Mathematics: A triangle has vertices at (2, 3), (−2, 4), and (−4, 5). What are the coordinates of the vertices of the image after the translation (x, y) arrowright (x + 2, y − 1)?

A triangle has vertices at (2, 3), (−2, 4), and, №13638533, 30.03.2022 17:50

Mathematics

Question 11 pts (04.01 LC) The coordinates of point A on a grid are (2, -5). Point A is reflected across the x-axis to obtain point B. The coordinates of point B are (2, ___). (1 point) Flag this Question Question 21 pts (04.01 LC) The map shows the location of four places in a city: Anna s house is in the same quadrant as the museum. Which of the following could be the coordinates of Anna s house? (1 point) Group of answer choices (1, 4) (−1, 4) (1, −4) (−1, −4) Flag this Question Question 31 pts (04.01 MC) On a coordinate grid, point P is at

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11) (2, 5);
21) (−1, 4);
31) Q(5, −8), S(−1, 3);
41) S is R reflected across the x-axis; only the signs of the y-coordinates of R and S are different;
51) Coordinate grid shown from negative 2 to positive 2 on the x-axis and negative 2 to positive 2 on the y-axis in increments of 1 over 4. Only the whole numbers are labeled. A point A is shown at the intersection of 2 grid lines to the right of the y-axis and 5 grid lines below the x-axis;
61) -4/8;
71) A coordinate grid from negative 3 to positive 3 on both axes is drawn in increments of 1 over 2. Point A is plotted 2 grid lines to the right of the y-axis and 4 grid lines above the x-axis. Point B is plotted at 6 grid lines to the right of the y-axis and 1 grid line below the x-axis. Point C is plotted at 6 grid lines to the left of the y-axis and 1 grid line below the x-axis;
81) (0.4, −1.8);
91) 7;
101) On the 3rd graph;
111) If the number of cups of raisins is 60, then the number of cups of cereal will be 20.

Mathematics

1. a triangle with vertices at a(20, –30), b(10, –15), and c(5, –20) has been dilated with a center of dilation at the origin. the image of b, point b , has the coordinates (2, –3). what is the scale factor of the dilation? 110 15 5 10 2. a triangle has vertices at a(–2, 4), b(–2, 8), and c(6, 4). if a has coordinates of (–0.25, 0.5) after the triangle has been dilated with a center of dilation about the origin, which statements are true? check all that apply. the coordinates of c are (0.75, 0.5). the coordinates of c are (1.5, 1). the scale factor

Step-by-step answer

P Answered by Master

27

Question (1):
The scale factor is $\frac{1}{5}$

Question (2):
scale factor = $\frac{1}{8}$
coordinates of C' are (0.75 , 0.5)
coordinates of B' are (-0.25,1)

Question (3):
The scale factor is 4

Question (4):
Rotation, translation, reflection

Exact steps and explanation for each question are shown in the attached images.

Hope this helps :)
1. a triangle with vertices at a(20, –30), b(10, –15), and c(5, –20) has been dilated with a center

1. a triangle with vertices at a(20, –30), b(10, –15), and c(5, –20) has been dilated with a center

Mathematics

Find the coordinates of the vertices of the figure after the given transformation, translation: (x, y)→(x - 4, y + 4) a) j (1, 0), n (5, 0), m (5, 3), p (2, 5) b) j (-2, 5), n (2, 5), m (2, 2), p (-1, 0) c) j (-3, 4), n (1, 4), m (1, 1), p (-2, -1) d) j (-1, 0), n (-5, 0), m (-5, -3), p (-2, -5)

Step-by-step answer

P Answered by PhD

5

Option C J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

Step-by-step explanation:

we know that

The rule of the translation is

(x,y) > (x-4,y+4)

That means > The translation is 4 units at left and 4 units up

Find out the coordinates of the pre-image

Observing the figure

J(1,0),N(5,0),M(5,-3),P(2,-5)

Find out the coordinates of the image

Applying the rule of the translation at the coordinates of pre-image

J(1,0) > J'(1-4,0+4)

J(1,0) > J'(-3,4)

N(5,0) ---> N'(5-4,0+4)

N(5,0) ---> N'(1,4)

M(5,-3)> M'(5-4,-3+4)

M(5,-3)> M'(1,1)

P(2,-5) > P'(2-4,-5+4)

P(2,-5) > P'(-2,-1)

therefore

The coordinates of the image (after the transformation) are

J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

Mathematics

Find the coordinates of the vertices of the figure after the given transformation, translation: (x, y)→(x - 4, y + 4) a) j (1, 0), n (5, 0), m (5, 3), p (2, 5) b) j (-2, 5), n (2, 5), m (2, 2), p (-1, 0) c) j (-3, 4), n (1, 4), m (1, 1), p (-2, -1) d) j (-1, 0), n (-5, 0), m (-5, -3), p (-2, -5)

Step-by-step answer

P Answered by PhD

5

Option C J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

Step-by-step explanation:

we know that

The rule of the translation is

(x,y) > (x-4,y+4)

That means > The translation is 4 units at left and 4 units up

Find out the coordinates of the pre-image

Observing the figure

J(1,0),N(5,0),M(5,-3),P(2,-5)

Find out the coordinates of the image

Applying the rule of the translation at the coordinates of pre-image

J(1,0) > J'(1-4,0+4)

J(1,0) > J'(-3,4)

N(5,0) ---> N'(5-4,0+4)

N(5,0) ---> N'(1,4)

M(5,-3)> M'(5-4,-3+4)

M(5,-3)> M'(1,1)

P(2,-5) > P'(2-4,-5+4)

P(2,-5) > P'(-2,-1)

therefore

The coordinates of the image (after the transformation) are

J'(-3, 4), N'(1, 4), M'(1, 1), P'(-2, -1)

Mathematics

1. a triangle with vertices at a(20, –30), b(10, –15), and c(5, –20) has been dilated with a center of dilation at the origin. the image of b, point b , has the coordinates (2, –3). what is the scale factor of the dilation? 110 15 5 10 2. a triangle has vertices at a(–2, 4), b(–2, 8), and c(6, 4). if a has coordinates of (–0.25, 0.5) after the triangle has been dilated with a center of dilation about the origin, which statements are true? check all that apply. the coordinates of c are (0.75, 0.5). the coordinates of c are (1.5, 1). the scale factor

Step-by-step answer

P Answered by Specialist

27

Question (1):
The scale factor is $\frac{1}{5}$

Question (2):
scale factor = $\frac{1}{8}$
coordinates of C' are (0.75 , 0.5)
coordinates of B' are (-0.25,1)

Question (3):
The scale factor is 4

Question (4):
Rotation, translation, reflection

Exact steps and explanation for each question are shown in the attached images.

Hope this helps :)
1. a triangle with vertices at a(20, –30), b(10, –15), and c(5, –20) has been dilated with a center

Mathematics

[3 marks) 9. find the value of k, if the points a(2,3) b(4, k) & c(6,3) are collinear. 10. find the perimeter of a triangle whose vertices have the coordinates i. (3, 10) (5,2) & (4, 12) ii. (-2, 1) (4,6) & (6,3) 11. find the area of a triangle whose vertices are i. (1, 1), (-4,6) & (-3,-5) ii. a (2,3), b(4,4) & c(2,6) iii. (2,-2), (-2,1) & (5,2)

Step-by-step answer

P Answered by PhD

1

9. We would normally use point-point form to determine three points on a line. Here A and C have the same y coordinate, so we skip all that and just write K=3.

10. Perimeters given coordinates can get messy because of the square roots

i. A(3,10), B(5,2), C(4,12)

AB² = (5-3)² + (2-10)² = 68 = 2²·17

BC²= (4-5)² + (12-2)² = 101

AC² = (4-3)² + (12-10)² = 5

Perimeter: 2√17 + √101 + √5

ii. A(-2,1), B(4,6), C(6,3)

AB² = (4 - - 2)² + (6-1)² = 61

BC²= (6-4)² + (3-6)² = 13

AC² = (6 - -2)² + (3-1)² = 68

Perimeter: √61 + √13 + 2√17

I'll leave any calculator approximating to you; I prefer exact answers.

11.

When the triangles have a common x or y coordinate we can just do half base times height. The first and the third question here don't work that way. We can use the Shoelace Formula which says the signed area is half the sum of the cross products of the sides, true for any polygon.

Actually let's translate the first vertex to the origin because the area of triangle with vertices (0,0), (a,b), (c,d) is (1/2)(ad - bc)

i. Translating (1,1), (-4,6), (-3,-5) becomes (0,0), (-5,5), (-4,-6) so area

(1/2)( -5(-6) - 5(-4) ) = (1/2)(50) = 25

25

ii. We have common coordinate, one side parallel to an axis, which makes things easier. We have a base along x=2 of length b=6-3=3. There's an altitude h=4 - 2 = 2, so area is (1/2)(3)(2) = 3.

iii. (2,-2), (-2,1) & (5,2)

No common coordinate. Let's try the full shoelace:

Δ = (1/2)( 2(1) - -2(-2) + (-2)(2) - 1(5) + 5(-2) - 2(2) ) = -25/2

The Shoelace Formula gives signed area, which may be negative. The area is the absolute value: 25/2

Mathematics

Trapezoid wklx has vertices w(2, −3), k(4, −3), l(5, −2), and x(1, −2). trapezoid wklx was reflected across the y-axis to produce trapezoid wꞌkꞌlꞌxꞌ. which coordinates describe the vertices of the image? a. wꞌ(2, 3), kꞌ(4, 3), lꞌ(5, 2), and xꞌ(1, 2) b. wꞌ(−2, −3), kꞌ(−4, −3), lꞌ(−5, −2), and xꞌ(−1, −2) c. wꞌ(−3, −2), kꞌ(−3, −4), lꞌ(−2, −5), and xꞌ(−2, −1) d. wꞌ(–3, 2), kꞌ(4, –3), lꞌ(–2, 5), and xꞌ(–2, 1)

Step-by-step answer

P Answered by Master

13

B because when you reflect over the y axis, u make the X value the opposite sign (if negative make positive, if positive make negative)

Mathematics

[3 marks) 9. find the value of k, if the points a(2,3) b(4, k) & c(6,3) are collinear. 10. find the perimeter of a triangle whose vertices have the coordinates i. (3, 10) (5,2) & (4, 12) ii. (-2, 1) (4,6) & (6,3) 11. find the area of a triangle whose vertices are i. (1, 1), (-4,6) & (-3,-5) ii. a (2,3), b(4,4) & c(2,6) iii. (2,-2), (-2,1) & (5,2)

Step-by-step answer

P Answered by PhD

1

9. We would normally use point-point form to determine three points on a line. Here A and C have the same y coordinate, so we skip all that and just write K=3.

10. Perimeters given coordinates can get messy because of the square roots

i. A(3,10), B(5,2), C(4,12)

AB² = (5-3)² + (2-10)² = 68 = 2²·17

BC²= (4-5)² + (12-2)² = 101

AC² = (4-3)² + (12-10)² = 5

Perimeter: 2√17 + √101 + √5

ii. A(-2,1), B(4,6), C(6,3)

AB² = (4 - - 2)² + (6-1)² = 61

BC²= (6-4)² + (3-6)² = 13

AC² = (6 - -2)² + (3-1)² = 68

Perimeter: √61 + √13 + 2√17

I'll leave any calculator approximating to you; I prefer exact answers.

11.

When the triangles have a common x or y coordinate we can just do half base times height. The first and the third question here don't work that way. We can use the Shoelace Formula which says the signed area is half the sum of the cross products of the sides, true for any polygon.

Actually let's translate the first vertex to the origin because the area of triangle with vertices (0,0), (a,b), (c,d) is (1/2)(ad - bc)

i. Translating (1,1), (-4,6), (-3,-5) becomes (0,0), (-5,5), (-4,-6) so area

(1/2)( -5(-6) - 5(-4) ) = (1/2)(50) = 25

25

ii. We have common coordinate, one side parallel to an axis, which makes things easier. We have a base along x=2 of length b=6-3=3. There's an altitude h=4 - 2 = 2, so area is (1/2)(3)(2) = 3.

iii. (2,-2), (-2,1) & (5,2)

No common coordinate. Let's try the full shoelace:

Δ = (1/2)( 2(1) - -2(-2) + (-2)(2) - 1(5) + 5(-2) - 2(2) ) = -25/2

The Shoelace Formula gives signed area, which may be negative. The area is the absolute value: 25/2

Mathematics

Trapezoid wklx has vertices w(2, −3), k(4, −3), l(5, −2), and x(1, −2). trapezoid wklx was reflected across the y-axis to produce trapezoid wꞌkꞌlꞌxꞌ. which coordinates describe the vertices of the image? a. wꞌ(2, 3), kꞌ(4, 3), lꞌ(5, 2), and xꞌ(1, 2) b. wꞌ(−2, −3), kꞌ(−4, −3), lꞌ(−5, −2), and xꞌ(−1, −2) c. wꞌ(−3, −2), kꞌ(−3, −4), lꞌ(−2, −5), and xꞌ(−2, −1) d. wꞌ(–3, 2), kꞌ(4, –3), lꞌ(–2, 5), and xꞌ(–2, 1)

Step-by-step answer

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7

The correct answer for the question that is being presented above is this one: "C. Wꞌ(−3, −2), Kꞌ(−3, −4), Lꞌ(−2, −5), and Xꞌ(−2, −1)." Trapezoid WKLX has vertices W(2, −3), K(4, −3), L(5, −2), and X(1, −2). Trapezoid WKLX was reflected across the y-axis to produce trapezoid WꞌKꞌLꞌXꞌ.

Mathematics

Trapezoid wklx has vertices w(2, −3), k(4, −3), l(5, −2), and x(1, −2). trapezoid wklx was reflected across the y-axis to produce trapezoid wꞌkꞌlꞌxꞌ. which coordinates describe the vertices of the image? a. wꞌ(2, 3), kꞌ(4, 3), lꞌ(5, 2), and xꞌ(1, 2) b. wꞌ(−2, −3), kꞌ(−4, −3), lꞌ(−5, −2), and xꞌ(−1, −2) c. wꞌ(−3, −2), kꞌ(−3, −4), lꞌ(−2, −5), and xꞌ(−2, −1) d. wꞌ(–3, 2), kꞌ(4, –3), lꞌ(–2, 5), and xꞌ(–2, 1)

Step-by-step answer

P Answered by Specialist

13

B because when you reflect over the y axis, u make the X value the opposite sign (if negative make positive, if positive make negative)

A triangle has vertices at (2, 3), (−2, 4), and (−4, 5). What are the coordinates of the vertices of the image after the translation (x, y) arrowright (x + 2, y − 1)?

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