Coordinate Geometry

273 soru

Soru 221Soru

In the standard (x,y)(x, y) coordinate plane, a line segment has endpoints A(2,3)A(-2, 3) and B(4,y)B(4, y). If the slope of the line passing through AA and BB is 23-\frac{2}{3}, what is the value of yy?

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Cevap: -1

Cevap

The value of yy is 1-1.
The correct option is the one showing 1-1. By applying the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} to the points A(2,3)A(-2, 3) and B(4,y)B(4, y) with slope 23-\frac{2}{3}, we set up the equation y34(2)=23\frac{y - 3}{4 - (-2)} = -\frac{2}{3}. Simplifying the denominator yields y36=23\frac{y - 3}{6} = -\frac{2}{3}. Multiplying both sides by 66 gives y3=4y - 3 = -4, and solving for yy yields y=1y = -1.

Adım Adım Çözüm

1
Write down the slope formula and substitute the given values.
y34(2)=23\frac{y - 3}{4 - (-2)} = -\frac{2}{3}
The slope mm of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is defined as m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Simplify the denominator on the left side of the equation.
y36=23\frac{y - 3}{6} = -\frac{2}{3}
Subtracting a negative number is equivalent to addition: 4(2)=4+2=64 - (-2) = 4 + 2 = 6.
3
Multiply both sides by 6 to isolate the numerator.
y3=4y - 3 = -4
Multiplying 23-\frac{2}{3} by 66 yields 4-4.
4
Solve for yy by adding 3 to both sides of the equation.
y=1y = -1
Isolating yy gives y=4+3=1y = -4 + 3 = -1.

Anahtar Kavram

Using the slope formula to find a missing coordinate
Tahmini Süre:1m 0s
Soru 222Soru

In the standard (x,y)(x, y) coordinate plane, a line passes through the points (12,23)(-\frac{1}{2}, \frac{2}{3}) and (34,12)(\frac{3}{4}, -\frac{1}{2}). What is the slope of this line?

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Cevap: 1415-\frac{14}{15}

Cevap

The slope of the line is 1415-\frac{14}{15}.
The slope of the line is found by using the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the given points (12,23)(-\frac{1}{2}, \frac{2}{3}) and (34,12)(\frac{3}{4}, -\frac{1}{2}) into the formula, we find the change in yy is 76-\frac{7}{6} and the change in xx is 54\frac{5}{4}. Dividing the change in yy by the change in xx yields 1415-\frac{14}{15}.

Adım Adım Çözüm

1
Identify the coordinates of the two points as (x1,y1)=(12,23)(x_1, y_1) = (-\frac{1}{2}, \frac{2}{3}) and (x2,y2)=(34,12)(x_2, y_2) = (\frac{3}{4}, -\frac{1}{2}).
x1=12x_1 = -\frac{1}{2}, y1=23y_1 = \frac{2}{3}, x2=34x_2 = \frac{3}{4}, and y2=12y_2 = -\frac{1}{2}
To use the slope formula, we must first map the given points to their respective coordinate variables.
2
Apply the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} to calculate the change in yy (rise) and the change in xx (run).
Rise = 1223=76-\frac{1}{2} - \frac{2}{3} = -\frac{7}{6}; Run = \frac{3}{4} - (-\frac{1}{2}) = \frac{5}{4}
The slope is the ratio of the vertical change to the horizontal change.
3
Divide the change in yy by the change in xx and simplify the resulting fraction.
m=7654=76×45=2830=1415m = \frac{-\frac{7}{6}}{\frac{5}{4}} = -\frac{7}{6} \times \frac{4}{5} = -\frac{28}{30} = -\frac{14}{15}
Dividing fractions is performed by multiplying the numerator fraction by the reciprocal of the denominator fraction.

Anahtar Kavram

Calculating the slope of a line between two points containing positive and negative fractional coordinates.
Soru 223Soru

A line with a slope of 2.52.5 passes through the point (3,4)(3, 4) in the coordinate plane. What is the y-coordinate of the point on this line whose x-coordinate is 1-1?

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Cevap: -6

Cevap

-6
By applying the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} with the slope equal to 2.52.5, the known point at (3,4)(3, 4), and the target point at (1,y)(-1, y), we obtain the linear equation 2.5=y4132.5 = \frac{y - 4}{-1 - 3}. Simplifying the denominator gives 2.5=y442.5 = \frac{y - 4}{-4}. Multiplying both sides by 4-4 yields 10=y4-10 = y - 4. Finally, adding 44 to both sides gives y=6y = -6.

Adım Adım Çözüm

1
Set up the slope formula relation
2.5=y4132.5 = \frac{y - 4}{-1 - 3}
The slope mm of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is defined as the change in y-values divided by the change in x-values.
2
Solve for the unknown y-coordinate
y=6y = -6
Simplifying the denominator to 4-4, multiplying both sides by 4-4 gives 10=y4-10 = y - 4, and adding 44 to both sides isolates yy.

Anahtar Kavram

Finding a missing coordinate on a line given its slope and another coordinate point.
Soru 224Soru

A kite has vertices K(2,8)K(2, 8), I(6,3)I(6, 3), T(2,4)T(2, -4), and E(2,3)E(-2, 3) in the standard (x,y)(x, y) coordinate plane. What is the area of kite KITEKITE, in square units?

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Cevap: 48

Cevap

The area of kite KITEKITE is 48 square units.
The area of kite KITEKITE is 48 because the horizontal diagonal IEIE has a length of 8 units, the vertical diagonal KTKT has a length of 12 units, and the area of a kite is calculated as half the product of its diagonal lengths: 12×8×12=48\frac{1}{2} \times 8 \times 12 = 48.

Adım Adım Çözüm

1
Calculate the lengths of the vertical diagonal KTKT and the horizontal diagonal IEIE.
The length of KTKT is 12 and the length of IEIE is 8.
The vertices K(2,8)K(2, 8) and T(2,4)T(2, -4) share the same xx-coordinate, so the diagonal is vertical with length 8(4)=128 - (-4) = 12. The vertices I(6,3)I(6, 3) and E(2,3)E(-2, 3) share the same yy-coordinate, so the diagonal is horizontal with length 6(2)=86 - (-2) = 8.
2
Apply the area formula for a kite: Area=12d1d2\text{Area} = \frac{1}{2} d_1 d_2.
Area = 48
Since the diagonals of a kite are perpendicular, the area is half the product of the lengths of the diagonals: 12×12×8=48\frac{1}{2} \times 12 \times 8 = 48.

Anahtar Kavram

Finding the area of a geometric figure on the coordinate plane by using diagonal lengths.
Soru 225Soru

In the standard (x,y)(x, y) coordinate plane, a triangle has vertices at D(1,2)D(1, -2), E(4,7)E(4, 7), and F(6,k)F(6, k). If the line containing the altitude from vertex FF to side DEDE has a yy-intercept of 88, what is the value of kk?

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Cevap: 6

Cevap

The value of kk is 66.
The slope of side DEDE is calculated as 7(2)41=3\frac{7 - (-2)}{4 - 1} = 3. An altitude is perpendicular to the side it intersects, so the slope of the altitude is the negative reciprocal of 33, which is 13-\frac{1}{3}. The line containing this altitude has a yy-intercept of 88, giving the equation y=13x+8y = -\frac{1}{3}x + 8. Since the vertex F(6,k)F(6, k) lies on this line, substituting x=6x = 6 and y=ky = k into the equation yields k=13(6)+8=2+8=6k = -\frac{1}{3}(6) + 8 = -2 + 8 = 6. This matches the correct value of 66.

Adım Adım Çözüm

1
Calculate the slope of side DEDE using the coordinates of D(1,2)D(1, -2) and E(4,7)E(4, 7).
The slope of DEDE is 33.
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the points gives mDE=7(2)41=93=3m_{DE} = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3.
2
Determine the slope of the altitude line, which is perpendicular to side DEDE.
The slope of the altitude is 13-\frac{1}{3}.
Perpendicular lines have slopes that are negative reciprocals of each other. The negative reciprocal of 33 is 13-\frac{1}{3}.
3
Write the equation of the line containing the altitude using its slope and the given yy-intercept of 88.
The equation of the altitude line is y=13x+8y = -\frac{1}{3}x + 8.
The slope-intercept form of a linear equation is y=mx+by = mx + b, where mm is the slope and bb is the yy-intercept.
4
Substitute the coordinates of point F(6,k)F(6, k) into the equation of the altitude to solve for kk.
k=6k = 6.
Since vertex FF lies on the line containing the altitude, its coordinates must satisfy the equation: k=13(6)+8=2+8=6k = -\frac{1}{3}(6) + 8 = -2 + 8 = 6.

Anahtar Kavram

Finding the equation of a line perpendicular to a given line segment and using it to find a missing coordinate of a point on that line.
Soru 226Soru

In the standard (x,y)(x, y) coordinate plane, if a line passes through the point (1,3)(1, 3) and has an undefined slope, then the point (1,5)(1, -5) lies on this line.

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Cevap: True

Cevap

The statement is true because a line with an undefined slope is a vertical line, meaning all points on the line share the same x-coordinate. Since the line passes through (1,3)(1, 3), the x-coordinate of all points on the line is 11, which includes the point (1,5)(1, -5).
A line with an undefined slope is a vertical line. Since it passes through (1,3)(1, 3), the x-coordinate of every point on this line must be 11. The point (1,5)(1, -5) has an x-coordinate of 11, so it lies on this vertical line, making the statement true.

Adım Adım Çözüm

1
Identify the geometric properties of a line with an undefined slope.
A line with an undefined slope is a vertical line.
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. The slope is undefined when the denominator, representing the change in xx, is equal to zero. This occurs only for vertical lines.
2
Determine the equation of the vertical line that passes through the point (1,3)(1, 3).
The equation of the line is x=1x = 1.
Since the line is vertical, all points on the line share the same x-coordinate as the given point (1,3)(1, 3).
3
Verify if the point (1,5)(1, -5) satisfies the equation of the line.
The x-coordinate of the point (1,5)(1, -5) is 11, which satisfies the equation x=1x = 1.
A point lies on a line if its coordinates satisfy the equation of the line.

Anahtar Kavram

Slope of a Line
Soru 227Soru

A line, L1L_1, contains the points (2,5)(2, 5) and (6,3)(6, -3) in a coordinate plane. Another line, L2L_2, is perpendicular to L1L_1 and is defined by the equation ax+2y=7ax + 2y = 7. What is the value of aa?

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Cevap: -1

Cevap

The value of aa is 1-1.
The slope of line L1L_1 is 2-2. A line perpendicular to it must have a slope that is the negative reciprocal, which is 12\frac{1}{2}. Rewriting ax+2y=7ax + 2y = 7 in slope-intercept form gives y=a2x+72y = -\frac{a}{2}x + \frac{7}{2}, where the slope is a2-\frac{a}{2}. Setting this slope equal to 12\frac{1}{2} gives a2=12-\frac{a}{2} = \frac{1}{2}, which simplifies to a=1a = -1.

Adım Adım Çözüm

1
Find the slope of line L1L_1 using the points (2,5)(2, 5) and (6,3)(6, -3).
m1=3562=84=2m_1 = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2.
The slope formula is the change in y divided by the change in x.
2
Determine the perpendicular slope for L2L_2.
m2=1m1=12m_2 = -\frac{1}{m_1} = \frac{1}{2}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Rewrite the equation of L2L_2, ax+2y=7ax + 2y = 7, in slope-intercept form to find its slope expression.
2y=ax+7y=a2x+722y = -ax + 7 \Rightarrow y = -\frac{a}{2}x + \frac{7}{2}. The slope is a2-\frac{a}{2}.
Slope-intercept form y=mx+by = mx + b allows direct identification of the slope coefficient.
4
Equate the slope of L2L_2 to the perpendicular slope and solve for aa.
a2=12a=1-\frac{a}{2} = \frac{1}{2} \Rightarrow a = -1.
Solving the equation gives the value of aa required for the lines to be perpendicular.

Anahtar Kavram

Perpendicular lines have slopes that are negative reciprocals of each other.
Soru 228Soru

In the standard (x,y)(x, y) coordinate plane, a line segment has endpoints at P(2,1)P(2, 1) and Q(8,5)Q(8, 5). Line LL is parallel to segment PQPQ and passes through the point (3,2)(3, -2). If the point (9,y)(9, y) also lies on line LL, what is the value of yy?

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Cevap: 2

Cevap

The value of yy is 22.
The slope of segment PQPQ is 5182=23\frac{5 - 1}{8 - 2} = \frac{2}{3}. Since line LL is parallel to segment PQPQ, its slope is also 23\frac{2}{3}. The slope of line LL through (3,2)(3, -2) and (9,y)(9, y) is given by y(2)93=y+26\frac{y - (-2)}{9 - 3} = \frac{y + 2}{6}. Equating the two slopes yields y+26=23\frac{y + 2}{6} = \frac{2}{3}, which simplifies to y+2=4y + 2 = 4, so y=2y = 2.

Adım Adım Çözüm

1
Calculate the slope of segment PQPQ.
Slope of PQ=23PQ = \frac{2}{3}
Since parallel lines have equal slopes, finding the slope of the reference segment PQPQ is the first step in determining the slope of line LL.
2
Set up the slope equation for line LL using its parallel relationship to segment PQPQ.
y(2)93=23\frac{y - (-2)}{9 - 3} = \frac{2}{3}
Because line LL is parallel to segment PQPQ, its slope must also be 23\frac{2}{3}. The slope of line LL is calculated using the points (3,2)(3, -2) and (9,y)(9, y).
3
Solve the equation for yy.
y=2y = 2
Simplifying the numerator gives y+2y + 2 and the denominator gives 66. Multiplying both sides of y+26=23\frac{y + 2}{6} = \frac{2}{3} by 66 yields y+2=4y + 2 = 4, which solves to y=2y = 2.

Anahtar Kavram

Parallel lines have equal slopes in the coordinate plane.
Soru 229Soru

In the standard (x,y)(x, y) coordinate plane, a right triangle has vertices A(2,3)A(-2, -3), B(6,3)B(6, -3), and C(6,y)C(6, y), where y>0y > 0. If the hypotenuse of the triangle has a length of 17 units, what is the value of yy?

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Cevap: 12

Cevap

The correct value of yy is 1212.
The horizontal leg ABAB has a length of 6(2)=86 - (-2) = 8 units. By the Pythagorean theorem, the vertical leg BCBC has a length of 17282=15\sqrt{17^2 - 8^2} = 15 units. Since the vertex BB is at (6,3)(6, -3) and CC is at (6,y)(6, y) with y>0y > 0, the vertical distance is y(3)=15y - (-3) = 15, which yields y=12y = 12.

Adım Adım Çözüm

1
Determine the length of the horizontal leg ABAB.
The length of ABAB is 88 units.
Since vertices A(2,3)A(-2, -3) and B(6,3)B(6, -3) share the same yy-coordinate, the segment is horizontal. The length is the positive difference between their xx-coordinates: 6(2)=86 - (-2) = 8.
2
Use the Pythagorean theorem to calculate the length of the vertical leg BCBC.
The length of BCBC is 1515 units.
Since ABAB is horizontal and BCBC is vertical (vertices BB and CC share the same xx-coordinate of 66), the angle at BB is a right angle. The hypotenuse is AC=17AC = 17. By the Pythagorean theorem, AB2+BC2=AC2    82+BC2=172    64+BC2=289    BC2=225    BC=15AB^2 + BC^2 = AC^2 \implies 8^2 + BC^2 = 17^2 \implies 64 + BC^2 = 289 \implies BC^2 = 225 \implies BC = 15.
3
Set up an equation using the coordinates to find yy.
y=12y = 12
The length of the vertical segment BCBC is the difference in yy-coordinates: y(3)=15|y - (-3)| = 15, which simplifies to y+3=15y + 3 = 15 since y>0y > 0. Solving for yy yields y=12y = 12.

Anahtar Kavram

Using coordinate differences and the Pythagorean theorem to determine unknown vertices of geometric figures in the coordinate plane.

Alternatif Yöntem

Alternatively, you can apply the distance formula directly between the vertices A(2,3)A(-2, -3) and C(6,y)C(6, y) with a distance of 17 units: (6(2))2+(y(3))2=17    82+(y+3)2=289    (y+3)2=225\sqrt{(6 - (-2))^2 + (y - (-3))^2} = 17 \implies 8^2 + (y + 3)^2 = 289 \implies (y + 3)^2 = 225, which yields y+3=15    y=12y + 3 = 15 \implies y = 12 since y>0y > 0.
Tahmini Süre:1m 30s
Soru 230Soru

In the standard (x,y)(x, y) coordinate plane, three vertices of a rectangle are A(4,1)A(-4, 1), B(2,9)B(2, 9), and C(6,6)C(6, 6). What is the area of the rectangle, in square units?

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Cevap: 50

Cevap

50
The length of side ABAB is calculated as (2(4))2+(91)2=36+64=10\sqrt{(2 - (-4))^2 + (9 - 1)^2} = \sqrt{36 + 64} = 10. The length of the adjacent side BCBC is calculated as (62)2+(69)2=16+9=5\sqrt{(6 - 2)^2 + (6 - 9)^2} = \sqrt{16 + 9} = 5. The area of the rectangle is the product of these two perpendicular side lengths, which is 10×5=5010 \times 5 = 50.

Adım Adım Çözüm

1
Calculate the length of side ABAB using the distance formula.
AB=10AB = 10
To find one of the side lengths of the rectangle.
2
Calculate the length of side BCBC using the distance formula.
BC=5BC = 5
To find the adjacent side length of the rectangle.
3
Multiply the two adjacent side lengths to find the area of the rectangle.
50
The area of a rectangle is equal to the product of its length and width.

Anahtar Kavram

Calculating the area of a geometric figure on the coordinate plane by determining its side lengths using the distance formula.
Soru 231Soru

A commercial drone starts at an altitude of 150150 meters and descends at a constant rate. After 1212 seconds of descent, its altitude is 114114 meters. If the altitude, aa, in meters, is modeled as a linear function of time, tt, in seconds, what is the slope of the line representing this relationship in the standard (t,a)(t, a) coordinate plane?

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Cevap: 3-3

Cevap

The slope of the line representing this relationship is 3-3.
The slope of the line represents the rate of change of the altitude with respect to time. By taking the initial point (0,150)(0, 150) and the point after 1212 seconds (12,114)(12, 114), the change in altitude is 114150=36114 - 150 = -36 meters. Dividing this by the change in time of 120=1212 - 0 = 12 seconds yields the correct slope of 3-3.

Adım Adım Çözüm

1
Identify the coordinate points (t,a)(t, a) from the given information.
The initial state corresponds to the point (0,150)(0, 150), and the state after 1212 seconds corresponds to (12,114)(12, 114).
Setting up coordinate pairs allows the direct application of the slope formula.
2
Apply the slope formula m=a2a1t2t1m = \frac{a_2 - a_1}{t_2 - t_1} to find the rate of change.
m=114150120m = \frac{114 - 150}{12 - 0}
The slope is the change in the vertical variable (altitude) divided by the change in the horizontal variable (time).
3
Calculate the difference and simplify the fraction.
m=3612=3m = \frac{-36}{12} = -3
Subtracting 150150 from 114114 gives the net change in altitude, which when divided by the duration of 1212 seconds yields the rate of change per second.

Anahtar Kavram

The slope of a line represents its constant rate of change, defined as the ratio of the change in the dependent variable to the change in the independent variable.
Soru 232Soru

In the standard (x,y)(x, y) coordinate plane, a parallelogram has vertices at P(2,1)P(-2, -1), Q(4,1)Q(4, -1), R(6,4)R(6, 4), and S(0,4)S(0, 4). What is the area of parallelogram PQRSPQRS, in square units?

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Cevap: 30

Cevap

30
The correct answer is 30. The area of a parallelogram is determined by multiplying its base by its perpendicular height. The base segment PQPQ is horizontal and extends from x=2x = -2 to x=4x = 4, giving a length of 4(2)=64 - (-2) = 6 units. The height is the vertical distance between the line containing the base PQPQ (y=1y = -1) and the line containing the opposite side SRSR (y=4y = 4). This distance is 4(1)=54 - (-1) = 5 units. Multiplying the base of 6 units by the height of 5 units yields an area of 30 square units.

Adım Adım Çözüm

1
Identify the base of the parallelogram by calculating the length of the horizontal side PQPQ.
PQ=4(2)=6PQ = 4 - (-2) = 6 units
The segment PQPQ lies on the horizontal line y=1y = -1, so its length is the difference between the x-coordinates of its endpoints.
2
Identify the height of the parallelogram by calculating the vertical distance between the parallel horizontal sides PQPQ (on y=1y = -1) and SRSR (on y=4y = 4).
height=4(1)=5height = 4 - (-1) = 5 units
The height of a parallelogram is the perpendicular distance between its parallel bases.
3
Calculate the area of the parallelogram using the formula Area=base×heightArea = \text{base} \times \text{height}.
Area=6×5=30Area = 6 \times 5 = 30 square units
Multiplying the base length by the vertical height gives the total area of the parallelogram.

Anahtar Kavram

Finding the area of a parallelogram on the coordinate plane using base and height calculations from coordinates.
Soru 233Soru

In the standard (x,y)(x, y) coordinate plane, the vertices of a right triangle are P(1,2)P(1, 2), Q(5,5)Q(5, 5), and R(k,9)R(k, 9). If the right angle of the triangle is at vertex QQ, what is the value of the constant kk?

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Cevap: 2

Cevap

The value of kk is 22.
Because the right angle of the triangle is at vertex QQ, segment PQPQ must be perpendicular to segment QRQR. The slope of PQPQ is 5251=34\frac{5 - 2}{5 - 1} = \frac{3}{4}. The slope of a perpendicular line is the negative reciprocal, so the slope of QRQR must be 43-\frac{4}{3}. Expressing the slope of QRQR using the coordinates of Q(5,5)Q(5, 5) and R(k,9)R(k, 9) gives 95k5=4k5\frac{9 - 5}{k - 5} = \frac{4}{k - 5}. Setting this equal to 43-\frac{4}{3} and solving for kk yields k5=3k - 5 = -3, which means k=2k = 2.

Adım Adım Çözüm

1
Use the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} to calculate the slope of the line segment PQPQ with endpoints P(1,2)P(1, 2) and Q(5,5)Q(5, 5).
mPQ=5251=34m_{PQ} = \frac{5 - 2}{5 - 1} = \frac{3}{4}
This establishes the direction of the first leg of the right triangle.
2
Find the slope of segment QRQR. Because the right angle is at vertex QQ, the segment PQPQ is perpendicular to segment QRQR.
mQR=43m_{QR} = -\frac{4}{3}
Perpendicular lines have slopes that are negative reciprocals of each other (m1m2=1m_1 \cdot m_2 = -1).
3
Write the slope of segment QRQR in terms of kk using coordinates Q(5,5)Q(5, 5) and R(k,9)R(k, 9).
mQR=95k5=4k5m_{QR} = \frac{9 - 5}{k - 5} = \frac{4}{k - 5}
This sets up an equation to find the unknown coordinate value.
4
Equate the two expressions for the slope of QRQR and solve for kk.
4k5=43k5=3k=2\frac{4}{k - 5} = -\frac{4}{3} \Rightarrow k - 5 = -3 \Rightarrow k = 2
Solving the rational equation yields the correct coordinate parameter.

Anahtar Kavram

Perpendicular lines in a coordinate plane have slopes that are negative reciprocals of each other.
Tahmini Süre:1m 30s
Soru 234Soru

In the standard (x,y)(x, y) coordinate plane, a line passes through the points (a,2)(a, 2) and (10,a1)(10, a - 1). If the slope of the line is 13-\frac{1}{3}, what is the value of aa?

Cevabı ve açıklamayı göster

Cevap: -0.5

Cevap

The value of aa is 0.5-0.5 (or 12-\frac{1}{2})
Applying the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} to the points (a,2)(a, 2) and (10,a1)(10, a - 1) with slope 13-\frac{1}{3} yields the equation 13=a310a-\frac{1}{3} = \frac{a - 3}{10 - a}. Solving this linear equation correctly yields a=0.5a = -0.5.

Adım Adım Çözüm

1
Apply the slope formula using the given coordinates.
13=(a1)210a-\frac{1}{3} = \frac{(a - 1) - 2}{10 - a}
The slope of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is defined as m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Simplify the numerator.
13=a310a-\frac{1}{3} = \frac{a - 3}{10 - a}
Subtracting 22 from a1a - 1 simplifies the numerator to a3a - 3.
3
Cross-multiply to solve the rational equation.
1(10a)=3(a3)-1(10 - a) = 3(a - 3)
Multiplying both sides by the denominators eliminates the fractions.
4
Solve the linear equation for aa.
a=0.5a = -0.5
Distributing on both sides gives 10+a=3a9-10 + a = 3a - 9. Rearranging terms yields 2a=12a = -1, which simplifies to a=0.5a = -0.5.

Anahtar Kavram

Slope of a Line
Soru 235Soru

Three vertices of a parallelogram are A(1,3)A(1, 3), B(2,1)B(-2, -1), and C(4,1)C(4, -1) in the standard (x,y)(x, y) coordinate plane. If the fourth vertex, DD, is located in the fourth quadrant, what is the yy-coordinate of DD?

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Cevap: -5

Cevap

The yy-coordinate of the fourth vertex of the parallelogram in the fourth quadrant is 5-5.
By applying the midpoint formula to the diagonals of the three possible parallelograms formed by the vertices A(1,3)A(1, 3), B(2,1)B(-2, -1), and C(4,1)C(4, -1), we find the candidate points for the fourth vertex DD to be (7,3)(7, 3), (5,3)(-5, 3), and (1,5)(1, -5). The point (1,5)(1, -5) is the only candidate that lies in the fourth quadrant, where x>0x > 0 and y<0y < 0. Therefore, the yy-coordinate of the fourth vertex is 5-5.

Adım Adım Çözüm

1
Set up equations based on the property that the diagonals of a parallelogram bisect each other (have the same midpoint).
Three possible configurations of diagonals lead to three potential sets of coordinates for the fourth vertex D(x,y)D(x, y): (7,3)(7, 3), (5,3)(-5, 3), and (1,5)(1, -5).
Three points in a coordinate plane can form three distinct parallelograms depending on which pairs are connected as diagonals.
2
Identify the signs of the coordinates for each potential vertex to determine which quadrant it lies in.
The point (7,3)(7, 3) is in Quadrant I (x>0,y>0x > 0, y > 0). The point (5,3)(-5, 3) is in Quadrant II (x<0,y>0x < 0, y > 0). The point (1,5)(1, -5) is in Quadrant IV (x>0,y<0x > 0, y < 0).
The fourth quadrant is defined by positive xx-values and negative yy-values.
3
Select the correct coordinate value corresponding to the question's requirement.
The yy-coordinate of the vertex in the fourth quadrant, (1,5)(1, -5), is 5-5.
The question asks specifically for the yy-coordinate of the fourth vertex DD.

Anahtar Kavram

Finding a missing vertex of a parallelogram on the coordinate plane using midpoint properties
Soru 236Soru

In the standard (x,y)(x, y) coordinate plane, line L1L_1 passes through the points (1,3)(1, 3) and (4,8)(4, 8). Line L2L_2 is perpendicular to line L1L_1. If line L2L_2 passes through the points (5,k)(5, k) and (10,1)(10, 1), what is the value of kk?

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Cevap: 4

Cevap

The value of kk is 44.
The correct answer is 44. First, determine the slope of line L1L_1 using the points (1,3)(1, 3) and (4,8)(4, 8), which is 8341=53\frac{8 - 3}{4 - 1} = \frac{5}{3}. Because line L2L_2 is perpendicular to L1L_1, its slope must be the negative reciprocal of 53\frac{5}{3}, which is 35-\frac{3}{5}. Next, set up the slope equation for L2L_2 with the points (5,k)(5, k) and (10,1)(10, 1), yielding 1k105=35\frac{1 - k}{10 - 5} = -\frac{3}{5}. Simplifying the equation gives 1k5=35\frac{1 - k}{5} = -\frac{3}{5}, which reduces to 1k=31 - k = -3. Solving for kk gives k=4k = 4.

Adım Adım Çözüm

1
Calculate the slope of line L1L_1 using the coordinates of the two given points, (1,3)(1, 3) and (4,8)(4, 8).
The slope of line L1L_1 is 53\frac{5}{3}.
Using the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}, we find m1=8341=53m_1 = \frac{8 - 3}{4 - 1} = \frac{5}{3}.
2
Determine the slope of line L2L_2 based on the perpendicular relationship between L1L_1 and L2L_2.
The slope of line L2L_2 is 35-\frac{3}{5}.
Perpendicular lines have slopes that are negative reciprocals of each other. The negative reciprocal of 53\frac{5}{3} is 35-\frac{3}{5}.
3
Use the coordinates (5,k)(5, k) and (10,1)(10, 1) on line L2L_2 to write an expression for its slope, set it equal to 35-\frac{3}{5}, and solve for kk.
k=4k = 4
The slope expression is 1k105=1k5\frac{1 - k}{10 - 5} = \frac{1 - k}{5}. Setting this equal to the perpendicular slope gives 1k5=35\frac{1 - k}{5} = -\frac{3}{5}. Multiplying both sides by 55 results in 1k=31 - k = -3. Adding kk to both sides and adding 33 to both sides yields k=4k = 4.

Anahtar Kavram

The slopes of perpendicular lines in a coordinate plane are negative reciprocals of each other, meaning their product is 1-1 (m1m2=1m_1 \cdot m_2 = -1).
Tahmini Süre:1m 15s
Soru 237Soru

In the standard (x,y)(x, y) coordinate plane, what is the yy-intercept of the perpendicular bisector of the line segment with endpoints (0,1)(0, 1) and (2,5)(2, 5)?

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Cevap: 72\frac{7}{2}

Cevap

72\frac{7}{2}
To find the perpendicular bisector, we first locate the midpoint of the segment with endpoints (0,1)(0, 1) and (2,5)(2, 5), which is (0+22,1+52)=(1,3)\left(\frac{0 + 2}{2}, \frac{1 + 5}{2}\right) = (1, 3). The slope of this segment is m=5120=2m = \frac{5 - 1}{2 - 0} = 2. The slope of the perpendicular bisector is the negative reciprocal, 12-\frac{1}{2}. Using the point-slope form with the midpoint (1,3)(1, 3) and slope 12-\frac{1}{2}, the equation of the perpendicular bisector is y3=12(x1)y - 3 = -\frac{1}{2}(x - 1). Setting x=0x = 0 to find the yy-intercept gives y=3+12=72y = 3 + \frac{1}{2} = \frac{7}{2}.

Adım Adım Çözüm

1
Find the midpoint of the line segment with endpoints (0,1)(0, 1) and (2,5)(2, 5).
The midpoint is M=(0+22,1+52)=(1,3)M = \left(\frac{0 + 2}{2}, \frac{1 + 5}{2}\right) = (1, 3).
A perpendicular bisector must pass through the midpoint of the segment it bisects.
2
Calculate the slope of the original line segment.
The slope is m=5120=42=2m = \frac{5 - 1}{2 - 0} = \frac{4}{2} = 2.
The slope of the segment is needed to find the slope of the line perpendicular to it.
3
Determine the slope of the perpendicular bisector.
The perpendicular slope is m=1m=12m_{\perp} = -\frac{1}{m} = -\frac{1}{2}.
The slope of a perpendicular line is the negative reciprocal of the original line's slope.
4
Write the equation of the perpendicular bisector and find the yy-intercept.
Using the point-slope form with point (1,3)(1, 3) and slope 12-\frac{1}{2} gives the equation y3=12(x1)y - 3 = -\frac{1}{2}(x - 1). Setting x=0x = 0 to find the yy-intercept yields y3=12(01)=12y - 3 = -\frac{1}{2}(0 - 1) = \frac{1}{2}, which simplifies to y=3+12=72y = 3 + \frac{1}{2} = \frac{7}{2}.
The yy-intercept is the value of the function when x=0x = 0.

Anahtar Kavram

Perpendicular Bisectors in the Coordinate Plane
Tahmini Süre:1m 30s
Soru 238Soru

On a coordinate map of a state park, a straight hiking trail begins at a campsite located at (2,3)(2, -3) and ends at a lookout point. A hiker walks along the trail at a constant pace, and after 33 hours, reaches a trail marker located at (11,9)(11, 9). What is the slope of the line on the coordinate map that represents this straight trail?

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Cevap: 43\frac{4}{3}

Cevap

43\frac{4}{3}
To find the slope of the line representing the trail, we identify the two points through which the line passes: (2,3)(2, -3) and (11,9)(11, 9). Applying the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}, we get m=9(3)112=129m = \frac{9 - (-3)}{11 - 2} = \frac{12}{9}. Simplifying the fraction by dividing the numerator and denominator by 33 yields the correct slope of 43\frac{4}{3}.

Adım Adım Çözüm

1
Identify the coordinates of the two points on the line representing the trail.
The two points are the campsite at (2,3)(2, -3) and the trail marker at (11,9)(11, 9).
To find the slope of a straight line, we need the coordinates of any two points that lie on the line.
2
Set up the slope formula.
m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}
The slope of a line is defined as the change in the yy-coordinates (rise) divided by the change in the xx-coordinates (run).
3
Substitute the coordinate values into the formula and simplify.
m=9(3)112=9+39=129m = \frac{9 - (-3)}{11 - 2} = \frac{9 + 3}{9} = \frac{12}{9}
Substituting the coordinates correctly handles the subtraction of the negative coordinate.
4
Reduce the fraction to simplest form.
m=43m = \frac{4}{3}
Dividing the numerator and the denominator by their greatest common divisor, 33, yields the final simplified slope value.

Anahtar Kavram

Calculating the slope of a line given two points on a coordinate plane
Soru 239Soru

An isosceles trapezoid ABCDABCD has vertices A(4,1)A(-4, -1), B(6,1)B(6, -1), and C(3,3)C(3, 3) in the standard (x,y)(x, y) coordinate plane. If the base ABAB is parallel to the x-axis, which of the following represents the coordinates of vertex DD?

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Cevap: (1,3)(-1, 3)

Cevap

(1,3)(-1, 3)
The correct answer is the coordinate pair representing (1,3)(-1, 3). Since the bases are parallel to the x-axis, they are horizontal, meaning DD must share the same y-coordinate as CC, which is 3. The vertical line of symmetry passes through the midpoint of the segment ABAB, which is at x=1x = 1. The vertex C(3,3)C(3, 3) is 2 units to the right of this line of symmetry, so the vertex DD must be 2 units to the left of the line of symmetry, giving an x-coordinate of 1-1.

Adım Adım Çözüm

1
Determine the orientation of the trapezoid and the y-coordinate of the missing vertex.
The base ABAB lies on the horizontal line y=1y = -1. Because the bases of a trapezoid are parallel, the second base CDCD must also be horizontal and lie on the line y=3y = 3. Therefore, the y-coordinate of vertex DD is 33.
Since the trapezoid's bases are parallel to the x-axis, they are horizontal lines, meaning vertices on the same base share the same y-coordinate.
2
Find the equation of the line of symmetry of the isosceles trapezoid.
The midpoint of the base ABAB has an x-coordinate of 4+62=1\frac{-4 + 6}{2} = 1. The vertical line of symmetry is x=1x = 1.
An isosceles trapezoid is symmetric with respect to the perpendicular bisector of its bases. For horizontal bases, this is a vertical line passing through the midpoint of the base.
3
Use the line of symmetry to find the x-coordinate of vertex DD.
The x-coordinate of C(3,3)C(3, 3) is 3, which is 31=23 - 1 = 2 units to the right of the line of symmetry x=1x = 1. Thus, the x-coordinate of vertex DD must be 2 units to the left of the line of symmetry: 12=11 - 2 = -1. Combining this with the y-coordinate gives D(1,3)D(-1, 3).
Symmetry requires that corresponding vertices on the opposite sides of the line of symmetry are equidistant from it.

Anahtar Kavram

Using symmetry and coordinate geometry properties of an isosceles trapezoid to determine the coordinates of a missing vertex.
Tahmini Süre:1m 30s
Soru 240Soru

In the standard (x,y)(x, y) coordinate plane, if a line LL is perpendicular to the line passing through the points (2,5)(2, 5) and (2,3)(2, -3), then the slope of line LL is 00.

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Cevap: True

Cevap

The statement is true because the line passing through the points (2,5)(2, 5) and (2,3)(2, -3) is vertical, and any line perpendicular to a vertical line is horizontal, which has a slope of 00.
The line passing through (2,5)(2, 5) and (2,3)(2, -3) has identical x-coordinates, which defines a vertical line. Any line perpendicular to a vertical line is horizontal, and horizontal lines have a slope of 00.

Adım Adım Çözüm

1
Identify the orientation of the line passing through the points (2,5)(2, 5) and (2,3)(2, -3).
Since both coordinates share the same x-value (x=2x = 2), the line is vertical.
In the standard coordinate plane, a line connecting points with identical x-coordinates is vertical and has an undefined slope.
2
Determine the orientation of line LL, which is perpendicular to this vertical line.
Line LL must be a horizontal line.
By geometric definition, a line perpendicular to a vertical line in the coordinate plane is horizontal.
3
Determine the slope of the horizontal line LL.
The slope of line LL is 00.
A horizontal line has no vertical change (rise = 00) as the horizontal position changes, resulting in a slope of 00.

Anahtar Kavram

The relationship between vertical and horizontal lines in the coordinate plane, specifically that perpendicular lines to vertical lines are horizontal and have a slope of 00.

Alternatif Yöntem

Write the equation of the line passing through (2,5)(2, 5) and (2,3)(2, -3), which is x=2x = 2. Any line perpendicular to x=2x = 2 must be in the form y=cy = c where cc is a constant. The slope of any line in the form y=cy = c is 00.
Tahmini Süre:1m 0s
ÖncekiSayfa 12 / 14Sonraki
Coordinate Geometry Alıştırma Soruları — ACT — Sayfa 12 | Examkin