Linear Functions and Graphs

71 soru

Soru 1Soru

Two lines, L1L_1 and L2L_2, are graphed in the xyxy-plane. Line L1L_1 passes through the points (2,11)(2, 11) and (6,23)(6, 23). If line L2L_2 is perpendicular to line L1L_1 and contains the point (3,10)(3, 10), what is the xx-coordinate of the xx-intercept of line L2L_2?

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Cevap: 33

Cevap

33
To find the xx-coordinate of the xx-intercept of line L2L_2, first determine the slope of line L1L_1 from the given points (2,11)(2, 11) and (6,23)(6, 23) using the formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}, which results in m1=231162=3m_1 = \frac{23 - 11}{6 - 2} = 3. Since line L2L_2 is perpendicular to line L1L_1, its slope must be the negative reciprocal of 3, which is 13-\frac{1}{3}. Using the point-slope form with the point (3,10)(3, 10), the equation of line L2L_2 is y10=13(x3)y - 10 = -\frac{1}{3}(x - 3), which simplifies to y=13x+11y = -\frac{1}{3}x + 11. Setting y=0y = 0 to find the xx-intercept yields 0=13x+110 = -\frac{1}{3}x + 11, which simplifies to x=33x = 33.

Adım Adım Çözüm

1
Calculate the slope of line L1L_1 using the points (2,11)(2, 11) and (6,23)(6, 23) with the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
m1=231162=124=3m_1 = \frac{23 - 11}{6 - 2} = \frac{12}{4} = 3
The slope of a line represents its rate of change and is required to find the relationship with perpendicular lines.
2
Find the slope of line L2L_2 which is perpendicular to L1L_1.
m2=13m_2 = -\frac{1}{3}
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Write the linear equation for line L2L_2 using the point-slope formula yy1=m(xx1)y - y_1 = m(x - x_1) with the point (3,10)(3, 10) and slope m2=13m_2 = -\frac{1}{3}.
y10=13(x3)    y=13x+11y - 10 = -\frac{1}{3}(x - 3) \implies y = -\frac{1}{3}x + 11
Defining the equation of the line allows us to find the coordinates of any of its intercepts.
4
Set y=0y = 0 in the equation for line L2L_2 to find the xx-coordinate of the xx-intercept.
0=13x+11    13x=11    x=330 = -\frac{1}{3}x + 11 \implies \frac{1}{3}x = 11 \implies x = 33
The xx-intercept is the point where the line crosses the xx-axis, which mathematically corresponds to y=0y = 0.

Anahtar Kavram

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other. The xx-intercept of a linear graph represents the value of xx when y=0y = 0.
Soru 2Soru

A line representing the linear function ff is graphed in the xyxy-plane. The line intersects the yy-axis at (0,3)(0, 3) and also passes through the point (2,11)(2, 11). What is the slope of this line?

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Cevap: 4

Cevap

The slope of the line is 4.
The slope mm of a line passing through two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by the formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the given points (0,3)(0, 3) and (2,11)(2, 11) into the formula gives m=11320=82=4m = \frac{11 - 3}{2 - 0} = \frac{8}{2} = 4.

Adım Adım Çözüm

1
Identify two coordinates on the line from the given information.
The points are (0,3)(0, 3) and (2,11)(2, 11).
To find the slope of a line, we need at least two points on that line.
2
Apply the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
m=11320m = \frac{11 - 3}{2 - 0}
The slope formula calculates the ratio of the vertical change (rise) to the horizontal change (run).
3
Perform the subtraction and division to find the slope value.
m=82=4m = \frac{8}{2} = 4
Simplifying the fraction gives the final numerical slope.

Anahtar Kavram

Calculating the slope of a linear function given two points
Soru 3Soru

In the xyxy-plane, the graph of a linear function ff passes through the points (0,1)(0, 1) and (2,k)(2, k). The graph of another linear function gg is perpendicular to the graph of ff and passes through the points (4,8)(-4, 8) and (2,k)(2, k). Which of the following could be the value of kk?

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Cevap: 5

Cevap

5
The slope of the line representing function ff is calculated as mf=k12m_f = \frac{k - 1}{2}, and the slope of the line representing function gg is mg=k86m_g = \frac{k - 8}{6}. Since the lines are perpendicular, the product of their slopes must be 1-1. This gives the equation k12k86=1\frac{k - 1}{2} \cdot \frac{k - 8}{6} = -1, which simplifies to (k1)(k8)=12(k - 1)(k - 8) = -12. Expanding and setting the quadratic equation to zero gives k29k+20=0k^2 - 9k + 20 = 0. Factoring this expression yields (k4)(k5)=0(k - 4)(k - 5) = 0. Thus, kk can be either 44 or 55. Since 5 is the only option listed, the option containing 5 is the correct answer.

Adım Adım Çözüm

1
Express the slope of the linear function ff in terms of kk.
mf=k120=k12m_f = \frac{k - 1}{2 - 0} = \frac{k - 1}{2}
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Express the slope of the linear function gg in terms of kk.
mg=k82(4)=k86m_g = \frac{k - 8}{2 - (-4)} = \frac{k - 8}{6}
The slope formula is applied to the points (4,8)(-4, 8) and (2,k)(2, k).
3
Apply the perpendicular lines condition to set up an equation for kk.
mfmg=1    (k12)(k86)=1m_f \cdot m_g = -1 \implies \left(\frac{k - 1}{2}\right)\left(\frac{k - 8}{6}\right) = -1
Perpendicular lines have slopes that are negative reciprocals of each other, meaning their product is 1-1.
4
Solve the quadratic equation for kk.
(k1)(k8)=12    k29k+8=12    k29k+20=0    (k4)(k5)=0(k - 1)(k - 8) = -12 \implies k^2 - 9k + 8 = -12 \implies k^2 - 9k + 20 = 0 \implies (k - 4)(k - 5) = 0
Multiply both sides by 12, expand the product, move all terms to one side, and factor the quadratic expression.
5
Identify the possible values of kk and match with the options.
k=4k = 4 or k=5k = 5. Since 5 is the only value present in the options, the correct value is 5.
Both values satisfy the condition, but only one is listed among the multiple-choice options.

Anahtar Kavram

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other (their product is 1-1).
Soru 4Soru

A factory manufactures solar panels at a constant rate. At the start of a morning shift, the factory has already manufactured 8080 solar panels. The total number of solar panels manufactured tt hours after the shift begins is modeled by a linear function. If the factory has manufactured a total of 200200 solar panels 33 hours after the shift begins, how many hours after the shift begins will the factory have manufactured a total of 440440 solar panels?

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Cevap: 9

Cevap

9
To find the time when the total number of solar panels reaches 440440, we first determine the constant rate of production. The change in solar panels over the first 33 hours is 20080=120200 - 80 = 120 panels. Dividing by the 33 hours gives a constant rate of 4040 panels per hour. The linear equation representing the total panels is N(t)=40t+80N(t) = 40t + 80. Setting N(t)=440N(t) = 440 gives 440=40t+80440 = 40t + 80, which simplifies to 360=40t360 = 40t. Solving for tt yields t=9t = 9.

Adım Adım Çözüm

1
Calculate the constant rate of production (slope) of the linear function.
The rate is 4040 solar panels per hour.
The factory starts with 8080 solar panels and reaches 200200 solar panels in 33 hours. The rate of change is the change in the number of panels divided by the change in time: 2008030=1203=40\frac{200 - 80}{3 - 0} = \frac{120}{3} = 40 panels per hour.
2
Write the linear equation representing the total number of solar panels manufactured, N(t)N(t), after tt hours.
N(t)=40t+80N(t) = 40t + 80
Since the initial quantity is 8080 and the rate of production is 4040 panels per hour, the linear function is N(t)=40t+80N(t) = 40t + 80.
3
Solve for the time tt when the total number of solar panels is 440440.
t=9t = 9 hours
Set N(t)=440N(t) = 440 in the equation: 440=40t+80440 = 40t + 80. Subtract 8080 from both sides to get 360=40t360 = 40t. Divide both sides by 4040 to find t=9t = 9.

Anahtar Kavram

Linear Functions and Graphs
Tahmini Süre:1m 30s
Soru 5Soru

In the xyxy-plane, the graph of the linear function f(x)=px+qf(x) = px + q, where pp and qq are constants, passes through the point (2,7)(2, 7). Line LL is parallel to the graph of ff and has a yy-intercept that is 33 units below the yy-intercept of the graph of ff. If the xx-intercept of line LL is (6,0)(-6, 0), what is the value of pp?

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Cevap: 0.5

Cevap

The value of pp is 0.50.5 (or the fraction 1/21/2).
The correct answer is 0.50.5 (or 1/21/2). The graph of f(x)=px+qf(x) = px + q passes through (2,7)(2, 7), which means 7=2p+q7 = 2p + q, or q=72pq = 7 - 2p. Line LL is parallel to ff, so its slope is pp, and its yy-intercept is q3q - 3. Thus, the equation of line LL is y=px+q3y = px + q - 3. Since line LL has an xx-intercept at (6,0)(-6, 0), we can substitute x=6x = -6 and y=0y = 0 into its equation, yielding 0=6p+q30 = -6p + q - 3. Substituting q=72pq = 7 - 2p into this equation gives 0=6p+(72p)30 = -6p + (7 - 2p) - 3, which simplifies to 8p+4=0-8p + 4 = 0. Solving for pp gives 8p=48p = 4, or p=0.5p = 0.5.

Adım Adım Çözüm

1
Express the relationship between pp and qq using the given point (2,7)(2, 7) on the graph of ff.
q=72pq = 7 - 2p
The point (2,7)(2, 7) must satisfy the equation f(x)=px+qf(x) = px + q.
2
Formulate the equation of line LL using the parallel slope and the shifted yy-intercept.
y=px+q3y = px + q - 3
Parallel lines have equal slopes, and the yy-intercept of LL is 33 units below the yy-intercept of ff, which is qq.
3
Substitute the xx-intercept (6,0)(-6, 0) into the equation of line LL.
6p+q3=0-6p + q - 3 = 0
The xx-intercept is a point on the line where y=0y = 0.
4
Substitute the expression for qq from Step 1 into the equation from Step 3 and solve for pp.
p=0.5p = 0.5 (or 12\frac{1}{2})
Solving the linear equation 6p+(72p)3=0-6p + (7 - 2p) - 3 = 0 simplifies to 8p+4=0-8p + 4 = 0, giving p=0.5p = 0.5.

Anahtar Kavram

Understanding linear functions, their graphs, slopes of parallel lines, and intercepts.

Alternatif Yöntem

Instead of solving for qq first, you can use the point-slope form. Line LL passes through (6,0)(-6, 0) and has slope pp, so its equation is y=p(x+6)y = p(x + 6), or y=px+6py = px + 6p. The yy-intercept of LL is 6p6p. The yy-intercept of the graph of ff is qq. We are given that the yy-intercept of LL is 33 units below the yy-intercept of ff, so 6p=q36p = q - 3. Since ff passes through (2,7)(2, 7), we have 7=2p+q7 = 2p + q, which means q=72pq = 7 - 2p. Substituting this into 6p=q36p = q - 3 gives 6p=(72p)36p = (7 - 2p) - 3, or 6p=42p6p = 4 - 2p. Adding 2p2p to both sides gives 8p=48p = 4, which results in p=0.5p = 0.5.
Tahmini Süre:2m 30s
Soru 6Soru

In the xyxy-plane, the graph of the linear function ff has a slope of 3-3 and passes through the point (4,18)(4, 18). What is the yy-coordinate of the yy-intercept of the graph of ff?

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Cevap: 30

Cevap

The correct answer is 30.
The equation of a linear function can be written in slope-intercept form as f(x)=mx+bf(x) = mx + b, where mm is the slope and bb is the yy-coordinate of the yy-intercept. Substituting the given slope m=3m = -3 yields f(x)=3x+bf(x) = -3x + b. Since the point (4,18)(4, 18) lies on the graph of ff, we substitute x=4x = 4 and f(x)=18f(x) = 18 into the equation to get 18=3(4)+b18 = -3(4) + b, which simplifies to 18=12+b18 = -12 + b. Adding 1212 to both sides of the equation yields b=30b = 30. Thus, the yy-coordinate of the yy-intercept of the graph of ff is 3030.

Adım Adım Çözüm

1
Write the general slope-intercept form of a linear equation.
f(x)=3x+bf(x) = -3x + b
The slope of the line is given as 3-3, so we substitute m=3m = -3 into f(x)=mx+bf(x) = mx + b.
2
Substitute the point (4,18)(4, 18) into the equation.
18=3(4)+b18 = -3(4) + b
Since the graph of ff passes through (4,18)(4, 18), these coordinates must satisfy the function equation.
3
Solve for the yy-intercept bb.
b=30b = 30
Simplify to 18=12+b18 = -12 + b, and then add 1212 to both sides to isolate bb.

Anahtar Kavram

Finding the equation of a line using its slope and a point.
Soru 7Soru

The table below shows some values of the linear function gg.

xxg(x)g(x)
2255
441111
661717

What is the slope of the graph of gg in the xyxy-plane?

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Cevap: 33

Cevap

The slope of the graph is 33.
The slope of a linear function is the constant rate of change, calculated as the change in the function values divided by the change in the input values. Choosing the points (2,5)(2, 5) and (4,11)(4, 11), the slope is 11542=62=3\frac{11 - 5}{4 - 2} = \frac{6}{2} = 3. This same slope is obtained using any other pair of points from the table, such as (4,11)(4, 11) and (6,17)(6, 17), where the slope is 171164=62=3\frac{17 - 11}{6 - 4} = \frac{6}{2} = 3.

Adım Adım Çözüm

1
Select two points from the table to find the change in the input and output values.
Using the points (2,5)(2, 5) and (4,11)(4, 11), the change in the output g(x)g(x) is 115=611 - 5 = 6, and the change in the input xx is 42=24 - 2 = 2.
The slope is defined as the change in the vertical coordinate divided by the change in the horizontal coordinate.
2
Divide the change in g(x)g(x) by the change in xx to find the slope.
The slope is 62=3\frac{6}{2} = 3.
Dividing the vertical change by the horizontal change yields the constant rate of change of the linear function.

Anahtar Kavram

Finding the slope of a linear function from a table of values
Tahmini Süre:45s
Soru 8Soru

The graph of a linear function is shown in the xyxy-plane. The line passes through the points (2,3)(-2, -3) and (2,5)(2, 5). If the point (k,11)(k, 11) lies on the line, what is the value of kk?

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Cevap: 5

Cevap

5
The slope of the line is calculated as m=5(3)2(2)=2m = \frac{5 - (-3)}{2 - (-2)} = 2. Using the point-slope form with the point (2,5)(2, 5), the equation of the line is y5=2(x2)y - 5 = 2(x - 2), which simplifies to y=2x+1y = 2x + 1. Substituting the point (k,11)(k, 11) into this equation gives 11=2k+111 = 2k + 1, which solves to k=5k = 5.

Adım Adım Çözüm

1
Calculate the slope of the line using the points (2,3)(-2, -3) and (2,5)(2, 5)
m=2m = 2
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the given coordinates: m=5(3)2(2)=84=2m = \frac{5 - (-3)}{2 - (-2)} = \frac{8}{4} = 2.
2
Determine the equation of the line using point-slope form
y=2x+1y = 2x + 1
Using yy1=m(xx1)y - y_1 = m(x - x_1) with the point (2,5)(2, 5) and m=2m = 2, we get y5=2(x2)y - 5 = 2(x - 2), which simplifies to y=2x+1y = 2x + 1.
3
Substitute (k,11)(k, 11) into the line equation and solve for kk
k=5k = 5
Substituting x=kx = k and y=11y = 11 yields 11=2k+111 = 2k + 1. Subtracting 1 from both sides gives 10=2k10 = 2k, and dividing by 2 gives k=5k = 5.

Anahtar Kavram

Finding the linear equation from coordinates and solving for variables
Soru 9Soru

In the xyxy-plane, line kk has the equation y=2x+10y = -2x + 10. Line mm is parallel to line kk and passes through the point (6,8)(6, 8). Line ll is perpendicular to line kk and intersects the xx-axis at the point (a,0)(a, 0), where a>10a > 10. If the region bounded by lines kk, mm, ll, and the yy-axis has an area of 110110, what is the value of aa?

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Cevap: 25

Cevap

25
To find the value of aa, we determine the equations of the lines mm and ll based on their geometric relationships to line kk. Line mm is parallel to line kk (y=2x+10y = -2x + 10), so its slope is 2-2. Using the point (6,8)(6, 8), its equation is y=2x+20y = -2x + 20. Line ll is perpendicular to line kk, so its slope is 12\frac{1}{2}. It intersects the xx-axis at (a,0)(a, 0), giving the equation y=12x12ay = \frac{1}{2}x - \frac{1}{2}a. The bounded region formed by the parallel lines kk and mm, the perpendicular line ll, and the yy-axis is a trapezoid. Calculating the area of this trapezoid by dividing it into a parallelogram and a triangle yields the area formula 60+2a60 + 2a. Setting this equal to the given area of 110110 yields 60+2a=11060 + 2a = 110, which solves to a=25a = 25.

Adım Adım Çözüm

1
Determine the equation of line mm using the parallel slope and the given point.
Line mm has the equation y=2x+20y = -2x + 20.
Parallel lines have equal slopes. Since line kk has a slope of 2-2, line mm also has a slope of 2-2. Substituting the point (6,8)(6, 8) into the point-slope form gives y8=2(x6)y - 8 = -2(x - 6).
2
Determine the equation of line ll using the perpendicular slope and its xx-intercept.
Line ll has the equation y=12x12ay = \frac{1}{2}x - \frac{1}{2}a.
Perpendicular lines have negative reciprocal slopes. The negative reciprocal of 2-2 is 12\frac{1}{2}. Using the point (a,0)(a, 0) in the point-slope form gives y0=12(xa)y - 0 = \frac{1}{2}(x - a).
3
Calculate the vertices of the bounded region by finding the intersection points of the boundary lines.
The vertices of the bounded region are (0,20)(0, 20), (0,10)(0, 10), (4+0.2a,20.4a)(4 + 0.2a, 2 - 0.4a), and (8+0.2a,40.4a)(8 + 0.2a, 4 - 0.4a).
The region is bounded by the parallel lines kk and mm, the perpendicular line ll, and the yy-axis (x=0x = 0).
4
Find the area of the region as an algebraic expression in terms of aa.
The area is equal to 60+2a60 + 2a.
The region can be divided into a parallelogram with a vertical base of 1010 and width 4+0.2a4 + 0.2a, and a right triangle with a vertical base of 1010 and width 44. The sum of their areas is 10(4+0.2a)+12(10)(4)=40+2a+20=60+2a10(4 + 0.2a) + \frac{1}{2}(10)(4) = 40 + 2a + 20 = 60 + 2a.
5
Set the area expression equal to the given area of 110110 and solve for aa.
a=25a = 25
Setting the area equal to 110110 yields 60+2a=11060 + 2a = 110, which simplifies to 2a=502a = 50, or a=25a = 25.

Anahtar Kavram

Linear functions, parallel and perpendicular lines, finding line equations, and coordinate geometry area.

Alternatif Yöntem

The area can also be calculated using the geometric properties of a trapezoid. The height of the trapezoid is the perpendicular distance between the parallel lines kk and mm, which is 2010(2)2+12=25\frac{|20 - 10|}{\sqrt{(-2)^2 + 1^2}} = 2\sqrt{5}. The bases of the trapezoid are the segments of lines kk and mm from the yy-axis to their intersection points with line ll. The length of the base on line kk is 959\sqrt{5} and the length of the base on line mm is 13513\sqrt{5} (when a=25a = 25). Using the formula for the area of a trapezoid, Area=95+1352×25=115×25=110\text{Area} = \frac{9\sqrt{5} + 13\sqrt{5}}{2} \times 2\sqrt{5} = 11\sqrt{5} \times 2\sqrt{5} = 110.
Tahmini Süre:3m 0s
Soru 10Soru

A water tank contains 24 gallons of water. Water is being drained from the tank at a constant rate. After 8 minutes, the tank contains 14 gallons of water. If the volume of water in the tank, in gallons, is a linear function of the time in minutes, how many minutes will it take for the tank to contain exactly 4 gallons of water?

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Cevap: 16

Cevap

16
The volume of water in the tank decreases linearly from an initial value of 2424 gallons at t=0t = 0 to 1414 gallons at t=8t = 8. The constant rate of change (slope) is calculated by dividing the change in volume by the change in time: 142480=1.25\frac{14 - 24}{8 - 0} = -1.25 gallons per minute. Using the slope-intercept form, the volume VV at time tt is given by V=1.25t+24V = -1.25t + 24. Setting the volume V=4V = 4 gives the equation 4=1.25t+244 = -1.25t + 24. Solving for tt yields 20=1.25t-20 = -1.25t, which simplifies to t=16t = 16.

Adım Adım Çözüm

1
Determine the initial state and the state after 8 minutes as coordinate points.
The initial state is (0,24)(0, 24) and the state after 88 minutes is (8,14)(8, 14).
These coordinates represent the relationship between time and volume of water in the tank.
2
Calculate the slope (constant rate of change) of the linear function.
Slope m=142480=1.25m = \frac{14 - 24}{8 - 0} = -1.25
The slope represents the constant rate at which water is being drained from the tank.
3
Formulate the linear equation.
V(t)=1.25t+24V(t) = -1.25t + 24
Using the slope-intercept form V(t)=mt+bV(t) = mt + b, where b=24b = 24 is the vertical intercept representing the initial volume.
4
Solve for the time tt when the volume of water is 44 gallons.
4=1.25t+24    20=1.25t    t=164 = -1.25t + 24 \implies -20 = -1.25t \implies t = 16
To find the time at which the volume decreases to exactly 44 gallons.

Anahtar Kavram

Linear Functions and Graphs
Soru 11Soru

The table below shows some values of the linear function ff.

xxf(x)f(x)
221111
441717
662323

What is the yy-coordinate of the yy-intercept of the graph of y=f(x)y = f(x) in the xyxy-plane?

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Cevap: 5

Cevap

The correct answer is 5, representing the y-coordinate of the y-intercept of the graph of f.
To find the yy-intercept of the linear function, we first determine its slope using the points (2,11)(2, 11) and (4,17)(4, 17) from the table. The slope mm is 171142=62=3\frac{17 - 11}{4 - 2} = \frac{6}{2} = 3. Next, we use the slope-intercept equation f(x)=mx+bf(x) = mx + b. Substituting m=3m = 3 and the point (2,11)(2, 11) gives 11=3(2)+b11 = 3(2) + b, which simplifies to 11=6+b11 = 6 + b. Solving for bb yields 55. Therefore, the yy-coordinate of the yy-intercept of the graph of ff is 55.

Adım Adım Çözüm

1
Calculate the slope (mm) of the linear function using two coordinate pairs from the table.
The slope is m=3m = 3.
A linear function has a constant slope, which can be found using the formula m=f(x2)f(x1)x2x1m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.
2
Substitute the slope and one of the points into the slope-intercept equation f(x)=mx+bf(x) = mx + b to solve for the yy-intercept bb.
The yy-intercept bb is 55.
The yy-coordinate of the yy-intercept of the graph of y=f(x)y = f(x) is the value of bb in the equation f(x)=mx+bf(x) = mx + b.

Anahtar Kavram

Finding the y-intercept of a linear function from a table of values.
Tahmini Süre:45s
Soru 12Soru

In the xyxy-plane, line l1l_1 passes through the origin and has a positive slope. Line l2l_2 is perpendicular to l1l_1 and intersects the xx-axis at (d,0)(d, 0), where d>0d > 0. The two lines intersect at the point (a,b)(a, b) in the first quadrant. If the ratio of aa to bb is 33 to 44, and the area of the triangle bounded by l1l_1, l2l_2, and the xx-axis is AA, which of the following expressions represents dd in terms of AA?

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Cevap: 56A6\frac{5\sqrt{6A}}{6}

Cevap

The expression 56A6\frac{5\sqrt{6A}}{6} represents dd in terms of AA.
The correct expression is 56A6\frac{5\sqrt{6A}}{6}. The slope of line l1l_1 passing through the origin and (a,b)(a,b) is given by m1=bam_1 = \frac{b}{a}. Since the ratio of aa to bb is 33 to 44, we have ab=34\frac{a}{b} = \frac{3}{4}, which implies the slope m1=43m_1 = \frac{4}{3}. Because line l2l_2 is perpendicular to l1l_1, its slope is m2=34m_2 = -\frac{3}{4}. The equation of l2l_2 passing through (d,0)(d,0) is y=34(xd)y = -\frac{3}{4}(x - d). Solving the system of equations gives the intersection coordinates a=925da = \frac{9}{25}d and b=1225db = \frac{12}{25}d. The area of the triangle bounded by the two lines and the xx-axis is A=12×d×b=625d2A = \frac{1}{2} \times d \times b = \frac{6}{25}d^2. Solving for dd yields d=25A6=56A6d = \sqrt{\frac{25A}{6}} = \frac{5\sqrt{6A}}{6}.

Adım Adım Çözüm

1
Determine the slope and equation of line l1l_1.
Slope m1=43m_1 = \frac{4}{3}, and the equation of the line is y=43xy = \frac{4}{3}x.
Since l1l_1 passes through (0,0)(0,0) and the point (a,b)(a,b) in the first quadrant, its slope is m1=bam_1 = \frac{b}{a}. Since the ratio of aa to bb is 33 to 44, we have ab=34    ba=43\frac{a}{b} = \frac{3}{4} \implies \frac{b}{a} = \frac{4}{3}.
2
Determine the equation of line l2l_2.
The equation of the line is y=34(xd)y = -\frac{3}{4}(x - d).
Line l2l_2 is perpendicular to l1l_1, so its slope is the negative reciprocal of m1m_1, which is m2=34m_2 = -\frac{3}{4}. Using the point-slope form with the xx-intercept (d,0)(d,0), the equation is y0=34(xd)y - 0 = -\frac{3}{4}(x - d).
3
Find the coordinates of the intersection point (a,b)(a,b) in terms of dd.
a=925da = \frac{9}{25}d and b=1225db = \frac{12}{25}d.
Set the two equations equal to find the xx-coordinate of the intersection: 43x=34(xd)    43x=34x+34d    2512x=34d    x=925d\frac{4}{3}x = -\frac{3}{4}(x - d) \implies \frac{4}{3}x = -\frac{3}{4}x + \frac{3}{4}d \implies \frac{25}{12}x = \frac{3}{4}d \implies x = \frac{9}{25}d. Substituting this back into the equation of l1l_1 yields the yy-coordinate: y=43(925d)=1225dy = \frac{4}{3}\left(\frac{9}{25}d\right) = \frac{12}{25}d.
4
Express the area of the triangle in terms of dd.
A=625d2A = \frac{6}{25}d^2.
The base of the triangle along the xx-axis is dd (from x=0x=0 to x=dx=d). The height of the triangle is the yy-coordinate of the intersection point, b=1225db = \frac{12}{25}d. The area of the triangle is given by A=12×base×height=12d(1225d)=625d2A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}d\left(\frac{12}{25}d\right) = \frac{6}{25}d^2.
5
Solve for dd in terms of AA.
d=56A6d = \frac{5\sqrt{6A}}{6}.
Isolate d2d^2 to get d2=25A6d^2 = \frac{25A}{6}. Taking the square root of both sides gives d=25A6=5A6=56A6d = \sqrt{\frac{25A}{6}} = \frac{5\sqrt{A}}{\sqrt{6}} = \frac{5\sqrt{6A}}{6}.

Anahtar Kavram

Using the properties of perpendicular lines, setting up equations from coordinate parameters, and applying geometric formulas to relate parameters in linear systems.

Alternatif Yöntem

Using the geometric mean theorem (altitude rule) in a right triangle, the altitude bb divides the hypotenuse dd into segments aa and dad-a, such that b2=a(da)b^2 = a(d-a). Since b=43ab = \frac{4}{3}a, we can substitute this to find a=925da = \frac{9}{25}d and b=1225db = \frac{12}{25}d directly without finding the line equations. Then, A=12db=625d2A = \frac{1}{2} d b = \frac{6}{25}d^2, which solves to d=56A6d = \frac{5\sqrt{6A}}{6}.
Tahmini Süre:3m 0s
Soru 13Soru

In the xyxy-plane, the graph of the linear function ff is perpendicular to the line with equation 3x+4y=483x + 4y = 48. The graph of ff intersects the xx-axis at the point (p,0)(p, 0) and the yy-axis at the point (0,q)(0, q), where q>0q > 0. If the distance between the points (p,0)(p, 0) and (0,q)(0, q) is 1515, what is the value of qq?

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Cevap: 12

Cevap

12
First, find the slope of the given line by rewriting 3x+4y=483x + 4y = 48 in slope-intercept form: y=34x+12y = -\frac{3}{4}x + 12. The slope is 34-\frac{3}{4}. The graph of the linear function ff is perpendicular to this line, so its slope is the negative reciprocal, 43\frac{4}{3}. With a yy-intercept of (0,q)(0, q), the equation of ff is y=43x+qy = \frac{4}{3}x + q. Setting y=0y = 0 gives the xx-intercept (p,0)=(34q,0)(p, 0) = (-\frac{3}{4}q, 0). The distance between these intercepts is (34q)2+q2=2516q2=54q\sqrt{(-\frac{3}{4}q)^2 + q^2} = \sqrt{\frac{25}{16}q^2} = \frac{5}{4}q (since q>0q > 0). Given that the distance is 1515, we solve 54q=15\frac{5}{4}q = 15 to find q=12q = 12.

Adım Adım Çözüm

1
Find the slope of the line 3x+4y=483x + 4y = 48.
The slope of the line is 34-\frac{3}{4}.
To find the slope of the perpendicular line ff, we first need the slope of the given line. Rewriting 3x+4y=483x + 4y = 48 in slope-intercept form gives y=34x+12y = -\frac{3}{4}x + 12.
2
Determine the slope of ff.
The slope of ff is 43\frac{4}{3}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Find the xx-intercept of ff in terms of qq.
The xx-intercept of ff is (34q,0)(-\frac{3}{4}q, 0).
Since the yy-intercept of ff is (0,q)(0, q), the equation of ff is y=43x+qy = \frac{4}{3}x + q. Setting y=0y = 0 gives 0=43x+q0 = \frac{4}{3}x + q, which simplifies to x=34qx = -\frac{3}{4}q.
4
Use the distance formula between the intercepts to solve for qq.
The value of qq is 1212.
The distance between (34q,0)(-\frac{3}{4}q, 0) and (0,q)(0, q) is (34q)2+q2=2516q2=54q\sqrt{(-\frac{3}{4}q)^2 + q^2} = \sqrt{\frac{25}{16}q^2} = \frac{5}{4}q since q>0q > 0. Setting this distance to 1515 gives 54q=15\frac{5}{4}q = 15, which yields q=12q = 12.

Anahtar Kavram

Properties of perpendicular lines, finding intercepts, and utilizing the distance formula in coordinate geometry.
Soru 14Soru

In the xyxy-plane, the graph of a linear function ff has a yy-intercept of (0,r)(0, r) and an xx-intercept of (s,0)(s, 0), where rr and ss are nonzero constants. A second line, gg, is perpendicular to the graph of ff and passes through the point (s,r)(s, r). Which of the following expressions represents the xx-intercept of line gg?

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Cevap: s2r2s\frac{s^2 - r^2}{s}

Cevap

The expression s2r2s\frac{s^2 - r^2}{s}
The slope of line ff is found using its intercepts (0,r)(0, r) and (s,0)(s, 0), which gives a slope of mf=rsm_f = -\frac{r}{s}. Since line gg is perpendicular to line ff, its slope is the negative reciprocal, mg=srm_g = \frac{s}{r}. Using the point-slope form with the point (s,r)(s, r), the equation for line gg is yr=sr(xs)y - r = \frac{s}{r}(x - s). Setting y=0y = 0 to find the xx-intercept yields x=s2r2sx = \frac{s^2 - r^2}{s}.

Adım Adım Çözüm

1
Determine the slope of the linear function ff using its intercepts.
The slope of ff is mf=0rs0=rsm_f = \frac{0 - r}{s - 0} = -\frac{r}{s}.
The slope of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Find the slope of line gg, which is perpendicular to the graph of ff.
The slope of line gg is mg=1mf=1r/s=srm_g = -\frac{1}{m_f} = -\frac{1}{-r/s} = \frac{s}{r}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Write the equation of line gg in point-slope form using the point (s,r)(s, r).
The equation of line gg is yr=sr(xs)y - r = \frac{s}{r}(x - s).
The point-slope form of a linear equation is yy1=m(xx1)y - y_1 = m(x - x_1).
4
Solve for the xx-intercept of line gg by setting y=0y = 0.
Setting y=0y = 0 gives r=sr(xs)    r2s=xs    x=sr2s=s2r2s-r = \frac{s}{r}(x - s) \implies -\frac{r^2}{s} = x - s \implies x = s - \frac{r^2}{s} = \frac{s^2 - r^2}{s}.
The xx-intercept of a line is the value of xx when y=0y = 0.

Anahtar Kavram

Finding the equation of a perpendicular line and its intercepts using algebraic representations.
Soru 15Soru

In the xyxy-plane, the graph of the linear function ff passes through the points (2,c)(-2, c) and (c,7)(c, 7), where cc is a constant. If the slope of the graph of ff is 12\frac{1}{2}, what is the value of cc?

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Cevap: 4

Cevap

The value of cc is 4.
Using the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} with the points (2,c)(-2, c) and (c,7)(c, 7) and the slope m=12m = \frac{1}{2} gives 7cc(2)=12\frac{7 - c}{c - (-2)} = \frac{1}{2}. Simplifying the denominator yields 7cc+2=12\frac{7 - c}{c + 2} = \frac{1}{2}. Cross-multiplying gives 2(7c)=c+22(7 - c) = c + 2, which expands to 142c=c+214 - 2c = c + 2. Adding 2c2c to both sides and subtracting 22 from both sides gives 12=3c12 = 3c. Dividing both sides by 33 gives c=4c = 4.

Adım Adım Çözüm

1
Set up the slope formula for the line passing through (2,c)(-2, c) and (c,7)(c, 7).
m=7cc(2)m = \frac{7 - c}{c - (-2)}
The slope of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Substitute the given slope m=12m = \frac{1}{2} into the formula and simplify the denominator.
7cc+2=12\frac{7 - c}{c + 2} = \frac{1}{2}
We are given that the slope of the line is 12\frac{1}{2}.
3
Solve the equation for cc by cross-multiplying and isolating the variable.
2(7c)=c+2    142c=c+2    12=3c    c=42(7 - c) = c + 2 \implies 14 - 2c = c + 2 \implies 12 = 3c \implies c = 4
Cross-multiplying eliminates the fractions, allowing us to solve the linear equation for cc.

Anahtar Kavram

Finding the slope of a line given two points containing an unknown coordinate
Soru 16Soru

In the xyxy-plane, the graph of a linear function ff has a positive yy-intercept and a positive xx-intercept. The area of the triangular region in the first quadrant bounded by the graph of ff and the coordinate axes is 3636. If the graph of ff passes through the point (4,4)(4, 4) and has a slope less than 1-1, what is the yy-coordinate of the yy-intercept of the graph of ff?

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Cevap: 12

Cevap

The yy-coordinate of the yy-intercept of the graph of ff is 1212.
The correct answer is 1212 because the system of equations derived from the area constraint (qr=72qr = 72) and the point constraint (q+r=18q+r=18) yields two possible values for the yy-intercept: 66 or 1212. The condition that the slope must be less than 1-1 means that the line must be steeper than a slope of 1-1, which requires the yy-intercept to be larger than the xx-intercept (r>qr > q). Thus, the yy-intercept is 1212.

Adım Adım Çözüm

1
Express the equation of the line using intercept form.
xq+yr=1\frac{x}{q} + \frac{y}{r} = 1, where q>0q > 0 is the xx-intercept and r>0r > 0 is the yy-intercept.
Since the line intersects the positive axes, this form directly relates the intercepts to the coordinates of points on the line.
2
Use the area of the triangle to find a relationship between qq and rr.
qr=72qr = 72
The area of the right triangle formed by the axes and the intercepts is given by Area=12×base×height=12qr=36\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}qr = 36.
3
Substitute the point (4,4)(4, 4) into the equation of the line and simplify using the area relationship.
q+r=18q + r = 18
Plugging in x=4x = 4 and y=4y = 4 yields 4q+4r=1    4(q+r)=qr\frac{4}{q} + \frac{4}{r} = 1 \implies 4(q + r) = qr. Substituting qr=72qr = 72 gives 4(q+r)=72    q+r=184(q + r) = 72 \implies q + r = 18.
4
Solve the system of equations q+r=18q + r = 18 and qr=72qr = 72.
(q,r)=(6,12)(q, r) = (6, 12) or (q,r)=(12,6)(q, r) = (12, 6)
Substituting r=18qr = 18 - q into qr=72qr = 72 yields q(18q)=72    q218q+72=0    (q6)(q12)=0q(18 - q) = 72 \implies q^2 - 18q + 72 = 0 \implies (q - 6)(q - 12) = 0.
5
Apply the slope condition to determine the unique value of rr.
r=12r = 12
The slope of the line is m=rqm = -\frac{r}{q}. If q=12q = 12 and r=6r = 6, then m=0.5m = -0.5, which is not less than 1-1. If q=6q = 6 and r=12r = 12, then m=2m = -2, which is less than 1-1. Thus, r=12r = 12 is the correct yy-intercept.

Anahtar Kavram

Using intercepts and area to determine the equation of a linear function under constraints.
Soru 17Soru

The table below shows some values of xx and the corresponding values of f(x)f(x) for a linear function ff.

xxf(x)f(x)
37
612
917

What is the value of f(15)f(15)?

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Cevap: 27

Cevap

The value of the linear function evaluated at 15 is 27.
To find f(15)f(15) for the linear function, we determine its constant rate of change. Using points (3,7)(3, 7) and (6,12)(6, 12), the change in f(x)f(x) is 127=512 - 7 = 5 for a change in xx of 63=36 - 3 = 3. This gives a slope of 53\frac{5}{3}. Using point-slope form with (3,7)(3, 7) gives the equation f(x)7=53(x3)f(x) - 7 = \frac{5}{3}(x - 3), which simplifies to f(x)=53x+2f(x) = \frac{5}{3}x + 2. Substituting x=15x = 15 yields f(15)=53(15)+2=25+2=27f(15) = \frac{5}{3}(15) + 2 = 25 + 2 = 27. Alternatively, since xx increases by 6 from 9 to 15, which is twice the step size of 3, the value of f(x)f(x) must increase by 2×5=102 \times 5 = 10 from 1717, resulting in 17+10=2717 + 10 = 27.

Adım Adım Çözüm

1
Calculate the slope of the linear function
slope m=53m = \frac{5}{3}
Linear functions have a constant rate of change, which is the slope.
2
Determine the equation of the function
f(x)=53x+2f(x) = \frac{5}{3}x + 2
Using the slope and one point from the table helps define the function for all inputs.
3
Evaluate the function at x=15x = 15
f(15)=27f(15) = 27
Substitute the given input value into the function equation to find the corresponding output value.

Anahtar Kavram

Linear Functions and Graphs
Soru 18Soru

A line in the xyxy-plane has the equation 3y2x=123y - 2x = 12. What is the slope of this line?

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Cevap: 23\frac{2}{3}

Cevap

The slope of the line is 23\frac{2}{3}.
To find the slope, the equation 3y2x=123y - 2x = 12 is rewritten in slope-intercept form (y=mx+by = mx + b). Adding 2x2x to both sides gives 3y=2x+123y = 2x + 12. Dividing by 33 yields y=23x+4y = \frac{2}{3}x + 4. The slope is the coefficient of xx, which is 23\frac{2}{3}.

Adım Adım Çözüm

1
Isolate the term containing yy by adding 2x2x to both sides of the equation.
3y=2x+123y = 2x + 12
To express the equation in slope-intercept form (y=mx+by = mx + b).
2
Divide all terms in the equation by 33 to solve for yy.
y=23x+4y = \frac{2}{3}x + 4
To isolate yy and identify the slope, mm, which is the coefficient of xx.

Anahtar Kavram

Finding the slope of a line from its linear equation by rewriting it in slope-intercept form.
Soru 19Soru

A linear function ff has a slope of 23\frac{2}{3}. The function gg is defined by g(x)=f(x+4)3g(x) = f(x + 4) - 3. If the graph of y=g(x)y = g(x) in the xyxy-plane has an xx-intercept at (5,0)(5, 0), what is the xx-intercept of the graph of y=f(x)y = f(x)?

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Cevap: 4.5

Cevap

4.5
The correct answer is 4.54.5 (or 92\frac{9}{2}). By using the function transformation relation, we find g(5)=f(9)3=0g(5) = f(9) - 3 = 0, which yields f(9)=3f(9) = 3. With a slope of 23\frac{2}{3}, the equation of the line is f(x)=23x3f(x) = \frac{2}{3}x - 3. Setting f(x)=0f(x) = 0 yields the xx-intercept at x=4.5x = 4.5.

Adım Adım Çözüm

1
Use the definition of function gg and its given xx-intercept to find a point on the graph of ff.
f(9)=3f(9) = 3
Since the graph of y=g(x)y = g(x) has an xx-intercept at (5,0)(5, 0), we know g(5)=0g(5) = 0. Substituting x=5x = 5 into the definition g(x)=f(x+4)3g(x) = f(x + 4) - 3 gives g(5)=f(5+4)3=f(9)3g(5) = f(5 + 4) - 3 = f(9) - 3. Since g(5)=0g(5) = 0, it follows that f(9)3=0f(9) - 3 = 0, or f(9)=3f(9) = 3.
2
Determine the equation of the linear function ff using its slope and the point identified in Step 1.
f(x)=23x3f(x) = \frac{2}{3}x - 3
The function ff is linear with a slope of 23\frac{2}{3} and passes through the point (9,3)(9, 3). Using the point-slope formula, we get f(x)3=23(x9)f(x) - 3 = \frac{2}{3}(x - 9), which simplifies to f(x)=23x3f(x) = \frac{2}{3}x - 3.
3
Find the xx-intercept of the graph of y=f(x)y = f(x) by setting f(x)=0f(x) = 0.
x=4.5x = 4.5 (or 92\frac{9}{2})
To find the xx-intercept, set f(x)=0f(x) = 0. This gives 23x3=0\frac{2}{3}x - 3 = 0. Adding 33 to both sides and multiplying by 32\frac{3}{2} yields x=92x = \frac{9}{2}, which is equal to 4.54.5.

Anahtar Kavram

Linear Functions and Graphs

Alternatif Yöntem

Alternatively, we can write the equation of ff in slope-intercept form as f(x)=23x+bf(x) = \frac{2}{3}x + b. Then the definition of g(x)g(x) becomes g(x)=23(x+4)+b3=23x+83+b3=23x+b13g(x) = \frac{2}{3}(x + 4) + b - 3 = \frac{2}{3}x + \frac{8}{3} + b - 3 = \frac{2}{3}x + b - \frac{1}{3}. Since the graph of gg has an xx-intercept at (5,0)(5, 0), we substitute x=5x = 5 and g(5)=0g(5) = 0 to get 23(5)+b13=0\frac{2}{3}(5) + b - \frac{1}{3} = 0, which simplifies to 3+b=03 + b = 0, so b=3b = -3. This gives f(x)=23x3f(x) = \frac{2}{3}x - 3. Finally, the xx-intercept is found by solving 23x3=0\frac{2}{3}x - 3 = 0, resulting in x=4.5x = 4.5.
Tahmini Süre:2m 0s
Soru 20Soru

In the xyxy-plane, line LL passes through the points (0,4)(0, 4) and (6,0)(6, 0). A second line, MM, is perpendicular to line LL and intersects line LL at a point on the line y=xy = x. If line MM is represented by the equation y=px+qy = px + q, where pp and qq are constants, what is the value of p+qp + q?

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Cevap: 310\frac{3}{10}

Cevap

The correct answer is the value stating three-tenths
To find the sum of the slope pp and yy-intercept qq of line MM, we first determine the equation of line LL. Line LL passes through (0,4)(0, 4) and (6,0)(6, 0), so its slope is 0460=23\frac{0 - 4}{6 - 0} = -\frac{2}{3}. The equation of line LL is therefore y=23x+4y = -\frac{2}{3}x + 4. Setting y=xy = x gives x=23x+4x = -\frac{2}{3}x + 4, which yields x=2.4x = 2.4. The intersection point is (2.4,2.4)(2.4, 2.4). Because line MM is perpendicular to line LL, its slope is the negative reciprocal of 23-\frac{2}{3}, which is p=1.5p = 1.5. Using point-slope form with the intersection point (2.4,2.4)(2.4, 2.4), the equation of line MM is y2.4=1.5(x2.4)y - 2.4 = 1.5(x - 2.4), which simplifies to y=1.5x1.2y = 1.5x - 1.2. Thus, q=1.2q = -1.2. The sum p+qp + q is 1.51.2=0.31.5 - 1.2 = 0.3, or 310\frac{3}{10}.

Adım Adım Çözüm

1
Find the equation of line LL using the given points (0,4)(0, 4) and (6,0)(6, 0).
The slope of line LL is mL=0460=23m_L = \frac{0 - 4}{6 - 0} = -\frac{2}{3}. Since the yy-intercept is (0,4)(0, 4), the equation of line LL is y=23x+4y = -\frac{2}{3}x + 4.
To find where line LL intersects another line, we first need to establish its linear equation.
2
Find the intersection point of line LL and the line y=xy = x.
Substitute y=xy = x into the equation for line LL: x=23x+4    53x=4    x=2.4x = -\frac{2}{3}x + 4 \implies \frac{5}{3}x = 4 \implies x = 2.4. Thus, the intersection point is (2.4,2.4)(2.4, 2.4).
The problem states that line MM intersects line LL at a point on the line y=xy = x.
3
Determine the equation of line MM using the intersection point and the perpendicular relationship.
Since line MM is perpendicular to line LL, its slope pp is the negative reciprocal of 23-\frac{2}{3}, which is p=32=1.5p = \frac{3}{2} = 1.5. Using point-slope form at (2.4,2.4)(2.4, 2.4): y2.4=1.5(x2.4)    y=1.5x1.2y - 2.4 = 1.5(x - 2.4) \implies y = 1.5x - 1.2. This gives q=1.2q = -1.2.
Perpendicular lines in the coordinate plane have slopes that multiply to negative one. We then use the point-slope formula to define the equation of line MM.
4
Calculate the sum of pp and qq.
p+q=1.5+(1.2)=0.3=310p + q = 1.5 + (-1.2) = 0.3 = \frac{3}{10}.
The question asks for the value of the expression p+qp + q.

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Perpendicular lines and linear graph intersections
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