Coordinate Geometry Alıştırma Soruları — ACT — Sayfa 4

Soru 61Soru →

A system administrator monitors the temperatures of two servers, Server A and Server B, over a period of $t$ hours after midnight. The temperature of Server A, in degrees Celsius, is modeled by a linear function of time. At $t = 2$ hours, the temperature is $30^\circ\text{C}$ , and at $t = a$ hours (where $a > 2$ ), the temperature is $(30 + a^2 - 2a)^\circ\text{C}$ . The temperature of Server B, in degrees Celsius, is also modeled by a linear function of time. At $t = 1$ hour, Server B's temperature is $25^\circ\text{C}$ , and at $t = a$ hours, its temperature is $(a^2 + 30)^\circ\text{C}$ . If the temperature of Server B increases at a constant rate that is $1.5$ times the constant rate of temperature increase of Server A, what is the value of $a$ ?

2

3

5

11

13

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of

a

is 5.

To find the rate of change (slope) for each server, we apply the slope formula

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

. For Server A, the slope is

m_A = \frac{(30 + a^2 - 2a) - 30}{a - 2} = a

. For Server B, the slope is

m_B = \frac{(a^2 + 30) - 25}{a - 1} = \frac{a^2 + 5}{a - 1}

. We are given that

m_B = 1.5 m_A

, which yields the equation

\frac{a^2 + 5}{a - 1} = 1.5a

. Multiplying both sides by

a - 1

gives

a^2 + 5 = 1.5a(a - 1) = 1.5a^2 - 1.5a

. Grouping all terms on one side gives

0.5a^2 - 1.5a - 5 = 0

. Multiplying the equation by 2 results in

a^2 - 3a - 10 = 0

, which factors as

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

. Since

a > 2

, the only valid solution is the positive root, which is 5.

Adım Adım Çözüm

1

Calculate the constant rate of temperature increase (slope) for Server A,

m_A

, using the coordinates

(2, 30)

and

(a, 30 + a^2 - 2a)

.

m_A = \frac{(30 + a^2 - 2a) - 30}{a - 2} = \frac{a^2 - 2a}{a - 2} = \frac{a(a-2)}{a-2} = a

(since

a > 2

)

The constant rate of change of a linear temperature model is represented by the slope of the line passing through its data points.

2

Calculate the constant rate of temperature increase (slope) for Server B,

m_B

, using the coordinates

(1, 25)

and

(a, a^2 + 30)

.

m_B = \frac{(a^2 + 30) - 25}{a - 1} = \frac{a^2 + 5}{a - 1}

The constant rate of change of Server B is represented by the slope of the line passing through its data points.

3

Set up the equation using the given relationship that Server B's rate of increase is

1.5

times Server A's rate (

m_B = 1.5 m_A

).

1.5a = \frac{a^2 + 5}{a - 1}

Equating the rate of change of Server B to 1.5 times the rate of change of Server A allows us to write an equation in terms of the variable

a

.

4

Solve the equation for

a

by cross-multiplying and simplifying the resulting quadratic equation.

1.5a(a - 1) = a^2 + 5 \implies 1.5a^2 - 1.5a = a^2 + 5 \implies 0.5a^2 - 1.5a - 5 = 0 \implies a^2 - 3a - 10 = 0 \implies (a-5)(a+2) = 0

Cross-multiplying eliminates the fraction, and grouping all terms on one side forms a standard quadratic equation that can be factored.

5

Identify the valid value of

a

that satisfies the initial domain constraint.

a = 5

(since

a = -2

is rejected because the problem states

a > 2

)

The solution must satisfy the given condition that

a > 2

for the time intervals to be valid.

Anahtar Kavram

Calculating and equating rates of change (slopes) of linear functions using the slope formula

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

and solving the resulting quadratic equation.

Soru 62Soru →

On a coordinate map, two tracking stations are located at the points $(2, -3)$ and $(-6, 12)$ . What is the straight-line distance, in map units, between the two stations?

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

The straight-line distance between the two stations is 17 map units.

Applying the coordinate distance formula to the coordinates

(2, -3)

and

(-6, 12)

yields a distance of

\sqrt{(-6 - 2)^2 + (12 - (-3))^2} = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

.

Adım Adım Çözüm

1

Identify coordinates of the two stations.

(x_1, y_1) = (2, -3)

and

(x_2, y_2) = (-6, 12)

Defining the coordinate variables is necessary to apply the formula correctly.

2

Apply the distance formula.

d = \sqrt{(-6 - 2)^2 + (12 - (-3))^2}

The distance formula calculates the straight-line distance between two coordinates.

3

Simplify the arithmetic terms.

d = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289}

Simplify the differences and squares under the radical.

4

Calculate the final square root.

d = 17

The square root of 289 is 17.

Anahtar Kavram

Distance Formula

Soru 63Soru →

A circle is graphed in the standard $(x, y)$ coordinate plane such that it is tangent to the $x$ -axis at $(4, 0)$ and tangent to the $y$ -axis at $(0, 4)$ . A line segment has one of its endpoints at the center of this circle and its midpoint at the point $(1, -1)$ . What is the total length of this line segment?

2\sqrt{10}

\sqrt{34}

2\sqrt{34}

16

2\sqrt{2}

Cevabı ve açıklamayı göster

Cevap: $2\sqrt{34}$

Cevap

The total length of the line segment is

2\sqrt{34}

.

The center of the circle is determined to be

C(4, 4)

based on the points of tangency at

(4, 0)

and

(0, 4)

. The distance between the center

C(4, 4)

and the midpoint

M(1, -1)

is calculated using the distance formula to be

\sqrt{34}

. Since the midpoint divides the segment into two equal halves, the total length of the segment is twice this distance, which is

2\sqrt{34}

.

Adım Adım Çözüm

1

Determine the coordinates of the center of the circle.

The center of the circle is

C(4, 4)

.

Since the circle is tangent to the

x

-axis at

(4, 0)

, its center must lie on the vertical line

x = 4

. Since it is tangent to the

y

-axis at

(0, 4)

, its center must lie on the horizontal line

y = 4

. Their intersection gives the center of the circle at

C(4, 4)

.

2

Calculate the distance from the center of the circle to the midpoint of the segment.

The distance is

\sqrt{34}

.

Using the distance formula between the center

C(4, 4)

and the midpoint

M(1, -1)

:

d(C, M) = \sqrt{(1 - 4)^2 + (-1 - 4)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}

.

3

Find the total length of the line segment.

The total length is

2\sqrt{34}

.

Since the midpoint divides the segment into two equal parts, the total length is twice the distance from one endpoint to the midpoint:

\text{Length} = 2 \times d(C, M) = 2\sqrt{34}

.

Anahtar Kavram

Using circle tangencies to find the center, and applying the distance and midpoint formulas to determine segment properties.

Tahmini Süre:1m 30s

Soru 64Soru →

In the standard $(x, y)$ coordinate plane, if the slope of a line passing through the points $(k, k^2)$ and $(k+1, (k+1)^2)$ is an integer, then $k$ must be an integer. Is this statement true or false?

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

False

The statement is false because the simplified slope of the line is

2k + 1

. The slope

2k + 1

is an integer for any half-integer value of

k

(such as

k = 0.5

, which yields a slope of

2

), showing that

k

does not have to be an integer.

Adım Adım Çözüm

1

Use the slope formula

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

with the given coordinates

(k, k^2)

and

(k+1, (k+1)^2)

.

The slope is

m = \frac{(k+1)^2 - k^2}{(k+1) - k}

.

To express the slope of the line in terms of the variable

k

.

2

Simplify the numerator and denominator of the slope expression.

m = \frac{k^2 + 2k + 1 - k^2}{1} = 2k + 1

.

To obtain a simplified algebraic expression for the slope.

3

Set the slope equal to an arbitrary integer

n

and solve for

k

.

2k + 1 = n \implies k = \frac{n-1}{2}

.

To find the set of values for

k

that yield an integer slope.

4

Evaluate if

k

must be an integer by choosing an integer value for

n

.

If

n = 2

, then

k = \frac{2-1}{2} = 0.5

, which is not an integer.

To test the truth value of the statement by finding a counterexample.

Anahtar Kavram

Determining the slope of a line from two coordinate points and analyzing the algebraic properties of the slope expression.

Soru 65Soru →

A hyperbola is defined by the equation $9x^2 - 4y^2 - 36x - 8y - 4 = 0$ in the standard $(x, y)$ coordinate plane. What is the slope of the asymptote of this hyperbola that has a positive slope?

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The positive slope of the asymptotes is 1.5.

By converting the general form of the hyperbola equation into standard form, we determine that it is a horizontal hyperbola with

a = 2

and

b = 3

. The slopes of the asymptotes for a horizontal hyperbola are

\pm \frac{b}{a}

, making the positive slope equal to

\frac{3}{2} = 1.5

.

Adım Adım Çözüm

1

Group x-terms and y-terms, and move the constant to the other side.

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

This prepares the equation for completing the square.

2

Complete the square for the quadratic expressions in x and y.

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

Completing the square yields

9[(x-2)^2 - 4] - 4[(y+1)^2 - 1] = 4 \implies 9(x-2)^2 - 36 - 4(y+1)^2 + 4 = 4

.

3

Divide both sides by 36 to format the equation in standard hyperbola form.

\frac{(x-2)^2}{4} - \frac{(y+1)^2}{9} = 1

The standard form of a horizontal hyperbola centered at

(h, k)

is

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

.

4

Identify the values of a and b from the denominators.

a = 2

and

b = 3

Since

a^2 = 4

and

b^2 = 9

, taking the square roots gives

a = 2

and

b = 3

.

5

Determine the positive slope of the asymptotes using the formula for a horizontal hyperbola.

m = \frac{b}{a} = 1.5

The asymptotes for a horizontal hyperbola are given by

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

, so the positive slope is

\frac{b}{a} = \frac{3}{2} = 1.5

.

Anahtar Kavram

Rewriting a hyperbola equation from general form to standard form to find asymptote equations.

Tahmini Süre:1m 30s

Soru 66Soru →

In the standard $(x, y)$ coordinate plane, a line segment $AB$ has endpoints $A(2, 5)$ and $B(6, -1)$ . Segment $AB$ is reflected across the line $y = -x$ to form segment $A'B'$ . Then, segment $A'B'$ is dilated by a scale factor of $3$ with a center of dilation at $C(2, -1)$ to form segment $A''B''$ . What is the sum of the $x$ - and $y$ -coordinates of the midpoint of segment $A''B''$ ?

16

-8

-16

-18

-20

Cevabı ve açıklamayı göster

Cevap: -20

Cevap

-20

To find the coordinates of the midpoint of the transformed segment

A''B''

, we can track the midpoint of the original segment

AB

through each transformation. The midpoint of

AB

is

M(4, 2)

. Reflecting

M

across the line

y = -x

swaps and negates its coordinates, resulting in

M'(-2, -4)

. Dilating

M'

by a scale factor of

3

centered at

C(2, -1)

means finding the point

M''

such that the vector

\vec{CM''}

is

3

times the vector

\vec{CM'}

. This is computed as

M'' = (2 + 3(-2 - 2), -1 + 3(-4 - (-1))) = (-10, -10)

. The sum of these coordinates is

-10 + (-10) = -20

.

Adım Adım Çözüm

1

Find the midpoint of the original segment

AB

.

The midpoint

M

of segment

AB

is

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

.

Since rigid transformations (reflections) and similarity transformations (dilations) preserve midpoints, we can find the final midpoint by applying the sequence of transformations directly to the initial midpoint.

2

Reflect the midpoint

M(4, 2)

across the line

y = -x

to find the midpoint

M'

of segment

A'B'

.

Applying the reflection rule

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

to

M(4, 2)

yields

M'(-2, -4)

.

Reflecting a point across the line

y = -x

swaps its coordinates and negates both values.

3

Dilate the midpoint

M'(-2, -4)

by a scale factor of

3

with the center of dilation at

C(2, -1)

to find the midpoint

M''

of segment

A''B''

.

Using the dilation formula

(x'', y'') = (x_c + k(x' - x_c), y_c + k(y' - y_c))

with

C(2, -1)

and

k = 3

:

x'' = 2 + 3(-2 - 2) = -10

, and

y'' = -1 + 3(-4 - (-1)) = -10

. This gives the midpoint

M''(-10, -10)

.

A dilation centered at

(x_c, y_c)

scales the horizontal and vertical distances from the center to the point by the scale factor

k

.

4

Calculate the sum of the

x

- and

y

-coordinates of

M

.

The sum of the coordinates is

-10 + (-10) = -20

.

The question asks for the sum of the coordinates of the final midpoint.

Anahtar Kavram

Applying composite transformations in the coordinate plane to geometric figures and midpoints.

Soru 67Soru →

In the standard $(x, y)$ coordinate plane, line segment $AB$ has endpoints $A(-2, 4)$ and $B(4, 2)$ . The segment is rotated $90^\circ$ counterclockwise about the origin, then reflected across the line $y = -x$ , and finally translated $3$ units to the right and $2$ units down. What are the coordinates of the midpoint of the final image of the segment?

(-4, 5)

(-3, 6)

(4, -5)

(2, 1)

(0, -1)

Cevabı ve açıklamayı göster

Cevap: $(2, 1)$

Cevap

(2, 1)

The correct midpoint is

(2, 1)

. First, the midpoint of the original segment

AB

is calculated using the midpoint formula, yielding

(1, 3)

. Rotating this point

90^\circ

counterclockwise about the origin maps

(x, y)

to

(-y, x)

, which gives

(-3, 1)

. Reflecting

(-3, 1)

across the line

y = -x

maps

(x, y)

to

(-y, -x)

, resulting in

(-1, 3)

. Finally, translating this point

3

units to the right and

2

units down yields

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

.

Adım Adım Çözüm

1

Find the midpoint of the original line segment

AB

.

M(1, 3)

The midpoint formula is

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

. Substituting the coordinates of

A(-2, 4)

and

B(4, 2)

yields

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

.

2

Apply a

90^\circ

counterclockwise rotation about the origin to the midpoint.

M'(-3, 1)

A

90^\circ

counterclockwise rotation about the origin maps

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

. Applying this rule to

M(1, 3)

gives

(-3, 1)

.

3

Reflect the rotated point across the line

y = -x

.

M''(-1, 3)

A reflection across the line

y = -x

maps

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

. Applying this rule to

M'(-3, 1)

gives

(-1, 3)

.

4

Translate the reflected point

3

units right and

2

units down.

M'''(2, 1)

A translation of

3

units right and

2

units down maps

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

. Translating

M''(-1, 3)

yields

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

.

Anahtar Kavram

Composite transformations of geometric figures in the coordinate plane

Alternatif Yöntem

Instead of transforming the midpoint, you can transform the individual endpoints

A(-2, 4)

and

B(4, 2)

through all three steps. Point

A(-2, 4)

becomes

A'(-4, -2) \rightarrow A''(2, 4) \rightarrow A'''(5, 2)

. Point

B(4, 2)

becomes

B'(-2, 4) \rightarrow B''(-4, 2) \rightarrow B'''(-1, 0)

. The midpoint of

A'''

and

B'''

is then calculated to be

\left(\frac{5 + (-1)}{2}, \frac{2 + 0}{2}\right) = (2, 1)

.

Tahmini Süre:1m 30s

Soru 68Soru →

In the standard $(x,y)$ coordinate plane, line $q$ is perpendicular to the line with equation $y = \frac{3}{4}x - 2$ . If line $q$ passes through the point $(2, 1)$ , what is the $y$ -intercept of line $q$ ?

-\frac{5}{3}

\frac{9}{4}

\frac{11}{3}

\frac{2}{3}

-\frac{1}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{11}{3}$

Cevap

The

y

-intercept of line

q

is

\frac{11}{3}

.

The slope of the given line

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

is

\frac{3}{4}

. The slope of a perpendicular line is the negative reciprocal of the original slope, which is

-\frac{4}{3}

. Using the point-slope form of a linear equation with the point

(2, 1)

gives

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

. To find the

y

-intercept, set

x = 0

, which gives

y - 1 = -\frac{4}{3}(-2) = \frac{8}{3}

. Adding

1

to both sides yields

y = \frac{8}{3} + 1 = \frac{11}{3}

. Therefore, the correct

y

-intercept is

\frac{11}{3}

.

Adım Adım Çözüm

1

Determine the slope of the perpendicular line

q

.

The slope of the given line is

\frac{3}{4}

, so the slope of line

q

(the negative reciprocal) is

-\frac{4}{3}

.

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

2

Write the equation of line

q

using the point-slope form with the point

(2, 1)

.

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

The point-slope form

y - y_1 = m(x - x_1)

is used to find the equation of a line given its slope and a point on the line.

3

Find the

y

-intercept by setting

x = 0

and solving for

y

.

y - 1 = -\frac{4}{3}(0 - 2) \Rightarrow y - 1 = \frac{8}{3} \Rightarrow y = \frac{8}{3} + 1 = \frac{11}{3}

The

y

-intercept of a line is the value of

y

when

x = 0

.

Anahtar Kavram

The slope of a line perpendicular to a line with slope

m

is

-\frac{1}{m}

. The equation of a line can be written in point-slope form as

y - y_1 = m(x - x_1)

.

Soru 69Soru →

A toy car moves along a straight track. Its position in meters, $s$ , is plotted against time in seconds, $t$ , on a standard coordinate plane. At $t = 2$ seconds, the car is at a position of $-3$ meters. From $t = 2$ to $t = 6$ seconds, the position changes at a constant rate of $m$ meters per second. From $t = 6$ to $t = 8$ seconds, the position changes at a constant rate of $2m + 3$ meters per second. If the average rate of change of the car's position over the entire interval from $t = 2$ to $t = 8$ seconds is $\frac{5}{3}$ meters per second, what is the constant rate of change, in meters per second, from $t = 6$ to $t = 8$ seconds?

1/2

12/5

29/9

4

19/4

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The correct constant rate of change from

t = 6

to

t = 8

seconds is

4

meters per second.

The correct answer of

4

meters per second is found by setting up the displacement for each interval in terms of

m

. The displacement during the first interval is

4m

, and the displacement during the second interval is

2(2m + 3) = 4m + 6

. Adding these gives a total displacement of

8m + 6

over a total time of

6

seconds. Dividing total displacement by total time yields the average rate of change,

\frac{8m + 6}{6} = \frac{4m + 3}{3}

. Setting this equal to the given average rate of

\frac{5}{3}

yields

m = \frac{1}{2}

. Finally, substituting this back into the rate expression for the second interval,

2m + 3

, gives

2(1/2) + 3 = 4

.

Adım Adım Çözüm

1

Express the displacement in each time interval using the rate of change (slope) formula:

\Delta s = \text{slope} \times \Delta t

.

For

t = 2

to

t = 6

:

\Delta s_1 = m \times (6 - 2) = 4m

. For

t = 6

to

t = 8

:

\Delta s_2 = (2m + 3) \times (8 - 6) = 2(2m + 3) = 4m + 6

.

The constant rate of change in a position-time graph is the slope of the line, which relates time intervals to position displacements.

2

Calculate the total displacement over the entire interval from

t = 2

to

t = 8

seconds.

Total displacement

\Delta s_{\text{total}} = \Delta s_1 + \Delta s_2 = 4m + (4m + 6) = 8m + 6

.

The total displacement is the sum of the individual displacements over consecutive sub-intervals.

3

Set up the average rate of change equation using the total displacement and the total time elapsed (

\Delta t_{\text{total}} = 8 - 2 = 6

seconds).

Average rate of change =

\frac{8m + 6}{6} = \frac{4m + 3}{3}

.

The average rate of change is the net displacement divided by the total time elapsed.

4

Equate the expression for the average rate of change to the given value of

\frac{5}{3}

and solve for

m

.

\frac{4m + 3}{3} = \frac{5}{3} \Rightarrow 4m + 3 = 5 \Rightarrow 4m = 2 \Rightarrow m = \frac{1}{2}

.

Solving this linear equation yields the value of the parameter

m

that satisfies the average rate condition.

5

Calculate the rate of change for the second interval from

t = 6

to

t = 8

seconds by substituting

m = \frac{1}{2}

into

2m + 3

.

Rate of change =

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

.

The question asks for the rate of change during the second interval, which is defined in terms of

m

as

2m + 3

.

Anahtar Kavram

Slope as a constant rate of change in a piecewise linear model

Tahmini Süre:3m 0s

Soru 70Soru →

In the standard $(x, y)$ coordinate plane, a line $L_1$ passes through the points $(2, 5)$ and $(-1, -4)$ . A second line, $L_2$ , passes through the $y$ -intercept of $L_1$ and has a slope that is twice the slope of $L_1$ . What is the $x$ -intercept of $L_2$ ?

\frac{1}{6}

-\frac{1}{6}

\frac{1}{3}

-\frac{3}{2}

-\frac{13}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{6}$

Cevap

The

x

-intercept of

L_2

is

\frac{1}{6}

.

To find the

x

-intercept of

L_2

, the slope and

y

-intercept of

L_1

must first be found. The slope of

L_1

is

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

. Using the point-slope form with the point

(2, 5)

, the equation of

L_1

is

y - 5 = 3(x - 2) \implies y = 3x - 1

, meaning its

y

-intercept is

(0, -1)

. The second line,

L_2

, has a slope of

2 \times 3 = 6

and passes through the same

y

-intercept

(0, -1)

, giving the equation

y = 6x - 1

. Setting

y = 0

to find the

x

-intercept yields

6x - 1 = 0 \implies x = \frac{1}{6}

.

Adım Adım Çözüm

1

Calculate the slope of line

L_1

using the coordinates

(2, 5)

and

(-1, -4)

.

The slope of

L_1

is

m_1 = \frac{5 - (-4)}{2 - (-1)} = \frac{9}{3} = 3

.

The slope of a line containing

(x_1, y_1)

and

(x_2, y_2)

is defined as

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

.

2

Determine the equation and the

y

-intercept of line

L_1

.

The equation of

L_1

in point-slope form is

y - 5 = 3(x - 2) \implies y = 3x - 1

. The

y

-intercept of

L_1

is

(0, -1)

.

Writing the equation in slope-intercept form

y = mx + b

directly identifies the

y

-intercept at

(0, b)

.

3

Determine the equation of line

L_2

.

The slope of

L_2

is

2 \times 3 = 6

. Since

L_2

passes through the

y

-intercept of

L_1

at

(0, -1)

, its equation is

y = 6x - 1

.

Using the relationship for the doubled slope and the shared

y

-intercept, the slope-intercept form of the second line is determined.

4

Find the

x

-intercept of

L_2

by setting

y = 0

.

Setting

y = 0

in the equation

y = 6x - 1

gives

0 = 6x - 1 \implies 6x = 1 \implies x = \frac{1}{6}

.

The

x

-intercept occurs where the line crosses the

x

-axis, which corresponds to setting

y = 0

.

Anahtar Kavram

Determining linear equations from coordinates, finding slopes, and calculating coordinate intercepts.

Soru 71Soru →

A line segment in the standard $(x, y)$ coordinate plane has endpoints at $(1, a)$ and $(5, a^2)$ . The perpendicular bisector of this segment is parallel to the line defined by the equation $x + 3y = 6$ . What is the sum of all possible values of the constant $a$ ?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The sum of all possible values of the constant

a

is 1.

To find the sum of all possible values of the constant

a

, we first find the slope of the line

x + 3y = 6

by writing it in slope-intercept form:

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

. Since the perpendicular bisector is parallel to this line, its slope is also

-\frac{1}{3}

. The line segment is perpendicular to its perpendicular bisector, so the slope of the line segment is the negative reciprocal of

-\frac{1}{3}

, which is

3

. Setting the slope of the segment

\frac{a^2 - a}{5 - 1}

equal to

3

gives the equation

\frac{a^2 - a}{4} = 3

, which simplifies to

a^2 - a - 12 = 0

. Solving this quadratic equation yields

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

, giving the values

a = 4

and

a = -3

. The sum of these possible values is

4 + (-3) = 1

.

Adım Adım Çözüm

1

Find the slope of the given line

x + 3y = 6

.

The slope is

-\frac{1}{3}

.

Rewriting the equation in slope-intercept form (

y = mx + b

) gives

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

, showing the slope is

-\frac{1}{3}

.

2

Determine the slope of the line segment.

The slope is 3.

The line segment is perpendicular to its perpendicular bisector. Because the perpendicular bisector is parallel to the reference line, its slope is also

-\frac{1}{3}

. The line segment's slope is the negative reciprocal of

-\frac{1}{3}

, which is

3

.

3

Write the slope of the segment in terms of

a

and set it equal to 3.

\frac{a^2 - a}{4} = 3

Using the slope formula with endpoints

(1, a)

and

(5, a^2)

gives the expression

\frac{a^2 - a}{5 - 1}

.

4

Solve the quadratic equation for

a

.

a = 4

or

a = -3

Multiplying both sides by 4 yields

a^2 - a = 12

, which simplifies to the quadratic

a^2 - a - 12 = 0

. Factoring gives

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

.

5

Sum all possible values of

a

.

1

The sum of the values is

4 + (-3) = 1

.

Anahtar Kavram

Understanding that parallel lines have equal slopes, perpendicular lines have slopes that are negative reciprocals of each other, and applying the slope formula to solve for coordinate variables.

Soru 72Soru →

In the standard $(x, y)$ coordinate plane, the triangular region $R$ is bounded by the lines $y = 2x$ , $y = -x + 9$ , and the $x$ -axis. A vertical line $x = k$ (where $0 < k < 9$ ) divides region $R$ into two sub-regions. If the area of the sub-region to the right of the line $x = k$ is exactly $8$ , what is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5

The boundary lines intersect to form a triangle with vertices at

(0, 0)

,

(9, 0)

, and

(3, 6)

. Since the area to the right of

x = k

is

8

, and the total area of the triangle is

27

,

k

must be greater than

3

. The region to the right of

x = k

is a right triangle with a base of

9 - k

and a height of

9 - k

. Setting its area

\frac{1}{2}(9 - k)^2

equal to

8

yields

(9 - k)^2 = 16

, which gives

9 - k = 4

(since

k < 9

), and thus

k = 5

.

Adım Adım Çözüm

1

Find the vertices of the triangular region

R

by finding the intersection points of the boundary lines

y = 2x

,

y = -x + 9

, and

y = 0

.

The vertices of the triangle are

A(0, 0)

,

B(9, 0)

, and

C(3, 6)

.

This establishes the boundaries and shape of the triangular region.

2

Determine which side of the peak

x = 3

the vertical line

x = k

must lie. Calculate the total area and the area of the left portion.

The total area of the triangle is

27

. The area to the left of the peak

x = 3

is

9

. Since the area of the region to the right of

x = k

is

8

, which is less than

18

,

k

must be greater than or equal to

3

.

This determines the geometric shape of the sub-region to the right of

x = k

as a right triangle.

3

Set up the area formula for the right-hand triangle with vertices

(k, 0)

,

(9, 0)

, and

(k, -k + 9)

, and set it equal to

8

.

The area is

\frac{1}{2}(9 - k)^2 = 8

, which simplifies to

(9 - k)^2 = 16

.

This relates the given area to the unknown coordinate

k

.

4

Solve the equation

(9 - k)^2 = 16

for

k

, keeping in mind that

k < 9

.

Taking the square root gives

9 - k = 4 \implies k = 5

.

This yields the final value of

k

.

Anahtar Kavram

Finding the area of a region defined by linear boundary equations and dividing it with a vertical line.

Alternatif Yöntem

Using similar triangles: The right-hand triangle formed by the line

x = 3

, the line

y = -x + 9

, and the

x

-axis has vertices at

(3,0)

,

(9,0)

, and

(3,6)

, with an area of

\frac{1}{2} \times 6 \times 6 = 18

. The smaller triangle to the right of

x = k

has an area of

8

and is similar to the larger triangle. The ratio of their areas is

\frac{8}{18} = \frac{4}{9}

, which means the ratio of their linear dimensions is

\sqrt{\frac{4}{9}} = \frac{2}{3}

. The base of the larger triangle is

9 - 3 = 6

, so the base of the smaller triangle must be

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

. This gives

9 - k = 4 \implies k = 5

.

Tahmini Süre:2m 30s

Soru 73Soru →

A circle in the standard $(x, y)$ coordinate plane has center $(3, -4)$ and passes through the point $(6, 0)$ . Which of the following is an equation of this circle?

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

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

x^2 + y^2 - 6x + 8y + 20 = 0

x^2 + y^2 - 18 = 0

x^2 + y^2 - 6x + 8y - 24 = 0

Cevabı ve açıklamayı göster

Cevap: $x^2 + y^2 - 6x + 8y = 0$

Cevap

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

The standard equation of a circle is

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

, where

(h, k)

is the center and

r

is the radius. Given the center is

(3, -4)

, the equation becomes

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

. Since the circle passes through

(6, 0)

, we substitute these coordinates to find

r^2

:

(6 - 3)^2 + (0 + 4)^2 = 3^2 + 4^2 = 9 + 16 = 25

. The equation is therefore

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

. Expanding this equation gives

x^2 - 6x + 9 + y^2 + 8y + 16 = 25

. Combining constant terms yields

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

. Subtracting 25 from both sides results in the general form equation

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

.

Adım Adım Çözüm

1

Calculate the radius squared of the circle using the distance formula between the center

(3, -4)

and the point on the circle

(6, 0)

.

r^2 = (6 - 3)^2 + (0 - (-4))^2 = 3^2 + 4^2 = 9 + 16 = 25

The distance between the center and any point on the circle is equal to the radius of the circle.

2

Write the standard form of the circle's equation using the center

(h, k) = (3, -4)

and the radius squared

r^2 = 25

.

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

The standard equation of a circle is

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

.

3

Expand the squared binomials in the standard equation to convert it into general form.

x^2 - 6x + 9 + y^2 + 8y + 16 = 25

Expanding allows us to combine like terms and match the general form expressions in the options.

4

Simplify the expanded equation by combining constant terms and setting the equation to zero.

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

Subtracting 25 from both sides yields the final simplified general form equation of the circle.

Anahtar Kavram

Deriving and expanding the equation of a circle from its center and a point.

Soru 74Soru →

In the standard $(x, y)$ coordinate plane, line $j$ has the equation $y = -\frac{1}{4}x + 7$ . If line $k$ is perpendicular to line $j$ , what is the slope of line $k$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The slope of line

k

is 4.

The slope of the perpendicular line is

4

because the negative reciprocal of the given slope,

-\frac{1}{4}

, is

4

.

Adım Adım Çözüm

1

Identify the slope of line

j

.

The slope of line

j

is

-\frac{1}{4}

.

The equation of line

j

is given in slope-intercept form,

y = mx + b

, where the coefficient of

x

represents the slope.

2

Calculate the negative reciprocal of the slope of line

j

to find the slope of perpendicular line

k

.

The slope of line

k

is

4

.

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

-\frac{1}{4}

is

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

.

Anahtar Kavram

The slopes of perpendicular lines are negative reciprocals of each other.

Tahmini Süre:45s

Soru 75Soru →

In the standard $(x,y)$ coordinate plane, point $P$ has coordinates $(3, 2)$ . If point $P$ is reflected across the $y$ -axis and then translated $4$ units up to create point $P'$ , what are the coordinates of $P'$ ?

(-3, 6)

(-3, 2)

(-3, -2)

(1, 2)

(3, 2)

Cevabı ve açıklamayı göster

Cevap: $(-3, 6)$

Cevap

The coordinates

(-3, 6)

Reflecting the point

(3, 2)

across the

y

-axis changes the sign of the

x

-coordinate, producing

(-3, 2)

. Translating this resulting point

4

units up adds

4

to the

y

-coordinate, giving the final coordinates

(-3, 6)

.

Adım Adım Çözüm

1

Reflect the point

P(3, 2)

across the

y

-axis.

The

x

-coordinate changes sign, while the

y

-coordinate remains unchanged, yielding

(-3, 2)

.

A reflection across the

y

-axis maps any point

(x, y)

to

(-x, y)

.

2

Translate the point

(-3, 2)

up by

4

units.

Add

4

to the

y

-coordinate:

2 + 4 = 6

, yielding

(-3, 6)

.

A translation of

d

units upward maps any point

(x, y)

to

(x, y + d)

.

Anahtar Kavram

Applying a sequence of transformations to a point in the coordinate plane by first reflecting it across an axis and then translating it.

Soru 76Soru →

A linear model predicts the height of a plant, $H$ , in inches, based on the number of weeks, $w$ , since it was planted. According to the model, the height of the plant after $2$ weeks is $3\frac{1}{2}$ inches, and its height after $4$ weeks is $4\frac{1}{4}$ inches. What was the initial height of the plant, in inches, when it was planted?

1\frac{5}{6}

\frac{3}{8}

2

2\frac{3}{4}

\frac{9}{16}

Cevabı ve açıklamayı göster

Cevap: $2\frac{3}{4}$

Cevap

The initial height of the plant was

2\frac{3}{4}

inches.

The correct answer is

2\frac{3}{4}

inches. First, the two given data points are

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

and

(4, \frac{17}{4})

. The rate of growth (slope) is calculated as

m = \frac{\frac{17}{4} - \frac{7}{2}}{4 - 2} = \frac{3}{8}

inches per week. Substituting this into the slope-intercept form equation

H = mw + b

at

w = 2

gives

\frac{7}{2} = \frac{3}{8}(2) + b

, which simplifies to

b = \frac{7}{2} - \frac{3}{4} = 2\frac{3}{4}

inches.

Adım Adım Çözüm

1

Convert the mixed numbers for the plant heights into improper fractions.

The height at

2

weeks is

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

inches, and the height at

4

weeks is

4\frac{1}{4} = \frac{17}{4}

inches.

Improper fractions are easier to use in slope and linear equation calculations.

2

Calculate the slope (weekly growth rate) of the linear function using the formula

m = \frac{H_2 - H_1}{w_2 - w_1}

.

m = \frac{\frac{17}{4} - \frac{7}{2}}{4 - 2} = \frac{\frac{3}{4}}{2} = \frac{3}{8}

The slope represents the constant rate of height increase per week.

3

Use the slope-intercept form

H = mw + b

and substitute one of the points to solve for the y-intercept,

b

.

Using the point

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

:

\frac{7}{2} = \frac{3}{8}(2) + b \Rightarrow \frac{7}{2} = \frac{3}{4} + b \Rightarrow b = \frac{7}{2} - \frac{3}{4} = \frac{14}{4} - \frac{3}{4} = \frac{11}{4} = 2\frac{3}{4}

The y-intercept,

b

, represents the initial height of the plant at

w = 0

weeks.

Anahtar Kavram

Determining the equation of a line from two points and finding the y-intercept in a real-world context.

Soru 77Soru →

In the standard $(x, y)$ coordinate plane, line $L_1$ is defined by the equation $3x - ky = 8$ , where $k$ is a non-zero constant. Line $L_2$ is perpendicular to $L_1$ and passes through the points $(k, 2)$ and $(-1, k+5)$ . What is the sum of all possible real values of $k$ ?

-9

-6

-2

0

2

Cevabı ve açıklamayı göster

Cevap: $2$

Cevap

The sum of all possible real values of

k

is

2

.

The slope of line

L_1

is found by converting its equation to slope-intercept form, yielding

m_1 = \frac{3}{k}

. Since line

L_2

is perpendicular, its slope must be the negative reciprocal,

m_2 = -\frac{k}{3}

. Using the slope formula with the given points on line

L_2

, we also have

m_2 = \frac{(k+5) - 2}{-1 - k} = -\frac{k+3}{k+1}

. Setting these two expressions for

m_2

equal to each other and solving the resulting proportion gives the quadratic equation

k^2 - 2k - 9 = 0

. The discriminant of this equation is positive (

40 > 0

), confirming the existence of two real solutions. By Vieta's formulas, the sum of the roots is

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

.

Adım Adım Çözüm

1

Determine the slope of line

L_1

in terms of

k

.

The slope of line

L_1

is

m_1 = \frac{3}{k}

.

Rewriting the equation

3x - ky = 8

in slope-intercept form (

y = mx + b

) gives

ky = 3x - 8

, which simplifies to

y = \frac{3}{k}x - \frac{8}{k}

. The coefficient of

x

represents the slope of the line.

2

Find the perpendicular slope

m_2

of line

L_2

in terms of

k

.

The slope of line

L_2

is

m_2 = -\frac{k}{3}

.

Since line

L_2

is perpendicular to line

L_1

, its slope must be the negative reciprocal of

m_1

(i.e.,

m_2 = -\frac{1}{m_1}

).

3

Express the slope of line

L_2

using the two given points

(k, 2)

and

(-1, k+5)

.

The slope of line

L_2

is

m_2 = -\frac{k+3}{k+1}

.

Using the slope formula

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

with the points

(k, 2)

and

(-1, k+5)

, we get

m_2 = \frac{(k+5) - 2}{-1 - k} = \frac{k+3}{-(k+1)} = -\frac{k+3}{k+1}

.

4

Set the two expressions for the slope of line

L_2

equal to each other and solve for

k

.

The quadratic equation is

k^2 - 2k - 9 = 0

.

Equating the slopes gives

-\frac{k}{3} = -\frac{k+3}{k+1}

. Multiplying both sides by

-1

gives

\frac{k}{3} = \frac{k+3}{k+1}

. Cross-multiplying yields

k(k+1) = 3(k+3) \implies k^2 + k = 3k + 9 \implies k^2 - 2k - 9 = 0

.

5

Calculate the sum of all possible real values of

k

.

The sum of all possible real values of

k

is

2

.

The discriminant of the quadratic equation

k^2 - 2k - 9 = 0

is

(-2)^2 - 4(1)(-9) = 40 > 0

, confirming that two distinct real solutions for

k

exist. According to Vieta's formulas, the sum of the roots of a quadratic equation

ax^2 + bx + c = 0

is

-\frac{b}{a}

. Here,

a=1

and

b=-2

, so the sum is

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

.

Anahtar Kavram

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

m_1 \cdot m_2 = -1

).

Soru 78Soru →

A scientist maps the movement of a cell on a coordinate grid. The cell is initially located at the point $(2, -5)$ . The cell then moves to a new position after being translated 6 units to the left and 4 units up. What are the coordinates of the cell's new position?

(-4, -9)

(8, -1)

(-4, -1)

(8, -9)

(-1, -4)

Cevabı ve açıklamayı göster

Cevap: $(-4, -1)$

Cevap

The correct position of the cell is

(-4, -1)

.

To find the coordinates of the new position, apply the translation to the initial coordinates

(2, -5)

. A translation of 6 units to the left subtracts 6 from the x-coordinate:

2 - 6 = -4

. A translation of 4 units up adds 4 to the y-coordinate:

-5 + 4 = -1

. This results in the coordinates

(-4, -1)

.

Adım Adım Çözüm

1

Calculate the new x-coordinate by applying the horizontal translation.

The initial x-coordinate is

2

. Translating 6 units to the left means subtracting 6:

2 - 6 = -4

.

A horizontal shift to the left decreases the x-coordinate value.

2

Calculate the new y-coordinate by applying the vertical translation.

The initial y-coordinate is

-5

. Translating 4 units up means adding 4:

-5 + 4 = -1

.

A vertical shift upward increases the y-coordinate value.

3

Combine the new coordinates into an ordered pair

(x, y)

.

The new coordinates are

(-4, -1)

.

The coordinates are represented by the new horizontal position followed by the new vertical position.

Anahtar Kavram

Translating a point in the coordinate plane by modifying its coordinates:

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

, where

h

is the horizontal shift and

k

is the vertical shift.

Soru 79Soru →

In the standard $(x,y)$ coordinate plane, triangle $PQR$ has vertices $P(2, 3)$ , $Q(6, 3)$ , and $R(2, 6)$ . The triangle undergoes a sequence of three transformations:

1. A dilation centered at the point $(4, 2)$ with a scale factor of $-2$ .
2. A reflection across the line $y = -x$ .
3. A rotation of $90^\circ$ counterclockwise about the origin.

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

(6, -8)

(-4, 12)

(8, 6)

(-8, -6)

(0, -10)

Cevabı ve açıklamayı göster

Cevap: $(8, 6)$

Cevap

The coordinate pair

(8, 6)

Applying the dilation formula centered at

(4, 2)

with scale factor

-2

to

R(2, 6)

gives the point

R'(8, -6)

. Reflecting this point across the line

y = -x

negates and swaps the coordinates, yielding

R''(6, -8)

. Finally, a

90^\circ

counterclockwise rotation about the origin swaps the coordinates and negates the new

x

-coordinate, producing the final coordinates

(8, 6)

.

Adım Adım Çözüm

1

Perform the dilation of vertex

R(2, 6)

centered at

C(4, 2)

with a scale factor of

k = -2

.

The intermediate image is

R'(8, -6)

.

For a dilation centered at

(x_c, y_c)

with scale factor

k

, the coordinates of the image are

(x_c + k(x - x_c), y_c + k(y - y_c))

. Substituting

R(2, 6)

,

C(4, 2)

, and

k = -2

gives:

x' = 4 - 2(2 - 4) = 8

and

y' = 2 - 2(6 - 2) = -6

.

2

Reflect the point

R'(8, -6)

across the line

y = -x

.

The intermediate image is

R''(6, -8)

.

A reflection across the line

y = -x

maps any point

(x, y)

to

(-y, -x)

. Applying this rule to

R'(8, -6)

yields

R''(-(-6), -8) = (6, -8)

.

3

Rotate the point

R''(6, -8)

by

90^\circ

counterclockwise about the origin.

The final image is

R'''(8, 6)

.

A rotation of

90^\circ

counterclockwise about the origin maps any point

(x, y)

to

(-y, x)

. Applying this rule to

R''(6, -8)

yields

R'''(-(-8), 6) = (8, 6)

.

Anahtar Kavram

Composite transformations in the coordinate plane combining dilation from a non-origin center, reflection across diagonal lines, and rotation about the origin.

Alternatif Yöntem

Instead of applying the transformations step-by-step to the point, we can track the transformations vectorially. For the dilation, the vector

\vec{CR} = R - C = (-2, 4)

is scaled by

-2

to get

-2\vec{CR} = (4, -8)

, which added back to

C(4, 2)

yields

R'(8, -6)

. Reflecting across

y = -x

exchanges the coordinates and negates them, giving

R''(6, -8)

. Rotating

90^\circ

counterclockwise about the origin maps

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

, yielding

R'''(8, 6)

.

Tahmini Süre:3m 0s

Soru 80Soru →

A square $PQRS$ has vertices at $P(2, 2)$ , $Q(6, 2)$ , $R(6, 6)$ , and $S(2, 6)$ in the coordinate plane. The square is first rotated $90^\circ$ counterclockwise about the origin, and then translated such that the final image of vertex $R$ is located at $(1, 5)$ . What are the coordinates of the final image of vertex $P$ after this sequence of transformations?

(-3, 9)

(5, 1)

(-9, 3)

(5, 9)

(-3, 1)

Cevabı ve açıklamayı göster

Cevap: $(5, 1)$

Cevap

(5, 1)

To find the coordinates of the final image of vertex P, we perform the transformations step-by-step. First, a

90^\circ

counterclockwise rotation about the origin maps any point

(x, y)

to

(-y, x)

. Under this rotation,

P(2, 2)

maps to

P'(-2, 2)

and

R(6, 6)

maps to

R'(-6, 6)

. Second, we find the translation vector by comparing the rotated vertex

R'(-6, 6)

to its final position

R''(1, 5)

. The change in the x-coordinate is

1 - (-6) = 7

, and the change in the y-coordinate is

5 - 6 = -1

. This represents a translation of

7

units right and

1

unit down, or the vector

(7, -1)

. Finally, applying this translation to the intermediate point

P'(-2, 2)

gives

P''(-2 + 7, 2 - 1) = (5, 1)

.

Adım Adım Çözüm

1

Apply the

90^\circ

counterclockwise rotation about the origin to the vertices

P(2, 2)

and

R(6, 6)

.

Using the rotation rule

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

, vertex

P(2, 2)

maps to

P'(-2, 2)

and vertex

R(6, 6)

maps to

R'(-6, 6)

.

This establishes the intermediate positions of the vertices before the translation is applied.

2

Determine the translation vector

(h, k)

that maps the intermediate point

R'(-6, 6)

to the final point

R''(1, 5)

.

Solving the equations

-6 + h = 1

and

6 + k = 5

gives

h = 7

and

k = -1

. The translation vector is

(7, -1)

.

The translation shifts all points in the plane by the same distance and direction.

3

Apply the translation vector

(7, -1)

to the intermediate vertex

P'(-2, 2)

to find the final image

P''

.

P'' = (-2 + 7, 2 - 1) = (5, 1)

.

Applying the same translation to the rotated position of vertex P yields its final coordinates.

Anahtar Kavram

Composite transformations in the coordinate plane

Tahmini Süre:2m 0s

Coordinate Geometry

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