Systems of Linear and Non-Linear Equations Alıştırma Soruları — ACT — Sayfa 2

Soru 21Soru →

A system of equations consists of the linear equation $y = x - 2$ and the quadratic equation $y = (x - 3)^2 - 5$ . If $(x_1, y_1)$ and $(x_2, y_2)$ are the two distinct real solutions to this system, what is the value of the product $y_1 y_2$ ?

-10

-4

3

4

6

Cevabı ve açıklamayı göster

Cevap: -4

Cevap

The product of the two

y

-coordinates of the intersection points is

-4

.

Substituting the linear expression for

y

into the quadratic equation yields the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation gives the

x

-coordinates

6

and

1

. Substituting these values back into the linear equation

y = x - 2

yields the corresponding

y

-coordinates

4

and

-1

. The product of these

y

-coordinates is

4 \times (-1) = -4

.

Adım Adım Çözüm

1

Equate the two expressions for

y

to solve for the

x

-coordinates of the intersection points.

x - 2 = (x - 3)^2 - 5

At the points of intersection, the

y

-values of both equations are equal.

2

Expand the squared binomial and simplify the equation into standard quadratic form.

x^2 - 7x + 6 = 0

Expanding

(x - 3)^2

yields

x^2 - 6x + 9

. Substituting this back gives

x - 2 = x^2 - 6x + 4

. Subtracting

x

and adding

2

to both sides yields the standard quadratic equation.

3

Factor the quadratic equation to find the solutions for

x

.

x = 6

or

x = 1

The equation factors as

(x - 6)(x - 1) = 0

, giving the two

x

-values.

4

Substitute each

x

-value back into the linear equation

y = x - 2

to find the corresponding

y

-coordinates.

y_1 = 4

and

y_2 = -1

For

x = 6

,

y = 6 - 2 = 4

. For

x = 1

,

y = 1 - 2 = -1

.

5

Calculate the product of the two

y

-coordinates.

y_1 y_2 = 4 \times (-1) = -4

The question asks for the product of the two

y

-coordinates.

Anahtar Kavram

Solving a system of linear and quadratic equations using substitution and factoring.

Alternatif Yöntem

Alternatively, you can expand the quadratic equation first to

y = x^2 - 6x + 4

and set it equal to the linear equation

y = x - 2

. Solving for

x

yields

x^2 - 7x + 6 = 0

, from which you can find the coordinates and calculate their product.

Tahmini Süre:1m 30s

Soru 22Soru →

A system of equations consists of the linear equation $y = x + 5$ and the quadratic equation $y = (x - 3)^2 + 2$ . If the graphs of these two equations intersect at two distinct points, $(x_1, y_1)$ and $(x_2, y_2)$ , in the standard $(x, y)$ coordinate plane, what is the value of $y_1 + y_2$ ?

3

7

11

15

17

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

17

To find the points of intersection, we set the two equations equal:

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

. Expanding the right side results in

x + 5 = x^2 - 6x + 9 + 2

, which simplifies to

x + 5 = x^2 - 6x + 11

. Subtracting

x + 5

from both sides gives the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation yields

(x - 1)(x - 6) = 0

, giving

x

-coordinates of

1

and

6

. Substituting these values into the linear equation

y = x + 5

gives the

y

-coordinates:

y = 1 + 5 = 6

and

y = 6 + 5 = 11

. The sum of these

y

-coordinates is

6 + 11 = 17

.

Adım Adım Çözüm

1

Expand the quadratic equation to its standard polynomial form.

y = x^2 - 6x + 11

Expanding

(x - 3)^2 + 2

allows for easier substitution and algebraic manipulation.

2

Equate the linear and quadratic expressions for

y

to find the intersection points.

x + 5 = x^2 - 6x + 11

Setting the two equations equal to each other identifies the

x

-coordinates where the graphs intersect.

3

Rearrange the equation into standard quadratic form and solve for

x

by factoring.

x^2 - 7x + 6 = 0 \implies (x - 1)(x - 6) = 0

, which gives

x = 1

and

x = 6

.

Factoring allows us to find the roots of the quadratic equation representing the intersection

x

-coordinates.

4

Substitute the

x

-coordinates back into the linear equation

y = x + 5

to determine the corresponding

y

-coordinates.

For

x = 1

,

y = 6

. For

x = 6

,

y = 11

.

Finding the

y

-coordinates corresponding to the intersection points.

5

Calculate the sum of the two

y

-coordinates.

6 + 11 = 17

The question asks specifically for the value of

y_1 + y_2

.

Anahtar Kavram

Solving systems of linear and quadratic equations by substitution and factoring

Soru 23Soru →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $(x - 2)^2 + y^2 = 20$ , and a line is defined by the equation $y = 2x - 4$ . If the circle and the line intersect at two points, $(x_1, y_1)$ and $(x_2, y_2)$ , what is the value of $x_1 + x_2 + y_1 + y_2$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of

x_1 + x_2 + y_1 + y_2

is 4.

Substituting the equation of the line into the circle's equation gives

(x-2)^2 + (2x-4)^2 = 20

. Since

(2x-4) = 2(x-2)

, we can write this as

(x-2)^2 + 4(x-2)^2 = 20

, which simplifies to

5(x-2)^2 = 20

, and further to

(x-2)^2 = 4

. Solving for

x

yields

x - 2 = 2

or

x - 2 = -2

, so

x = 4

or

x = 0

. Substituting these values into the linear equation gives the points

(4, 4)

and

(0, -4)

. The sum of all coordinates is

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

.

Adım Adım Çözüm

1

Substitute the linear equation into the circle equation.

(x - 2)^2 + (2x - 4)^2 = 20

To find the coordinates of the intersection points by reducing the system to a single variable equation.

2

Simplify the equation using algebraic substitution.

5(x - 2)^2 = 20

Factoring 2 out of the term

(2x-4)

yields

2(x-2)

. Squaring it results in

4(x-2)^2

. Combining this with the first term

(x-2)^2

simplifies the expression.

3

Solve for the two possible x-coordinates.

x = 0

and

x = 4

Dividing both sides by 5 yields

(x-2)^2 = 4

, which means

x-2 = \pm 2

.

4

Substitute the x-coordinates back into the linear equation to find the corresponding y-coordinates.

For

x = 0

,

y = -4

, giving the point

(0, -4)

. For

x = 4

,

y = 4

, giving the point

(4, 4)

.

To determine the coordinates of both intersection points.

5

Calculate the sum of all coordinates:

x_1 + x_2 + y_1 + y_2

.

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

To find the final value requested by the question.

Anahtar Kavram

Solving systems of linear and quadratic (circular) equations by substitution and factoring.

Soru 24Soru →

A software company's weekly revenue, $R$ , in thousands of dollars, is modeled by the function $R(x) = -x^2 + 8x$ , where $x$ represents the price of the software in dollars. The company's weekly operating cost, $C$ , in thousands of dollars, is modeled by the function $C(x) = 2x + 5$ . The company breaks even when its weekly revenue equals its weekly operating cost. What is the sum of the possible software prices, in dollars, at which the company breaks even?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The sum of the possible software prices at which the company breaks even is 6.

To find the software prices at which the company breaks even, we set the revenue function

R(x)

equal to the cost function

C(x)

, which gives

-x^2 + 8x = 2x + 5

. Rearranging this equation into standard quadratic form gives

x^2 - 6x + 5 = 0

. Factoring this expression yields

(x - 5)(x - 1) = 0

, meaning the individual solutions are

x = 5

and

x = 1

. The sum of these possible break-even software prices is

5 + 1 = 6

.

Adım Adım Çözüm

1

Set the revenue function equal to the cost function.

-x^2 + 8x = 2x + 5

The break-even points occur where the revenue and cost are equal.

2

Rearrange the terms to set the equation to zero.

x^2 - 6x + 5 = 0

Moving all terms to one side of the equation puts it in standard quadratic form, allowing us to solve for

x

.

3

Factor the quadratic equation.

(x - 5)(x - 1) = 0

Factoring is a standard method to find the roots of a quadratic equation of the form

ax^2 + bx + c = 0

.

4

Find the individual roots by setting each factor to zero.

x = 5

and

x = 1

If the product of two factors is zero, at least one of the factors must be zero.

5

Calculate the sum of the solutions.

5 + 1 = 6

The question specifically asks for the sum of the possible prices.

Anahtar Kavram

Solving systems of linear and non-linear equations by setting the equations equal to each other and solving the resulting quadratic equation.

Soru 25Soru →

In the standard $(x, y)$ coordinate plane, the circle defined by $x^2 + y^2 = 26$ and the line defined by $y = x - 4$ intersect at two points. What is the distance between these two points of intersection?

2\sqrt{10}

4\sqrt{2}

6

6\sqrt{2}

12

Cevabı ve açıklamayı göster

Cevap: $6\sqrt{2}$

Cevap

6\sqrt{2}

Solving the system of equations by substituting

y = x - 4

into

x^2 + y^2 = 26

yields

2x^2 - 8x - 10 = 0

, which simplifies to

x^2 - 4x - 5 = 0

. Factoring gives

(x-5)(x+1) = 0

, leading to the points

(5, 1)

and

(-1, -5)

. The distance between these points is

\sqrt{6^2 + 6^2} = 6\sqrt{2}

.

Adım Adım Çözüm

1

Substitute the linear equation into the circle equation.

x^2 + (x - 4)^2 = 26

This reduces the system to a single equation in terms of

x

.

2

Expand the squared binomial and simplify the quadratic equation.

2x^2 - 8x - 10 = 0

, which simplifies to

x^2 - 4x - 5 = 0

.

Expanding

(x - 4)^2

yields

x^2 - 8x + 16

. Setting the equation to zero allows us to solve for

x

.

3

Factor the quadratic equation to find the

x

-coordinates.

x = 5

or

x = -1

.

The factored form is

(x - 5)(x + 1) = 0

.

4

Substitute the

x

-values back into the linear equation to find the corresponding

y

-coordinates.

The intersection points are

(5, 1)

and

(-1, -5)

.

For

x = 5

,

y = 5 - 4 = 1

. For

x = -1

,

y = -1 - 4 = -5

.

5

Apply the distance formula to find the straight-line distance between the two points.

6\sqrt{2}

The distance is

\sqrt{(5 - (-1))^2 + (1 - (-5))^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}

.

Anahtar Kavram

Solving systems of linear and non-linear (circular) equations by substitution and finding the distance between intersection points.

Soru 26Soru →

A parabola is defined by the equation $y = x^2 - 4x + 3$ , and a line is defined by the equation $y = -x + 7$ . The parabola and the line intersect at two points in the standard $(x, y)$ coordinate plane. What is the distance between these two points of intersection?

5

\sqrt{34}

5\sqrt{2}

3\sqrt{2}

50

Cevabı ve açıklamayı göster

Cevap: $5\sqrt{2}$

Cevap

5\sqrt{2}

The correct answer is the value representing the straight-line distance between the two intersection points. Setting the equations equal to each other gives

x^2 - 3x - 4 = 0

, which factors to

(x - 4)(x + 1) = 0

. This yields

x = 4

and

x = -1

. Evaluating these in

y = -x + 7

yields the points

(4, 3)

and

(-1, 8)

. The distance between them is

\sqrt{(4 - (-1))^2 + (3 - 8)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}

.

Adım Adım Çözüm

1

Set the equations of the parabola and line equal to each other to find the

x

-coordinates of the intersection points.

x^2 - 4x + 3 = -x + 7

Intersection points must satisfy both equations simultaneously.

2

Rearrange the equation into standard quadratic form and factor it to solve for

x

.

x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0 \implies x = 4

or

x = -1

Factoring the quadratic equation gives the roots which correspond to the

x

-coordinates of the intersection points.

3

Substitute the

x

-values back into the linear equation to determine the corresponding

y

-coordinates.

For

x = 4

,

y = -(4) + 7 = 3

, yielding point

(4, 3)

. For

x = -1

,

y = -(-1) + 7 = 8

, yielding point

(-1, 8)

.

Substituting into the simpler linear equation provides the

y

-coordinates of the intersection points.

4

Apply the distance formula to calculate the distance between

(4, 3)

and

(-1, 8)

.

d = \sqrt{(4 - (-1))^2 + (3 - 8)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}

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

Anahtar Kavram

Systems of Linear and Non-Linear Equations

Soru 27Soru →

In the standard $(x, y)$ coordinate plane, the graph of the parabola $y = x^2 - 5x + 2$ intersects the line $2x - y = 4$ at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the value of $x_1 x_2 + y_1 y_2$ ?

50

22

-10

6

102

Cevabı ve açıklamayı göster

Cevap: -10

Cevap

The value of

x_1 x_2 + y_1 y_2

is

-10

.

The correct answer is

-10

. To find the intersection points, substitute the linear equation

y = 2x - 4

into the quadratic equation

y = x^2 - 5x + 2

, which simplifies to

x^2 - 7x + 6 = 0

. Factoring gives

(x-1)(x-6) = 0

, yielding

x

-coordinates

x_1 = 1

and

x_2 = 6

. Substituting these back into the linear equation gives the corresponding

y

-coordinates

y_1 = -2

and

y_2 = 8

. Evaluating the requested expression gives

(1)(6) + (-2)(8) = 6 - 16 = -10

.

Adım Adım Çözüm

1

Isolate

y

in the linear equation.

y = 2x - 4

This allows for substitution into the quadratic equation.

2

Substitute

y = 2x - 4

into the quadratic equation

y = x^2 - 5x + 2

.

2x - 4 = x^2 - 5x + 2

Setting the two equations equal to each other helps find the x-coordinates of the intersection points.

3

Rearrange the equation into standard quadratic form.

x^2 - 7x + 6 = 0

This sets up the equation for factoring.

4

Factor the quadratic equation.

(x-1)(x-6) = 0

Factoring reveals the solutions for the x-coordinates.

5

Solve for the two x-coordinates.

x_1 = 1

and

x_2 = 6

These are the x-coordinates of the two intersection points.

6

Find the corresponding y-coordinates by substituting the x-values back into the linear equation.

For

x_1 = 1

:

y_1 = 2(1) - 4 = -2

. For

x_2 = 6

:

y_2 = 2(6) - 4 = 8

.

This gives the full coordinate pairs:

(1, -2)

and

(6, 8)

.

7

Calculate the value of the expression

x_1 x_2 + y_1 y_2

.

(1)(6) + (-2)(8) = 6 - 16 = -10

This provides the final required value.

Anahtar Kavram

Solving systems of linear and quadratic equations by substitution and evaluating expressions of their coordinate solutions.

Tahmini Süre:1m 30s

Soru 28Soru →

A line is defined by the equation $y = 2x + k$ , where $k$ is a constant. A parabola is defined by the equation $y = x^2 - 4x + 14$ . If the line and the parabola intersect at exactly one point in the standard $(x, y)$ coordinate plane, what is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of the constant

k

must be 5.

To find the intersection of the line and the parabola, set their equations equal to each other:

x^2 - 4x + 14 = 2x + k

. Rearranging this into standard quadratic form gives

x^2 - 6x + (14 - k) = 0

. For the system to have exactly one solution, the discriminant of this quadratic equation must be zero. The discriminant is

b^2 - 4ac = (-6)^2 - 4(1)(14 - k) = 36 - 56 + 4k = 4k - 20

. Setting

4k - 20 = 0

yields

k = 5

.

Adım Adım Çözüm

1

Equate the linear and quadratic expressions to find their intersection.

x^2 - 4x + 14 = 2x + k

Setting the two equations equal to each other allows us to find the

x

-coordinates of any intersection points.

2

Rearrange the terms to write the equation in standard quadratic form,

ax^2 + bx + c = 0

.

x^2 - 6x + (14 - k) = 0

Subtracting

2x

and

k

from both sides simplifies the equation into a quadratic form where

a = 1

,

b = -6

, and

c = 14 - k

.

3

Apply the condition for exactly one intersection point by setting the discriminant to zero.

(-6)^2 - 4(1)(14 - k) = 0

A quadratic equation has exactly one real root if and only if its discriminant,

b^2 - 4ac

, is equal to zero.

4

Solve the linear equation for

k

.

k = 5

Expanding the equation yields

36 - 56 + 4k = 0

, which simplifies to

4k - 20 = 0

, leading to

k = 5

.

Anahtar Kavram

Determining the condition for a linear equation to be tangent to a quadratic equation by setting the discriminant of their intersection equation to zero.

Soru 29Soru →

In the standard $(x, y)$ coordinate plane, a circular running track is modeled by the equation $(x - 2)^2 + (y - 3)^2 = 25$ . A straight pathway, modeled by the line $y = 2x - 1$ , cuts through the track. What is the distance, in coordinate units, between the two points where the pathway intersects the track?

5

2\sqrt{5}

10

20

50

Cevabı ve açıklamayı göster

Cevap: $10$

Cevap

The distance between the two intersection points is

10

.

The circle

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

has a center of

(2, 3)

and a radius of

r = 5

. Since the line

y = 2x - 1

passes through

(2, 3)

, the line contains a diameter of the circle. The distance between the two intersection points is therefore the diameter of the circle, which is

2r = 10

.

Adım Adım Çözüm

1

Identify the center and radius of the circle from its equation

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

.

The center is

(2, 3)

and the radius is

r = \sqrt{25} = 5

.

To understand the geometry of the circle and find its radius.

2

Determine if the line

y = 2x - 1

passes through the center of the circle

(2, 3)

by substituting the coordinates into the equation.

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

, which is true.

If the line passes through the center, the distance between the intersection points is simply the diameter of the circle.

3

Calculate the diameter of the circle.

\text{Diameter} = 2r = 2(5) = 10

.

The distance between two opposite points on a circle passing through the center is equal to the diameter.

Anahtar Kavram

Systems of Linear and Non-Linear Equations

Soru 30Soru →

A circle in the standard $(x, y)$ coordinate plane is described by the equation $x^2 + y^2 = 25$ . A line is described by the equation $3x - 4y = c$ , where $c$ is a positive constant. If the system of these two equations has exactly one real solution for $(x, y)$ , what is the value of $c$ ?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The value of

c

is

25

.

The correct answer is

25

. The equation

x^2 + y^2 = 25

represents a circle centered at

(0, 0)

with a radius of

5

. For the system of equations to have exactly one real solution, the line

3x - 4y = c

must be tangent to the circle. The perpendicular distance from the center

(0,0)

to the line

3x - 4y - c = 0

is given by

\frac{|3(0) - 4(0) - c|}{\sqrt{3^2 + (-4)^2}} = \frac{|c|}{5}

. Setting this distance equal to the radius of the circle yields

\frac{|c|}{5} = 5

, which gives

|c| = 25

. Since

c

is specified as a positive constant,

c

must be

25

.

Adım Adım Çözüm

1

Find the center and radius of the circle.

Center is

(0, 0)

and radius is

r = 5

.

The circle equation

x^2 + y^2 = 25

is in the standard form

x^2 + y^2 = r^2

centered at the origin with radius

r = \sqrt{25} = 5

.

2

Set up the condition for tangency (exactly one real solution).

The perpendicular distance from the center

(0,0)

to the line

3x - 4y - c = 0

must equal the radius

5

.

A line intersects a circle at exactly one point if and only if the line is tangent to the circle.

3

Apply the point-to-line distance formula.

Distance

d = \frac{|3(0) - 4(0) - c|}{\sqrt{3^2 + (-4)^2}} = \frac{|c|}{5}

.

The formula for the distance from

(x_0, y_0)

to the line

Ax + By + C = 0

is

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

.

4

Solve for the positive constant

c

.

c = 25

Setting the distance

\frac{|c|}{5}

equal to the radius

5

gives

|c| = 25

. Since

c

is a positive constant,

c = 25

.

Anahtar Kavram

Determining conditions for tangency in a system of linear and circular equations.

Alternatif Yöntem

Instead of using the geometric distance formula, the system can be solved algebraically by substitution. Express

y

in terms of

x

from the linear equation:

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

. Substitute this expression into the circle's equation:

x^2 + \left(\frac{3x - c}{4}\right)^2 = 25

. Expand the terms and multiply by

16

to clear the denominator:

16x^2 + 9x^2 - 6cx + c^2 = 400

, which simplifies to the quadratic equation

25x^2 - 6cx + (c^2 - 400) = 0

. For the system to have exactly one solution, this quadratic equation must have a discriminant equal to zero. Calculate the discriminant:

D = (-6c)^2 - 4(25)(c^2 - 400) = 36c^2 - 100c^2 + 40000 = -64c^2 + 40000 = 0

. Solving for

c

yields

64c^2 = 40000 \implies c^2 = 625

. Since

c

is a positive constant,

c = 25

.

Tahmini Süre:1m 30s

Soru 31Soru →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 = 25$ and a line is defined by the equation $3x + 4y = 15$ . The line intersects the circle at two points, $A$ and $B$ . What is the distance between point $A$ and point $B$ ?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The distance between the two intersection points is 8.

The correct answer is 8. We can find the intersection points by substituting

y = \frac{15 - 3x}{4}

into the circle's equation

x^2 + y^2 = 25

, yielding the quadratic equation

5x^2 - 18x - 35 = 0

. Solving this gives

x = 5

and

x = -1.4

, with corresponding

y

-coordinates

y = 0

and

y = 4.8

. The distance between

(5, 0)

and

(-1.4, 4.8)

is

\sqrt{(-1.4 - 5)^2 + (4.8 - 0)^2} = \sqrt{6.4^2 + 4.8^2} = \sqrt{64} = 8

. Alternatively, we can use geometry: the distance from the center of the circle

(0,0)

to the line

3x + 4y - 15 = 0

is

d = \frac{|3(0) + 4(0) - 15|}{\sqrt{3^2 + 4^2}} = 3

. Since the radius of the circle is

r = 5

, the right triangle formed by the radius, the perpendicular segment, and half the chord has a half-chord length of

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

. Thus, the total chord length is

2 \times 4 = 8

.

Adım Adım Çözüm

1

Express the linear equation in terms of one variable

y = \frac{15 - 3x}{4}

This allows for substitution into the equation of the circle.

2

Substitute the expression into the circle's equation and simplify

x^2 + \left(\frac{15 - 3x}{4}\right)^2 = 25 \Rightarrow 16x^2 + (225 - 90x + 9x^2) = 400 \Rightarrow 25x^2 - 90x - 175 = 0 \Rightarrow 5x^2 - 18x - 35 = 0

To create a single quadratic equation in terms of

x

representing the intersection points.

3

Solve the quadratic equation for

x

(5x + 7)(x - 5) = 0 \Rightarrow x = 5

or

x = -1.4

To find the

x

-coordinates of the intersection points.

4

Calculate the corresponding

y

-coordinates

For

x = 5

,

y = 0

, giving point

A(5, 0)

. For

x = -1.4

,

y = 4.8

, giving point

B(-1.4, 4.8)

.

To determine the exact coordinates of both intersection points.

5

Apply the distance formula to find the length of the segment

AB

d = \sqrt{(-1.4 - 5)^2 + (4.8 - 0)^2} = \sqrt{(-6.4)^2 + 4.8^2} = \sqrt{40.96 + 23.04} = \sqrt{64} = 8

To compute the final distance between the two intersection points.

Anahtar Kavram

Solving systems of linear and quadratic equations to determine intersection points and calculating the distance between coordinates.

Soru 32Soru →

A circle in the standard $(x, y)$ coordinate plane is described by the equation $x^2 + y^2 = 68$ . A line is described by the equation $y = x - 6$ . The line intersects the circle at two points. What is the distance between these two points of intersection?

8\sqrt{2}

2\sqrt{17}

10\sqrt{2}

20

6\sqrt{2}

Cevabı ve açıklamayı göster

Cevap: $10\sqrt{2}$

Cevap

The distance between the two points of intersection is

10\sqrt{2}

.

To find the points of intersection, substitute

y = x - 6

into

x^2 + y^2 = 68

, which yields

x^2 + (x - 6)^2 = 68

. Expanding the binomial correctly results in

x^2 + x^2 - 12x + 36 = 68

. Combining like terms and writing the equation in standard form gives

2x^2 - 12x - 32 = 0

. Dividing the entire equation by 2 gives

x^2 - 6x - 16 = 0

. Factoring this quadratic equation yields

(x - 8)(x + 2) = 0

, giving solutions

x = 8

and

x = -2

. Substituting these values back into the linear equation gives the points

(8, 2)

and

(-2, -8)

. Finally, using the distance formula, the distance between the two points is

\sqrt{(8 - (-2))^2 + (2 - (-8))^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}

.

Adım Adım Çözüm

1

Substitute the linear equation into the circle equation.

x^2 + (x - 6)^2 = 68

To eliminate

y

and solve for the

x

-coordinates of the intersection points.

2

Expand the binomial and collect like terms.

2x^2 - 12x - 32 = 0

Expanding

(x - 6)^2

yields

x^2 - 12x + 36

, and combining it with

x^2

and subtracting

68

from both sides puts the equation in quadratic form.

3

Divide the quadratic equation by 2 and solve by factoring.

(x - 8)(x + 2) = 0

, so

x = 8

or

x = -2

Simplifying the equation to

x^2 - 6x - 16 = 0

makes it easy to find the roots of the quadratic equation.

4

Find the corresponding

y

-coordinates by substituting the

x

-values back into the linear equation.

For

x = 8

,

y = 2

, yielding point

(8, 2)

. For

x = -2

,

y = -8

, yielding point

(-2, -8)

.

To determine the full coordinates of the two intersection points.

5

Apply the distance formula to find the distance between

(8, 2)

and

(-2, -8)

.

d = \sqrt{(8 - (-2))^2 + (2 - (-8))^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}

To calculate the straight-line distance between the two coordinate points.

Anahtar Kavram

Solving systems of linear and non-linear equations by substitution and finding the distance between intersection points.

Tahmini Süre:1m 30s

Soru 33Soru →

In the standard $(x, y)$ coordinate system, a line given by the equation $y = 2x + 4$ intersects a parabola given by the equation $y = x^2 - 2x - 1$ at exactly two points. What is the sum of the $y$ -coordinates of these two points of intersection?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

16

The system of equations is solved by setting the equations equal to each other, yielding the quadratic equation

x^2 - 4x - 5 = 0

. Solving for

x

gives

x = 5

and

x = -1

. Substituting these values into the linear equation gives the y-coordinates

14

and

2

. The sum of these y-coordinates is

14 + 2 = 16

.

Adım Adım Çözüm

1

Equate the linear and quadratic equations to find the x-values of the intersection points.

x^2 - 2x - 1 = 2x + 4

At the points of intersection, the y-values of both equations must be equal.

2

Rearrange the equation to standard quadratic form.

x^2 - 4x - 5 = 0

Grouping all terms on one side allows the quadratic equation to be solved.

3

Factor the quadratic equation to find the x-coordinates.

(x - 5)(x + 1) = 0

, so

x = 5

or

x = -1

The numbers that multiply to

-5

and add up to

-4

are

-5

and

1

.

4

Substitute the x-coordinates into the linear equation to find the y-coordinates.

For

x = 5

:

y = 2(5) + 4 = 14

. For

x = -1

:

y = 2(-1) + 4 = 2

.

Evaluating the linear equation is simpler than evaluating the quadratic equation.

5

Find the sum of the y-coordinates.

14 + 2 = 16

The question asks for the sum of the y-coordinates of the two points of intersection.

Anahtar Kavram

Solving a system of linear and quadratic equations by substitution.

Alternatif Yöntem

Instead of solving for the individual intersection points, Vieta's formulas can be applied. The x-coordinates satisfy

x^2 - 4x - 5 = 0

, so their sum is

x_1 + x_2 = 4

. Since the points lie on the line

y = 2x + 4

, the sum of the y-coordinates is

y_1 + y_2 = (2x_1 + 4) + (2x_2 + 4) = 2(x_1 + x_2) + 8 = 2(4) + 8 = 16

.

Tahmini Süre:1m 30s

Soru 34Soru →

The path of a particle in the standard $(x, y)$ coordinate plane is described by the linear equation $3x - y = 2$ , and the path of another particle is described by the quadratic equation $y = x^2 - x - 7$ . If the two paths intersect at two locations, what is the sum of the $y$ -coordinates of these intersection points?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The sum of the

y

-coordinates of the intersection points is 8.

The correct answer is 8. Solving the system by setting

x^2 - x - 7 = 3x - 2

results in the quadratic equation

x^2 - 4x - 5 = 0

. Factoring gives

(x-5)(x+1) = 0

, which yields intersection

x

-coordinates of

x = 5

and

x = -1

. Substituting these back into the linear equation gives

y

-coordinates of

13

and

-5

. Adding these values together yields

13 + (-5) = 8

.

Adım Adım Çözüm

1

Express the linear equation in terms of

y

.

y = 3x - 2

This allows for direct substitution into the quadratic equation.

2

Equate the linear and quadratic expressions to solve for the

x

-coordinates of the intersection points.

x^2 - x - 7 = 3x - 2

Intersection points share the same coordinates for both equations.

3

Set the quadratic equation to zero.

x^2 - 4x - 5 = 0

This puts the equation in standard form so it can be solved by factoring.

4

Factor the quadratic equation.

(x - 5)(x + 1) = 0

, so

x = 5

or

x = -1

Factoring determines the

x

-coordinates of the intersection points.

5

Substitute the

x

-values into the linear equation to determine the

y

-coordinates.

For

x = 5

,

y = 13

. For

x = -1

,

y = -5

.

Finding the

y

-coordinates is necessary to compute their sum.

6

Add the

y

-coordinates together.

13 + (-5) = 8

The question asks for the sum of the

y

-coordinates of the intersection points.

Anahtar Kavram

Solving systems of linear and quadratic equations by substitution and factoring

Alternatif Yöntem

We can use Vieta's formulas to find the sum of the

y

-coordinates without calculating each individual coordinate. The sum of the

y

-coordinates is

y_1 + y_2 = (3x_1 - 2) + (3x_2 - 2) = 3(x_1 + x_2) - 4

. Since

x_1

and

x_2

are the roots of

x^2 - 4x - 5 = 0

, Vieta's formulas state that the sum of the roots is

x_1 + x_2 = -\frac{-4}{1} = 4

. Substituting this value into our sum expression yields

3(4) - 4 = 12 - 4 = 8

.

Tahmini Süre:1m 30s

Soru 35Soru →

A parabolic arch is modeled by the equation $y = (x - 3)^2 - 5$ in the standard $(x, y)$ coordinate plane. A straight pathway is modeled by a line where the $y$ -coordinate of any point is 2 less than its $x$ -coordinate. If the pathway intersects the arch at points $A$ and $B$ , what is the area, in square units, of the triangle with vertices at $A$ , $B$ , and the origin $(0, 0)$ ?

3

5

7

8

10

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The area of the triangle with vertices at the intersection points and the origin is 5 square units.

To find the area of the triangle, we first solve the system of equations. Substituting the pathway's equation y = x - 2 into the parabola's equation y = (x - 3)^2 - 5 yields x^2 - 7x + 6 = 0, which factors to (x - 6)(x - 1) = 0. This gives x = 6 and x = 1. The corresponding y-coordinates are y = 4 and y = -1, representing the intersection points (6, 4) and (1, -1). The area of the triangle with these vertices and the origin (0, 0) is calculated as 0.5 * |6(-1) - 4(1)| = 5.

Adım Adım Çözüm

1

Set up the system of equations by substituting the linear equation into the quadratic equation.

(x - 3)^2 - 5 = x - 2

The pathway is described as having a y-coordinate that is 2 less than the x-coordinate, which translates to the linear equation y = x - 2. Substituting this into the parabola's equation allows us to find the intersection points.

2

Expand the quadratic term and simplify the equation into standard quadratic form.

x^2 - 7x + 6 = 0

Expanding

(x - 3)^2

gives

x^2 - 6x + 9

. Subtracting

x

and adding

2

to both sides results in standard quadratic form.

3

Factor the quadratic equation to solve for the x-coordinates.

(x - 6)(x - 1) = 0

, so

x = 6

or

x = 1

Factoring allows us to find the x-values that satisfy the intersection condition.

4

Substitute the x-values back into the linear equation to find the corresponding y-coordinates.

For

x = 6

,

y = 4

giving point

(6, 4)

. For

x = 1

,

y = -1

giving point

(1, -1)

.

The intersection points must satisfy both equations in the system.

5

Calculate the area of the triangle with vertices

(0, 0)

,

(6, 4)

, and

(1, -1)

using the coordinate area formula.

Area =

\frac{1}{2} |6(-1) - 4(1)| = \frac{1}{2} |-10| = 5

square units.

The area of a triangle with one vertex at the origin and others at

(x_1, y_1)

and

(x_2, y_2)

is

\frac{1}{2} |x_1 y_2 - y_1 x_2|

.

Anahtar Kavram

Solving systems of linear and quadratic equations and finding the area of a triangle in the coordinate plane.

Soru 36Soru →

A circular search-and-rescue radar zone centered at a local station is modeled by the equation $x^2 + y^2 = 25$ in the standard $(x, y)$ coordinate plane, where coordinates are measured in miles. A rescue helicopter flies along a straight path modeled by the line $y = 2x - 5$ . What is the distance, in miles, the helicopter travels through the radar zone?

2\sqrt{5}

4\sqrt{5}

8

10

10\sqrt{2}

Cevabı ve açıklamayı göster

Cevap: $4\sqrt{5}$

Cevap

The distance the helicopter travels through the radar zone is

4\sqrt{5}

miles.

To find the distance the helicopter travels through the radar zone, we must determine the distance between the two points of intersection of the circular boundary

x^2 + y^2 = 25

and the line

y = 2x - 5

. Substituting the expression for

y

into the circular equation yields

x^2 + (2x-5)^2 = 25

. Expanding the binomial correctly gives

x^2 + 4x^2 - 20x + 25 = 25

, which simplifies to

5x^2 - 20x = 0

. Factoring this equation as

5x(x-4) = 0

gives

x = 0

and

x = 4

. Substituting these values back into the linear equation yields the points of intersection

(0, -5)

and

(4, 3)

. The distance between these two points is

\sqrt{(4-0)^2 + (3 - (-5))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}

miles.

Adım Adım Çözüm

1

Substitute the linear equation

y = 2x - 5

into the circular equation

x^2 + y^2 = 25

.

x^2 + (2x - 5)^2 = 25

To find the coordinates of the intersection points where the helicopter's path meets the boundary of the radar zone.

2

Expand the binomial

(2x - 5)^2

and simplify the quadratic equation.

x^2 + 4x^2 - 20x + 25 = 25 \implies 5x^2 - 20x = 0

To collect like terms and put the equation in a solvable quadratic form.

3

Factor the quadratic equation

5x^2 - 20x = 0

to solve for

x

.

5x(x - 4) = 0 \implies x = 0 \text{ or } x = 4

To determine the

x

-coordinates of the two intersection points.

4

Determine the corresponding

y

-coordinates by substituting the

x

-values into the linear equation

y = 2x - 5

.

For

x = 0

:

y = 2(0) - 5 = -5 \implies (0, -5)

. For

x = 4

:

y = 2(4) - 5 = 3 \implies (4, 3)

.

To obtain the exact coordinate pairs for the entry and exit points.

5

Use the distance formula to calculate the distance between the two points

(0, -5)

and

(4, 3)

.

d = \sqrt{(4 - 0)^2 + (3 - (-5))^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}

To compute the straight-line distance traveled by the helicopter through the radar zone.

Anahtar Kavram

Solving systems of linear and circular equations by substitution and finding the distance between their intersection points.

Soru 37Soru →

A circle and a line are graphed in the standard $(x, y)$ coordinate plane. The equations of the circle and the line are given by:

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

y = x + 1

The line intersects the circle at two points, $P$ and $Q$ . What is the distance between $P$ and $Q$ ?

2\sqrt{10}

3\sqrt{2}

\sqrt{22}

7\sqrt{2}

\sqrt{5}

Cevabı ve açıklamayı göster

Cevap: $3\sqrt{2}$

Cevap

The distance between the intersection points is

3\sqrt{2}

.

To find the points of intersection, substitute the linear equation into the circle's equation. This results in the quadratic equation

x^2 - 5x + 4 = 0

, which yields the solutions

x = 1

and

x = 4

. Substituting these values into the linear equation gives the points

(1, 2)

and

(4, 5)

. The distance between these points is computed using the distance formula, which gives

3\sqrt{2}

.

Adım Adım Çözüm

1

Substitute the equation of the line into the equation of the circle.

(x-3)^2 + ((x+1)-3)^2 = 5 \implies (x-3)^2 + (x-2)^2 = 5

This reduces the system of two equations with two variables to a single quadratic equation in terms of

x

.

2

Expand the squared binomials and simplify the quadratic equation.

(x^2 - 6x + 9) + (x^2 - 4x + 4) = 5 \implies 2x^2 - 10x + 13 = 5 \implies 2x^2 - 10x + 8 = 0 \implies x^2 - 5x + 4 = 0

Expanding the terms allows us to combine like terms and set the quadratic equation to zero.

3

Factor the quadratic equation to find the

x

-coordinates of the intersection points.

(x-1)(x-4) = 0 \implies x = 1 \text{ or } x = 4

Factoring is the most direct method to solve the simplified quadratic equation.

4

Find the corresponding

y

-coordinates by substituting the

x

-values back into the linear equation

y = x + 1

.

\text{For } x = 1: y = 1 + 1 = 2 \implies P(1, 2)

\text{For } x = 4: y = 4 + 1 = 5 \implies Q(4, 5)

This determines the coordinates of the two intersection points.

5

Use the distance formula to find the distance between the two points

P(1, 2)

and

Q(4, 5)

.

d = \sqrt{(4-1)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}

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

Anahtar Kavram

Solving systems of linear and quadratic equations by substitution and finding the distance between intersection points.

Alternatif Yöntem

Find the distance geometrically: The center of the circle is

(3, 3)

and the radius is

r = \sqrt{5}

. The distance

d

from the center to the line

x - y + 1 = 0

is

d = \frac{|3 - 3 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}

. Using a right triangle formed by the radius, the distance from the center, and half of the chord length

h

, we have

h = \sqrt{r^2 - d^2} = \sqrt{5 - \frac{1}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}

. The total distance between the intersection points is the full chord length,

2h = 2 \times \frac{3}{\sqrt{2}} = 3\sqrt{2}

.

Tahmini Süre:3m 0s

Soru 38Soru →

A small drone's path in a vertical plane is modeled by the equation $y = 3x^2 - 4x + 2$ , where $x$ is the horizontal distance in meters and $y$ is the height in meters. A laser beam travels along a straight line in the same plane such that the sum of twice its horizontal distance and its height is a constant $c$ , where both are in meters. The laser beam intersects the drone's path at two distinct points. If the distance between these two intersection points is $\frac{5\sqrt{2}}{3}$ meters, what is the value of $2c$ ?

6

10

5

-\frac{1}{3}

\frac{5}{3}

Cevabı ve açıklamayı göster

Cevap: $5$

Cevap

The correct value of

2c

is

5

.

The correct value of

2c

is

5

. Substituting the linear equation

y = c - 2x

into the quadratic equation

y = 3x^2 - 4x + 2

yields

3x^2 - 2x + (2-c) = 0

. The distance between the intersection points is

d = |x_2 - x_1|\sqrt{5} = \frac{5\sqrt{2}}{3}

, which simplifies to

(x_2 - x_1)^2 = \frac{10}{9}

. Using the identity

(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2

and Vieta's formulas, we find the equation

\frac{4}{9} - \frac{8-4c}{3} = \frac{10}{9}

. Solving this equation yields

c = \frac{5}{2}

, and thus

2c = 5

.

Adım Adım Çözüm

1

Set up the system of equations.

The drone's path is

y = 3x^2 - 4x + 2

and the laser's path is

2x + y = c \implies y = -2x + c

.

This represents the mathematical formulation of both paths in the vertical plane.

2

Equate the equations to find the x-coordinates of the intersection points.

3x^2 - 4x + 2 = -2x + c \implies 3x^2 - 2x + (2-c) = 0

.

The intersection points satisfy both equations, so we can solve for

x

by substitution.

3

Express the distance between the intersection points

P(x_1, y_1)

and

Q(x_2, y_2)

using the slope.

d = |x_2 - x_1|\sqrt{5}

.

Since the points lie on the line with slope

-2

, we have

y_2 - y_1 = -2(x_2 - x_1)

. The distance formula becomes

d = \sqrt{(x_2 - x_1)^2 + (-2(x_2 - x_1))^2} = \sqrt{5(x_2 - x_1)^2} = |x_2 - x_1|\sqrt{5}

.

4

Equate the distance expression to the given distance to find

(x_2 - x_1)^2

.

(x_2 - x_1)^2 = \frac{10}{9}

.

We are given

d = \frac{5\sqrt{2}}{3}

. Setting

|x_2 - x_1|\sqrt{5} = \frac{5\sqrt{2}}{3}

and squaring both sides gives

5(x_2 - x_1)^2 = \frac{50}{9}

, which simplifies to

(x_2 - x_1)^2 = \frac{10}{9}

.

5

Apply Vieta's formulas and the algebraic identity for

(x_2 - x_1)^2

.

\frac{10}{9} = \frac{4}{9} - \frac{8-4c}{3}

.

For the quadratic equation

3x^2 - 2x + (2-c) = 0

, we have

x_1 + x_2 = \frac{2}{3}

and

x_1 x_2 = \frac{2-c}{3}

. We use the identity

(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2

.

6

Solve for

c

and find

2c

.

c = \frac{5}{2} \implies 2c = 5

.

Multiplying the equation by

9

gives

10 = 4 - 3(8-4c) \implies 10 = -20 + 12c \implies 12c = 30 \implies c = \frac{5}{2}

. Therefore,

2c = 5

.

Anahtar Kavram

Solving systems of linear and non-linear equations using substitution, coordinate geometry distance formula, and quadratic root relationships.

Tahmini Süre:3m 0s

Soru 39Soru →

The circle $x^2 + y^2 = 25$ and the line $y = 2x - 5$ intersect at two points. What is the sum of the $y$ -coordinates of these two intersection points?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The sum of the

y

-coordinates of the intersection points is

-2

.

Substituting

y = 2x - 5

into the circular equation

x^2 + y^2 = 25

yields the quadratic equation

5x^2 - 20x = 0

. Factoring this equation gives

x = 0

and

x = 4

. Evaluating the linear equation at these values gives the

y

-coordinates

-5

and

3

. The sum of these coordinates is

-5 + 3 = -2

.

Adım Adım Çözüm

1

Substitute

y = 2x - 5

into the circle equation

x^2 + y^2 = 25

.

x^2 + (2x - 5)^2 = 25

To find the points of intersection, we solve the system of equations by substitution.

2

Expand and simplify the resulting equation.

5x^2 - 20x = 0

Expanding

(2x - 5)^2

gives

4x^2 - 20x + 25

. Combining like terms and subtracting 25 from both sides simplifies the equation.

3

Factor the quadratic equation to solve for

x

.

x = 0

or

x = 4

Factoring out

5x

gives

5x(x - 4) = 0

, which yields the roots

x = 0

and

x = 4

.

4

Find the corresponding

y

-coordinates by substituting the

x

-values back into

y = 2x - 5

.

The intersection points are

(0, -5)

and

(4, 3)

.

For

x = 0

,

y = 2(0) - 5 = -5

. For

x = 4

,

y = 2(4) - 5 = 3

.

5

Calculate the sum of the

y

-coordinates.

-2

Adding the

y

-coordinates

-5

and

3

gives

-5 + 3 = -2

.

Anahtar Kavram

Systems of Linear and Non-Linear Equations

Soru 40Soru →

Consider the system of equations consisting of the quadratic function $f(x) = (x - 2)^2 - 3$ and the linear function $g(x) = 3x - 9$ . If the graphs of these functions intersect at two distinct points, $(x_1, y_1)$ and $(x_2, y_2)$ , what is the value of $x_1 y_2 + x_2 y_1$ ?

-15

-8

-3

15

24

Cevabı ve açıklamayı göster

Cevap: -3

Cevap

The correct answer is

-3

. The intersection points of the two functions are

(2, -3)

and

(5, 6)

, and evaluating the expression

x_1 y_2 + x_2 y_1

gives

-3

.

To find the points where the graphs of the functions intersect, we set their expressions equal to each other:

(x - 2)^2 - 3 = 3x - 9

. Expanding the squared term gives

x^2 - 4x + 4 - 3 = 3x - 9

, which simplifies to

x^2 - 4x + 1 = 3x - 9

. Moving all terms to the left side yields

x^2 - 7x + 10 = 0

. Factoring this quadratic equation gives

(x - 2)(x - 5) = 0

, so the

x

-coordinates of the intersection points are

2

and

5

. Substituting these back into the linear equation gives the corresponding

y

-coordinates:

y = 3(2) - 9 = -3

and

y = 3(5) - 9 = 6

, resulting in the intersection points

(2, -3)

and

(5, 6)

. Finally, evaluating the requested expression gives

(2)(6) + (5)(-3) = 12 - 15 = -3

.

Adım Adım Çözüm

1

Set the quadratic and linear functions equal to find the

x

-coordinates of their intersection points.

(x - 2)^2 - 3 = 3x - 9

At the points of intersection, the values of

f(x)

and

g(x)

must be equal.

2

Expand the binomial squared term

(x - 2)^2

.

x^2 - 4x + 4 - 3 = 3x - 9

Expanding the binomial is necessary to combine like terms and write the equation in standard quadratic form.

3

Move all terms to one side to set the equation to zero.

x^2 - 7x + 10 = 0

A quadratic equation must be set to zero before factoring or applying the quadratic formula.

4

Factor the quadratic equation.

(x - 2)(x - 5) = 0 \implies x_1 = 2 \text{ and } x_2 = 5

Factoring determines the

x

-coordinates of the intersection points.

5

Substitute the

x

-values into the linear equation

g(x) = 3x - 9

to find the corresponding

y

-coordinates.

y_1 = 3(2) - 9 = -3

y_2 = 3(5) - 9 = 6

This yields the two intersection points:

(2, -3)

and

(5, 6)

.

6

Calculate the value of the expression

x_1 y_2 + x_2 y_1

.

(2)(6) + (5)(-3) = 12 - 15 = -3

This evaluates the requested secondary value using the intersection coordinates.

Anahtar Kavram

Solving systems of linear and non-linear equations by substitution and algebraic manipulation.

Systems of Linear and Non-Linear Equations

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