, 7)

into this equation gives

7 = \frac{5}{2}a - 3

. Adding

3

to both sides results in

10 = \frac{5}{2}a

, and solving for

a

yields

a = 4

.

Adım Adım Çözüm

1

Find the slope of line

p

.

The slope of line

p

is

-\frac{2}{5}

.

Rewriting the equation

2x + 5y = 10

in slope-intercept form (

y = mx + b

) gives

5y = -2x + 10

, which simplifies to

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

. The slope

m

is the coefficient of

x

, which is

-\frac{2}{5}

.

2

Determine the slope of line

q

.

The slope of line

q

is

\frac{5}{2}

.

Since line

q

is perpendicular to line

p

, its slope is the negative reciprocal of line

p

's slope:

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

.

3

Write the equation of line

q

.

The equation of line

q

is

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

.

Line

q

has a slope of

\frac{5}{2}

and a

y

-intercept of

-3

. Using the slope-intercept form

y = mx + b

gives

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

.

4

Substitute the point

(a, 7)

into the equation of line

q

and solve for

a

.

a = 4

Substituting

x = a

and

y = 7

into the equation gives

7 = \frac{5}{2}a - 3

. Adding

3

to both sides yields

10 = \frac{5}{2}a

. Multiplying both sides by

2

gives

20 = 5a

, and dividing by

5

results in

a = 4

.

Anahtar Kavram

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

Tahmini Süre:1m 30s

Soru 2Soru →

A line in the standard $(x,y)$ coordinate plane passes through the point $(0, 4)$ and has a slope of $3$ . Which of the following equations represents this line?

y = 3x + 4

y = 4x + 3

y = 3x - 4

y = -3x + 4

y = \frac{1}{3}x + 4

Cevabı ve açıklamayı göster

Cevap: $y = 3x + 4$

Cevap

y = 3x + 4

The slope-intercept form of a linear equation is

y = mx + b

. Here, the slope

m

is given as

3

, and the y-intercept

b

is given as

4

since the line crosses the y-axis at the point

(0,4)

. Substituting these values into the form gives the equation

y = 3x + 4

.

Adım Adım Çözüm

1

Identify the slope-intercept form of a linear equation.

The slope-intercept form is

y = mx + b

, where

m

is the slope and

b

is the

y

-intercept.

To write the equation of a line, we can substitute the known slope and y-intercept into this general form.

2

Determine the values of

m

and

b

from the given information.

The slope

m = 3

. The point

(0, 4)

lies on the y-axis, meaning the y-intercept

b = 4

.

The y-intercept is the y-coordinate of the point where the line crosses the y-axis, which occurs when

x = 0

.

3

Substitute the values of

m

and

b

into the slope-intercept equation.

y = 3x + 4

Replacing

m

with

3

and

b

with

4

yields the equation representing the line.

Anahtar Kavram

Writing linear equations in slope-intercept form using the slope and a point.

Soru 3Soru →

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

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The

x

-intercept of the line is

4

.

Adım Adım Çözüm

1

Substitute

y = 0

into the equation to find the point where the line intersects the

x

-axis.

3x + 2(0) = 12

By definition, the

x

-intercept occurs where the

y

-coordinate is equal to

0

.

2

Simplify the equation and solve for the variable

x

.

3x = 12 \Rightarrow x = 4

Simplifying

2(0)

to

0

leaves

3x = 12

. Dividing both sides of the equation by

3

isolates

x

, giving

x = 4

.

Anahtar Kavram

Determining the

x

-intercept of a linear equation by evaluating it at

y = 0

.

Tahmini Süre:45s

Soru 4Soru →

The slope of a line is $2$ , and the line passes through the point $(3, 1)$ . What is the $y$ -intercept of this line in the standard $(x,y)$ coordinate plane?

-5

-4

1

2.5

5

Cevabı ve açıklamayı göster

Cevap: $-5$

Cevap

The

y

-intercept of the line is

-5

.

The correct answer is found by substituting the slope

m = 2

and the point

(3, 1)

into the slope-intercept equation

y = mx + b

. This gives

1 = 2(3) + b

, which simplifies to

1 = 6 + b

. Subtracting

6

from both sides isolates

b

, giving

b = -5

.

Adım Adım Çözüm

1

Recall the slope-intercept form of a linear equation.

y = mx + b

, where

m

is the slope and

b

is the

y

-intercept.

This formula allows us to solve for the

y

-intercept using the given slope and a point on the line.

2

Substitute the given slope

m = 2

and the coordinates of the point

(3, 1)

into the slope-intercept equation.

1 = 2(3) + b

The point

(3, 1)

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

3

Simplify the equation and solve for

b

.

1 = 6 + b \Rightarrow b = -5

Subtracting

6

from both sides isolates

b

, which represents the

y

-intercept of the line.

Anahtar Kavram

Linear Equations and Graphing

Soru 5Soru →

A line in the standard $(x,y)$ coordinate plane is defined by the equation $kx + 5y = 20$ , where $k$ is a constant. If the $x$ -intercept of this line is $6$ units greater than its $y$ -intercept, what is the value of $k$ ?

\frac{1}{2}

2

\frac{10}{3}

10

-10

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

2

To find the

y

-intercept of the line

kx + 5y = 20

, set

x = 0

, which yields

5y = 20

, or

y = 4

. The problem states that the

x

-intercept is

6

units greater than the

y

-intercept, so the

x

-intercept is

4 + 6 = 10

. Substituting the point

(10, 0)

back into the original equation gives

k(10) + 5(0) = 20

. Solving for

k

gives

10k = 20

, which simplifies to

k = 2

.

Adım Adım Çözüm

1

Find the

y

-intercept of the line.

The

y

-coordinate of the

y

-intercept is

4

.

To find the

y

-intercept, set

x = 0

in the equation

kx + 5y = 20

, which gives

5y = 20

, so

y = 4

.

2

Determine the

x

-intercept of the line based on the given relationship.

The

x

-coordinate of the

x

-intercept is

10

.

The problem states that the

x

-intercept is

6

units greater than the

y

-intercept. Since the

y

-intercept value is

4

, the

x

-intercept value is

4 + 6 = 10

.

3

Substitute the

x

-intercept coordinates into the equation to solve for

k

.

k = 2

The

x

-intercept is the point

(10, 0)

. Substituting

x = 10

and

y = 0

into the equation

kx + 5y = 20

gives

k(10) + 5(0) = 20

, which simplifies to

10k = 20

. Dividing both sides by

10

yields

k = 2

.

Anahtar Kavram

Finding intercepts of a linear equation in standard form and using coordinate substitution to solve for an unknown coefficient.

Soru 6Soru →

A line in the standard $(x,y)$ coordinate plane passes through the point $(4, 1)$ and has a negative slope $m$ . The line and the coordinate axes bound a region in the first quadrant. If the area of this region is $8$ square units, which of the following is the value of $m$ ?

-4

-\frac{9}{16}

-\frac{1}{4}

-\frac{3}{4}

-\frac{9}{8}

Cevabı ve açıklamayı göster

Cevap: $-\frac{1}{4}$

Cevap

-\frac{1}{4}

The correct answer is

-\frac{1}{4}

. The line passes through

(4, 1)

with slope

m

. Using the point-slope formula, the equation of the line is

y - 1 = m(x - 4)

, which simplifies to

y = mx - 4m + 1

. The

y

-intercept is found by setting

x = 0

, giving

1 - 4m

. The

x

-intercept is found by setting

y = 0

, giving

\frac{4m-1}{m}

. The area of the right-triangular region in the first quadrant bounded by the axes is

\frac{1}{2} \cdot (\text{base}) \cdot (\text{height}) = 8

. Substituting the intercepts, we get

\frac{1}{2} \cdot \left(\frac{4m-1}{m}\right) \cdot (1-4m) = 8

. Multiplying by

2m

(which is negative, so we maintain positive side lengths) yields

-(1-4m)^2 = 16m

, which simplifies to

16m^2 + 8m + 1 = 0

. Factoring the quadratic expression gives

(4m+1)^2 = 0

, which has the single solution

m = -\frac{1}{4}

.

Adım Adım Çözüm

1

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

(4, 1)

and slope

m

.

y - 1 = m(x - 4) \implies y = mx - 4m + 1

The point-slope formula

y - y_1 = m(x - x_1)

defines any line passing through a given point with a specific slope.

2

Find the

x

-intercept and

y

-intercept of the line.

y\text{-intercept} = 1 - 4m

(when

x=0

), and

x\text{-intercept} = \frac{4m-1}{m}

(when

y=0

)

The intercepts represent the vertices of the right triangle formed by the line and the coordinate axes.

3

Set up the area equation for the right triangle in the first quadrant, noting that since

m < 0

, both intercepts are positive.

Area = \frac{1}{2} \cdot \text{base} \cdot \text{height} \implies \frac{1}{2} \cdot \left(\frac{4m-1}{m}\right) \cdot (1-4m) = 8

The area of the region bounded by the axes and the line is a right triangle whose legs are the intercepts.

4

Solve the algebraic equation for

m

.

(4m-1)(1-4m) = 16m \implies 16m^2 + 8m + 1 = 0 \implies (4m+1)^2 = 0 \implies m = -\frac{1}{4}

Multiplying both sides by

2m

and expanding the terms leads to a quadratic equation in terms of

m

, which resolves to a single real root.

Anahtar Kavram

Using linear equation forms and coordinate geometry to represent boundary lines and compute bounded areas.

Alternatif Yöntem

Instead of using the point-slope form, write the line in intercept form:

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

, where

a

and

b

are the

x

- and

y

-intercepts. The area of the region is

\frac{1}{2}ab = 8 \implies ab = 16 \implies b = \frac{16}{a}

. Substitute this back into the intercept form to get

\frac{x}{a} + \frac{y}{16/a} = 1 \implies \frac{x}{a} + \frac{ay}{16} = 1

. Since the line passes through

(4, 1)

, substitute these coordinates:

\frac{4}{a} + \frac{a(1)}{16} = 1

. Multiply the entire equation by

16a

to clear the denominators:

64 + a^2 = 16a \implies a^2 - 16a + 64 = 0 \implies (a-8)^2 = 0 \implies a = 8

. Since

a = 8

, the

y

-intercept is

b = \frac{16}{8} = 2

. Using the intercepts

(8, 0)

and

(0, 2)

, the slope is

m = \frac{2 - 0}{0 - 8} = -\frac{2}{8} = -\frac{1}{4}

.

Tahmini Süre:2m 30s

Soru 7Soru →

In the standard $(x,y)$ coordinate plane, the graph of the linear equation $y - 7 = -2(x + 3)$ crosses the $y$ -axis. What is the $y$ -coordinate of this intersection point?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The

y

-coordinate of the intersection point is

1

.

To find where the graph of the linear equation crosses the

y

-axis, substitute

x = 0

into the equation

y - 7 = -2(x + 3)

. This gives

y - 7 = -2(0 + 3)

, which simplifies to

y - 7 = -6

. Adding

7

to both sides results in

y = 1

. Thus, the

y

-coordinate of the intersection point is

1

.

Adım Adım Çözüm

1

Substitute

x = 0

into the linear equation.

y - 7 = -2(0 + 3)

The graph of an equation crosses the

y

-axis (which defines the

y

-intercept) when the

x

-coordinate is equal to

0

.

2

Simplify the expression on the right side of the equation.

y - 7 = -6

Evaluating

-2(0 + 3)

yields

-2(3) = -6

.

3

Solve the simplified linear equation for

y

.

y = 1

Adding

7

to both sides isolates

y

, giving

y = -6 + 7 = 1

.

Anahtar Kavram

Finding the

y

-intercept of a line from its point-slope form equation

Soru 8Soru →

A line in the standard $(x,y)$ coordinate plane passes through the points shown in the table below:

$x$	$y$
$0$	$5$
$2$	$9$
$4$	$13$

Which of the following equations represents this line?

y = 5x + 2

y = 2x - 5

y = 2x + 5

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

y = 4x + 5

Cevabı ve açıklamayı göster

Cevap: $y = 2x + 5$

Cevap

y = 2x + 5

The correct equation is

y = 2x + 5

because the slope of the line is

2

, calculated from the points

(0, 5)

and

(2, 9)

as

\frac{9 - 5}{2 - 0} = 2

, and the

y

-intercept is

5

, corresponding to the point

(0, 5)

.

Adım Adım Çözüm

1

Identify the

y

-intercept from the table.

The point

(0, 5)

indicates that the

y

-intercept is

5

, which means

b = 5

in the slope-intercept form

y = mx + b

.

The

y

-intercept is the point where the line crosses the

y

-axis, which occurs when

x = 0

.

2

Calculate the slope using two points from the table.

Using

(0, 5)

and

(2, 9)

, the slope

m = \frac{9 - 5}{2 - 0} = \frac{4}{2} = 2

.

The slope formula is the change in

y

divided by the change in

x

.

3

Substitute the slope and

y

-intercept into the slope-intercept equation.

Substituting

m = 2

and

b = 5

into

y = mx + b

yields

y = 2x + 5

.

This represents the linear relationship of the points.

Anahtar Kavram

Writing a linear equation in slope-intercept form from a table of values.

Tahmini Süre:45s

Soru 9Soru →

A taxi company charges a flat fee plus a constant rate per mile driven. The total cost, $C$ (in dollars), for a ride of $d$ miles is given by a linear relationship. If a 6-mile ride costs $18.50 and a 10-mile ride costs$ 26.50, what is the flat fee, in dollars, charged by the taxi company?

15.50

2.00

6.50

12.50

25.00

Cevabı ve açıklamayı göster

Cevap: 6.50

Cevap

The flat fee charged by the taxi company is 6.50 dollars.

The correct flat fee is 6.50 dollars. The linear relationship is of the form

C = md + b

, where

m

is the cost per mile and

b

is the flat fee. The slope is calculated as

m = \frac{26.50 - 18.50}{10 - 6} = 2.00

. Substituting the point

(6, 18.50)

into the equation gives

18.50 = 2(6) + b

, which simplifies to

18.50 = 12.00 + b

, yielding

b = 6.50

.

Adım Adım Çözüm

1

Identify the two coordinates from the word problem and calculate the slope (rate per mile).

The points are

(6, 18.50)

and

(10, 26.50)

. The slope

m = \frac{26.50 - 18.50}{10 - 6} = \frac{8.00}{4} = 2.00

.

In a linear relationship, the constant rate of change is represented by the slope of the line.

2

Use the slope-intercept form

C = md + b

and substitute one of the points to solve for the vertical intercept

b

(flat fee).

Using

(6, 18.50)

:

18.50 = 2.00(6) + b \implies 18.50 = 12.00 + b \implies b = 6.50

.

The flat fee corresponds to the cost when the distance is zero miles, which is the vertical intercept

b

.

Anahtar Kavram

Determining the equation of a line from two points in a real-world linear application.

Soru 10Soru →

A linear function $f(x) = mx + b$ satisfies the inequality $f(x + 3) - f(x) \geq 6$ for all real numbers $x$ , and its graph passes through the point $(1, -2)$ . If the region bounded by the graph of $f$ , the line $x = -2$ , the $x$ -axis, and the $y$ -axis has an area of $20$ square units, what is the value of $b$ ?

-10

-8

-6

-4

4

Cevabı ve açıklamayı göster

Cevap: -6

Cevap

The value of

b

is

-6

.

To find the value of

b

, we first determine the constraints on the slope

m

. Since

f(x) = mx + b

, the difference

f(x+3) - f(x) = 3m

. Given

3m \geq 6

, we find

m \geq 2

. The graph of

f

passes through

(1, -2)

, which gives

-2 = m + b

, or

b = -2 - m

. Since

m \geq 2

, the

y

-intercept

b

must be less than or equal to

-4

, meaning the function is negative for the entire interval

[-2, 0]

. The region bounded by the graph of

f

,

x = -2

, the

x

-axis, and the

y

-axis forms a trapezoid below the

x

-axis. The vertical parallel sides of this trapezoid have lengths equal to the absolute values of the

y

-coordinates at

x = -2

and

x = 0

, which are

2m - b

and

-b

, respectively. The width is

2

. The area of the trapezoid is

\frac{(2m - b) + (-b)}{2} \times 2 = 2m - 2b

. Setting the area to

20

square units gives

2m - 2b = 20

, which simplifies to

m - b = 10

. Solving the system of equations

m + b = -2

and

m - b = 10

yields

m = 4

and

b = -6

.

Adım Adım Çözüm

1

Find the constraint on the slope

m

using the given inequality

f(x+3) - f(x) \geq 6

.

3m \geq 6 \implies m \geq 2

.

The difference in function values over an interval of

3

for a linear function is

3

times the slope

m

.

2

Substitute the point

(1, -2)

into the equation

f(x) = mx + b

to find a relationship between

m

and

b

.

-2 = m(1) + b \implies b = -2 - m

.

The graph of the function must pass through the given coordinate point.

3

Determine the shape and boundaries of the region bounded by the graph of

f

,

x = -2

, the

x

-axis, and the

y

-axis.

A trapezoid below the

x

-axis with parallel vertical sides of lengths

2m - b

(at

x = -2

) and

-b

(at

x = 0

), and a horizontal width of

2

.

Since

m \geq 2

, the

y

-intercept

b

is at most

-4

, so the function is strictly negative on the interval

[-2, 0]

.

4

Set up the area formula for the trapezoid and set it equal to

20

to find another relation between

m

and

b

.

\frac{(2m-b) + (-b)}{2} \times 2 = 20 \implies 2m - 2b = 20 \implies m - b = 10

.

The area of a trapezoid is the average of the parallel side lengths multiplied by the width.

5

Solve the system of equations:

m + b = -2

and

m - b = 10

.

Adding the equations gives

2m = 8 \implies m = 4

. Substituting

m = 4

gives

b = -6

.

To find the specific value of

b

that satisfies both the point condition and the area condition.

Anahtar Kavram

Graphing linear equations, calculating area of bounded regions on the coordinate plane, and using linear slope and point-intercept forms.

Alternatif Yöntem

Instead of solving the system of equations algebraically, we can express the line in point-slope form as

y + 2 = m(x - 1)

. At

x = 0

,

y = -m - 2

, and at

x = -2

,

y = -3m - 2

. The heights of the trapezoid are

m + 2

and

3m + 2

(since they are below the

x

-axis and

m \geq 2

). The area of the trapezoid is

\frac{(m+2) + (3m+2)}{2} \times 2 = 4m + 4

. Setting

4m + 4 = 20

yields

4m = 16 \implies m = 4

, which gives

b = -m - 2 = -6

.

Tahmini Süre:3m 0s

Soru 11Soru →

A line graphed on the $(x, y)$ coordinate plane has a $y$ -intercept of $-1$ and passes through the point $(3, 8)$ . If the point $(a, 14)$ also lies on this line, what is the value of $a$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5

The line has a y-intercept of

-1

, which corresponds to the point

(0, -1)

. The slope is calculated as

m = \frac{8 - (-1)}{3 - 0} = 3

. The equation of the line is

y = 3x - 1

. Substituting

(a, 14)

into this equation gives

14 = 3a - 1

, which simplifies to

15 = 3a

, and solving for

a

yields

5

.

Adım Adım Çözüm

1

Identify the coordinate representation of the y-intercept.

The point is

(0, -1)

.

A y-intercept of

-1

means the line crosses the y-axis at the point where

x = 0

, which is

(0, -1)

.

2

Calculate the slope of the line.

m = 3

Using the slope formula

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

with the points

(0, -1)

and

(3, 8)

, we get

m = \frac{8 - (-1)}{3 - 0} = 3

.

3

Write the equation of the line and solve for

a

.

a = 5

The equation in slope-intercept form is

y = 3x - 1

. Substituting the point

(a, 14)

yields

14 = 3a - 1

. Adding 1 to both sides gives

15 = 3a

, so

a = 5

.

Anahtar Kavram

Finding the equation of a line from two points and solving for a missing coordinate.

Tahmini Süre:1m 30s

Soru 12Soru →

A circle in the standard $(x, y)$ coordinate plane has its center at $(5, -2)$ and a radius of $7$ units. Which of the following is an equation of this circle?

x^2 + y^2 - 10x + 4y = 20

x^2 + y^2 + 10x - 4y = 20

x^2 + y^2 - 10x + 4y = -22

x^2 + y^2 = 70

x^2 + y^2 - 10x + 4y = 7

Cevabı ve açıklamayı göster

Cevap: $x^2 + y^2 - 10x + 4y = 20$

Cevap

The equation

x^2 + y^2 - 10x + 4y = 20

The standard form equation of a circle is

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

, where

(h, k)

is the center and

r

is the radius. Substituting

h = 5

,

k = -2

, and

r = 7

yields

(x - 5)^2 + (y + 2)^2 = 49

. Expanding both binomials gives

x^2 - 10x + 25 + y^2 + 4y + 4 = 49

. Combining the constant terms on the left side gives

x^2 + y^2 - 10x + 4y + 29 = 49

. Subtracting

29

from both sides results in the equation

x^2 + y^2 - 10x + 4y = 20

.

Adım Adım Çözüm

1

Write down the standard equation of a circle.

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

, where

(h, k)

is the center and

r

is the radius.

This is the fundamental formula relating the geometric properties of a circle to its algebraic representation.

2

Substitute the given values into the standard equation.

(x - 5)^2 + (y - (-2))^2 = 7^2

, which simplifies to

(x - 5)^2 + (y + 2)^2 = 49

.

Substituting the center

(5, -2)

and radius

7

into the standard equation sets up the expression for expansion.

3

Expand the squared binomials.

x^2 - 10x + 25 + y^2 + 4y + 4 = 49

.

Expanding

(x - 5)^2

into

x^2 - 10x + 25

and

(y + 2)^2

into

y^2 + 4y + 4

allows conversion to the general form.

4

Simplify and rearrange the equation to match the form of the options.

x^2 + y^2 - 10x + 4y + 29 = 49

, which simplifies to

x^2 + y^2 - 10x + 4y = 20

.

Combining the constants and subtracting

29

from both sides of the equation yields the final simplified general form.

Anahtar Kavram

Equation of a Circle in General Form

Tahmini Süre:1m 0s

Soru 13Soru →

An ellipse in the standard coordinate plane is defined by the equation $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ . What is the distance between the two foci of this ellipse?

8

4

16

2\sqrt{34}

12

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8

The correct answer is 8. Rearranging the given equation

9x^2 + 25y^2 - 36x + 50y - 164 = 0

by completing the square gives the standard form

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

. In this standard horizontal ellipse equation, the semi-major axis squared is

a^2 = 25

and the semi-minor axis squared is

b^2 = 9

. The focal distance

c

from the center to each focus is found using the relation

c^2 = a^2 - b^2

, which yields

c = \sqrt{25 - 9} = 4

. Since the distance between the two foci is

2c

, the final distance is

2(4) = 8

.

Adım Adım Çözüm

1

Group the

x

and

y

terms and move the constant to the right-hand side.

(9x^2 - 36x) + (25y^2 + 50y) = 164

Grouping like variables allows us to factor out coefficients before completing the square.

2

Factor out the leading coefficients of the quadratic terms.

9(x^2 - 4x) + 25(y^2 + 2y) = 164

Completing the square requires the quadratic terms inside the parentheses to have a coefficient of 1.

3

Complete the square for both variables by adding the square of half the linear coefficients inside the parentheses, and balance the equation by adding the distributed values to the right side.

9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 9(4) + 25(1)

which simplifies to

9(x-2)^2 + 25(y+1)^2 = 225

.

This rewrites the quadratic expressions into perfect square binomials.

4

Divide both sides of the equation by 225 to write the equation in standard form.

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

The standard form of an ellipse equation is equal to 1.

5

Identify the values of

a^2

and

b^2

to calculate the distance

c

from the center to each focus.

a^2 = 25

and

b^2 = 9

. Using

c^2 = a^2 - b^2

, we get

c^2 = 25 - 9 = 16

, so

c = 4

.

For a horizontal ellipse, the larger denominator is

a^2

and the smaller is

b^2

, and the focal distance satisfies

c^2 = a^2 - b^2

.

6

Multiply the focal distance from the center by 2 to find the total distance between the two foci.

\text{Distance} = 2c = 2(4) = 8

.

The distance between the two foci is the length of the segment connecting them, which is centered at

(2, -1)

and extends

c

units in both horizontal directions.

Anahtar Kavram

Rewriting an ellipse equation in standard form to determine its key geometric features including its foci.

Tahmini Süre:1m 30s

Soru 14Soru →

A linear function contains the points shown in the table below:

$x$	$y$
$k - 2$	$5$
$k + 1$	$14$
$2k + 3$	$35$

What is the $y$ -intercept of the line representing this function?

Cevabı ve açıklamayı göster

Cevap: -4

Cevap

The

y

-intercept of the line representing this function is

-4

.

By using the first two coordinates, the constant slope of the linear function is determined to be

3

. Equating the slope between the second and third coordinates to

3

yields the parameter

k = 5

. Substituting this back into the coordinate expressions gives the points

(3, 5)

and

(6, 14)

. Using the point-slope form, the equation of the line is

y = 3x - 4

, meaning the

y

-intercept is

-4

.

Adım Adım Çözüm

1

Calculate the slope of the line using the first two points:

(k-2, 5)

and

(k+1, 14)

.

m = \frac{14 - 5}{(k+1) - (k-2)} = \frac{9}{3} = 3

Since the function is linear, the rate of change (slope) remains constant between any two points.

2

Express the slope using the second and third points,

(k+1, 14)

and

(2k+3, 35)

, set it equal to

3

, and solve for

k

.

\frac{35 - 14}{(2k+3) - (k+1)} = 3 \implies \frac{21}{k+2} = 3 \implies 3(k+2) = 21 \implies k+2 = 7 \implies k = 5

The slope between the second and third points must also equal the constant slope of

3

.

3

Substitute

k = 5

back into the coordinates to determine the actual points on the line.

The points are

(3, 5)

,

(6, 14)

, and

(13, 35)

.

This provides concrete coordinates that can be used to write the equation of the line.

4

Use the point-slope form with the point

(3, 5)

and slope

m = 3

to write the equation of the line, then convert to slope-intercept form.

y - 5 = 3(x - 3) \implies y - 5 = 3x - 9 \implies y = 3x - 4

Converting to slope-intercept form (

y = mx + b

) directly gives the

y

-intercept as the constant term

b

.

Anahtar Kavram

Determining the equation and intercepts of a line using the constant slope property of linear functions.

Alternatif Yöntem

Once the slope

m = 3

and a point such as

(3, 5)

are established, substitute these values directly into the slope-intercept equation

y = mx + b

to solve for

b

. This gives

5 = 3(3) + b \implies 5 = 9 + b \implies b = -4

.

Tahmini Süre:2m 30s

Soru 15Soru →

A parabola in the standard $(x, y)$ coordinate plane is defined by the equation $y = \frac{1}{12}x^2 - 2x + 13$ . What is the $y$ -coordinate of the focus of this parabola?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The y-coordinate of the focus is 4.

The standard form of the parabola is

y - 1 = \frac{1}{12}(x - 12)^2

. Comparing this to

y - k = \frac{1}{4p}(x - h)^2

gives the vertex

(h, k) = (12, 1)

and

4p = 12 \implies p = 3

. Since the parabola opens upward, the focus is at

(12, 1 + 3) = (12, 4)

, making the

y

-coordinate 4.

Adım Adım Çözüm

1

Complete the square to rewrite the equation in standard vertex form.

y = \frac{1}{12}(x - 12)^2 + 1

Rewriting the general quadratic equation into standard form allows us to directly identify the vertex and focal parameters.

2

Equate the coefficients to find the focal distance

p

and vertex

(h, k)

.

Vertex

(h, k) = (12, 1)

and

p = 3

The standard vertex form of a vertical parabola is

y - k = \frac{1}{4p}(x - h)^2

. Setting

\frac{1}{4p} = \frac{1}{12}

gives

p = 3

.

3

Calculate the focus coordinates

(h, k + p)

.

Focus

= (12, 4)

, so the

y

-coordinate is

4

For an upward-opening parabola, the focus is located

p

units directly above the vertex.

Anahtar Kavram

Rewriting a quadratic equation into standard vertex form to find the properties of a parabola, including its vertex and focus.

Soru 16Soru →

A parabola in the standard $(x, y)$ coordinate plane is defined by the equation $y^2 = 12x$ . What is the $x$ -coordinate of the focus of this parabola?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The

x

-coordinate of the focus is 3.

The equation

y^2 = 12x

represents a parabola with vertex at the origin opening to the right. The standard form for such a parabola is

y^2 = 4px

, where the focus is located at

(p, 0)

. By setting

4p = 12

, we find

p = 3

. Thus, the focus is

(3, 0)

, and its

x

-coordinate is 3.

Adım Adım Çözüm

1

Identify the standard form of the parabola's equation.

The equation

y^2 = 12x

fits the standard form of a horizontal parabola with its vertex at the origin,

y^2 = 4px

.

This allows us to relate the given equation to the coordinate of the focus, which is located at

(p, 0)

.

2

Solve for the parameter

p

.

4p = 12

, which gives

p = 3

.

By equating the coefficients of

x

from the given equation and the standard form, we can find the value of

p

.

3

Determine the focus coordinate.

The focus is at

(3, 0)

, so the

x

-coordinate is 3.

The focus of a parabola of the form

y^2 = 4px

has coordinates

(p, 0)

.

Anahtar Kavram

Focus of a Parabola

Soru 17Soru →

In the standard $(x,y)$ coordinate plane, two opposite vertices of a square are $(1, 2)$ and $(4, 6)$ . If all four vertices of the square lie in the first quadrant, what is the $x$ -coordinate of the vertex that is closest to the $y$ -axis?

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

The correct answer is

0.5

. The vertex closest to the

y

-axis is

(0.5, 5.5)

, which has an

x

-coordinate of

0.5

.

The diagonals of a square are perpendicular, equal in length, and bisect each other. Using the given opposite vertices

(1, 2)

and

(4, 6)

, we find the midpoint to be

(2.5, 4)

. The vector between them is

(3, 4)

with length

5

. A perpendicular vector of length

5

is

(-4, 3)

. Adding and subtracting half of this vector,

(2, -1.5)

, from the midpoint yields the other two vertices:

(0.5, 5.5)

and

(4.5, 2.5)

. Since all four vertices are in the first quadrant, we compare their

x

-coordinates:

1

,

4

,

0.5

, and

4.5

. The smallest

x

-coordinate is

0.5

, which represents the vertex closest to the

y

-axis.

Adım Adım Çözüm

1

Find the midpoint of the given diagonal.

The midpoint is

M(2.5, 4)

.

The diagonals of a square bisect each other at their common midpoint.

2

Determine the vector representing the given diagonal

AB

and its length.

\vec{AB} = (3, 4)

and its length is

5

.

The vector is found by subtracting coordinates:

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

, and its length is

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

.

3

Find a perpendicular vector of the same length to represent the other diagonal.

A perpendicular vector is

(-4, 3)

.

The dot product of

(3, 4)

and

(-4, 3)

is

3(-4) + 4(3) = 0

, and its length is

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

.

4

Calculate the coordinates of the other two vertices of the square.

The vertices are

C(0.5, 5.5)

and

D(4.5, 2.5)

.

The vertices are located at

M \pm \frac{1}{2}\vec{CD}

, which gives

(2.5, 4) \pm ( -2, 1.5 )

.

5

Determine which of the four vertices is closest to the

y

-axis and identify its

x

-coordinate.

The vertex closest to the

y

-axis is

C(0.5, 5.5)

, and its

x

-coordinate is

0.5

.

The distance to the

y

-axis is the

x

-coordinate of the point. Comparing the

x

-coordinates

1

,

4

,

0.5

, and

4.5

, the smallest value is

0.5

.

Anahtar Kavram

Properties of diagonals of a square on a coordinate plane, including midpoint and perpendicularity.

Alternatif Yöntem

Instead of using vectors, one can set up a system of equations. Let

(x, y)

be one of the unknown vertices. Since it forms a right isosceles triangle with the midpoint

(2.5, 4)

and has distance

2.5

from it along a line with slope

-3/4

, we can write the equation of the line as

y - 4 = -0.75(x - 2.5)

and use the distance formula

(x - 2.5)^2 + (y - 4)^2 = 2.5^2

to solve for

x

and

y

.

Tahmini Süre:3m 0s

Soru 18Soru →

In the coordinate plane, the line representing the linear equation $3x - 4y = 8$ contains the point $(-4, b)$ . What is the value of $b$ ?

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

The value of

b

is

-5

.

Substituting the coordinates

(-4, b)

into the equation

3x - 4y = 8

yields

3(-4) - 4b = 8

. Simplifying gives

-12 - 4b = 8

. Adding

12

to both sides results in

-4b = 20

. Finally, dividing by

-4

gives

b = -5

.

Adım Adım Çözüm

1

Substitute the point

(-4, b)

into the equation

3x - 4y = 8

.

3(-4) - 4(b) = 8

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

2

Simplify the constant term.

-12 - 4b = 8

Multiplying

3

by

-4

yields

-12

.

3

Isolate the variable term by adding

12

to both sides.

-4b = 20

Adding

12

to

8

gives

20

.

4

Solve for

b

by dividing both sides by

-4

.

b = -5

Dividing

20

by

-4

yields

-5

.

Anahtar Kavram

Determining an unknown coordinate of a point on a line by substitution into the linear equation.

Soru 19Soru →

In the standard $(x, y)$ coordinate plane, a circle passes through the points $A(-1, -2)$ and $B(3, 6)$ . The center of the circle, $C$ , lies on the line with the equation $y = 2x - 5$ . What is the radius of this circle?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The radius of the circle is 5.

The perpendicular bisector of the segment connecting

A(-1, -2)

and

B(3, 6)

passes through their midpoint

(1, 2)

with a slope of

-\frac{1}{2}

, giving the equation

x + 2y = 5

. Solving the system of equations with

y = 2x - 5

yields the center at

C(3, 1)

. The distance from

C(3, 1)

to

A(-1, -2)

is

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

.

Adım Adım Çözüm

1

Find the midpoint and slope of the segment

AB

connecting

A(-1, -2)

and

B(3, 6)

.

Midpoint

M = (1, 2)

and slope

m = 2

.

The center of any circle passing through

A

and

B

must lie on the perpendicular bisector of segment

AB

.

2

Determine the equation of the perpendicular bisector of

AB

.

x + 2y = 5

The perpendicular bisector passes through the midpoint

M(1, 2)

and has a slope that is the negative reciprocal of the slope of

AB

, which is

-\frac{1}{2}

.

3

Find the intersection point of the perpendicular bisector

x + 2y = 5

and the given line

y = 2x - 5

.

Center

C(3, 1)

The center of the circle lies on both the perpendicular bisector of

AB

and the line

y = 2x - 5

.

4

Calculate the distance from the center

C(3, 1)

to point

A(-1, -2)

using the distance formula.

Radius

r = 5

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

Anahtar Kavram

The perpendicular bisector of a chord of a circle passes through the center of that circle.

Soru 20Soru →

A hyperbola in the standard coordinate plane is represented by the equation $9x^2 - 16y^2 - 36x - 32y - 124 = 0$ . Which of the following equations represents one of the asymptotes of this hyperbola?

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

y = \frac{4}{3}x - \frac{11}{3}

y = \frac{3}{4}x - \frac{5}{2}

y = \frac{3}{4}x + \frac{5}{2}

y = \frac{4}{3}x - \frac{5}{3}

Cevabı ve açıklamayı göster

Cevap: $y = \frac{3}{4}x - \frac{5}{2}$

Cevap

y = \frac{3}{4}x - \frac{5}{2}

The correct equation is found by rewriting the hyperbola equation in standard form

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

through completing the square. This indicates a horizontal hyperbola centered at

(2, -1)

with

a = 4

and

b = 3

. The asymptotes are given by

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

. Substituting the values gives

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

. Simplifying the positive case results in the correct equation.

Adım Adım Çözüm

1

Group the

x

-terms and

y

-terms and factor out the coefficients of the squared terms.

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

Grouping prepares the algebraic expression for completing the square for both variables.

2

Complete the square for both the

x

and

y

expressions by adding the balanced constants to the right side.

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

Adding

9(4) = 36

and subtracting

16(1) = 16

to the right side balances the equation after completing the square.

3

Divide both sides by

144

to express the equation in the standard form of a hyperbola.

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

The standard form equation

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

reveals the center

(h, k)

and the semi-axes values

a

and

b

.

4

Identify the key parameters of the hyperbola and state the general formula for its asymptotes.

Center

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

,

a = 4

,

b = 3

. The asymptotes are given by

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

, which simplifies to

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

.

For a horizontal hyperbola, the rise-over-run slope of the asymptotes is governed by the ratio

\frac{b}{a}

.

5

Simplify the positive slope case to find the matching slope-intercept equation.

y = \frac{3}{4}x - \frac{5}{2}

Distributing the slope gives

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

, and subtracting

1

from both sides yields the final equation.

Anahtar Kavram

Rewriting a hyperbola equation using completing the square to find its asymptotes.

Alternatif Yöntem

Instead of completing the square entirely, find the center of the hyperbola by taking partial derivatives. The derivative with respect to

x

is

18x - 36 = 0 \implies x = 2

. The derivative with respect to

y

is

-32y - 32 = 0 \implies y = -1

. Thus, the center is

(2, -1)

. The slope of the asymptotes can be found from the ratio of the square roots of the coefficients of the quadratic terms:

m = \pm \sqrt{\frac{9}{16}} = \pm \frac{3}{4}

. Using the point-slope form with the center

(2, -1)

and slope

\frac{3}{4}

gives

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

, which simplifies directly to

y = \frac{3}{4}x - \frac{5}{2}

.

Tahmini Süre:2m 30s

Coordinate Geometry

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Alternatif Yöntem

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Alternatif Yöntem

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