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3 years ago

Given the preimage points: find the coordinates

Mathematics

1 answer:

m_a_m_a [10]3 years ago

8 0

Question 1)

Answer:

Please see the explanation.

Step-by-step explanation:

We know that the dilation of an original object shows how much bigger or smaller the image will really get, as compared to the original shape.

Also,

if the value of scale factor > 1, the image will get bigger.

if the value of scale factor < 1, the image will get smaller.

The new coordinates of the image can be determined by multiplying the scale factor with the coordinates of the original object.

Given the preimage points:

A(0, 5)

B(4, 9)

C(-1, -8)

The formula of the dilation by a scale factor of 3:

(x, y) → (3x, 3y)

As the scale factor > 1, so the image will be bigger.

Therefore, the coordinates of the dilated image after the dilation by a scale factor of 3 will be:

A(x, y) → (3x, 3y) = A(0, 5) → (3(0), 3(5)) = A'(0, 15)

B(x, y) → (3x, 3y) = B(4, 9) → (3(4), 3(9)) = B'(12, 27)

C(x, y) → (3x, 3y) = C(-1, -8) → (3(-1), 3(-8)) = C'(-3, -24)

Question 2)

Answer:

Please see the explanation.

Step-by-step explanation:

Given the preimage points:

A(0, 5)

B(4, 9)

C(-1, -8)

The formula of the dilation by a scale factor of 1/2:

(x, y) → (1/2x, 1/2y)

As the scale factor < 1, so the image will be smaller.

Therefore, the coordinates of the dilated image after the dilation by a scale factor of 1/2 will be:

A(x, y) → (3x, 3y) = A(0, 5) → (1/2(0), 1/2(5)) = A'(0, 5/2)

B(x, y) → (3x, 3y) = B(4, 9) → (1/2(4), 1/2(9)) = B'(2, 9/2)

C(x, y) → (3x, 3y) = C(-1, -8) → (1/2(-1), 1/2(-8)) = C'(-1/2, -4)

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Determine the coordinates of the vertices of the triangle to compute the area of the triangle using the distance formula (round

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Answer:

1. D. $50\text{ units}^2$

2. D. 45 units.

Step-by-step explanation:

We have been two graphs.

1. To find the area of our given triangle we will use distance formula.

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Upon substituting coordinates of base line of our triangle we will get,

$\text{Base length of triangle}=\sqrt{(15-5)^2+(5-15)^2}$

$\text{Base length of triangle}=\sqrt{(10)^2+(-10)^2}$

$\text{Base length of triangle}=\sqrt{100+100}$

$\text{Base length of triangle}=\sqrt{200}$

$\text{Base length of triangle}=10\sqrt{2}$

Now let us find the height of triangle similarly.

$\text{Height of triangle}=\sqrt{(20-15)^2+(10-5)^2}$

$\text{Height of triangle}=\sqrt{(5)^2+(5)^2}$

$\text{Height of triangle}=\sqrt{25+25}$

$\text{Height of triangle}=\sqrt{50}$

$\text{Height of triangle}=5\sqrt{2}$

$\text{Area of triangle}=\frac{\text{Base*Height}}{2}$

$\text{Area of triangle}=\frac{10\sqrt{2}*5\sqrt{2}}{2}$

$\text{Area of triangle}=\frac{50*2}{2}$

$\text{Area of triangle}=50$

Therefore, area of our given triangle is 50 square units and option D is the correct choice.

2. Using distance formula we will find the length of large side of triangle as:

$\text{Large side of rectangle}=\sqrt{(14-1)^2+(21-8)^2}$

$\text{Large side of rectangle}=\sqrt{(13)^2+(13)^2}$

$\text{Large side of rectangle}=\sqrt{169+169}$

$\text{Large side of rectangle}=\sqrt{338}$

$\text{Large side of rectangle}=13\sqrt{2}$

$\text{Small side of rectangle}=\sqrt{(4-1)^2+(5-8)^2}$

$\text{Small side of rectangle}=\sqrt{(3)^2+(-3)^2}$

$\text{Small side of rectangle}=\sqrt{9+9}$

$\text{Small side of rectangle}=\sqrt{18}$

$\text{Small side of rectangle}=3\sqrt{2}$

$\text{Perimeter of rectangle}=2(\text{Length + Width)}$

$\text{Perimeter of rectangle}=2(13\sqrt{2}+3\sqrt{2}}$

$\text{Perimeter of rectangle}=2(16\sqrt{2}}$

$\text{Perimeter of rectangle}=32\sqrt{2}$

$\text{Perimeter of rectangle}=45.2548339959390416\approx 45$

Therefore, the perimeter of our given rectangle is 45 units and option D is the correct choice.

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