Write the slope intercept form of a line that passes<br> through (3, -2) and is 1 to y = x + 3.

a) The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively.

b) The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively.

<h3>How to perform transformations with points</h3>

a) A dillation centered at the origin is defined by following operation:

$P'(x,y) = k\cdot P(x,y)$ (1)

Where:

$P(x,y)$ - Original point
$P'(x,y)$ - Dilated point.

If we know that $k = \frac{1}{3}$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ , then the new points of the quadrilateral are:

$A'(x,y) = \frac{1}{3}\cdot (0,6)$

$A'(x,y) = (0, 2)$

$B'(x,y) = \frac{1}{3} \cdot (-3,-6)$

$B'(x,y) = (-1, -2)$

$C'(x,y) = \frac{1}{3}\cdot (-9,-6)$

$C'(x,y) = \left(-3,-2\right)$

$D'(x,y) = \frac{1}{3}\cdot (-12,-3)$

$D'(x,y) = (-4, -1)$

The points of the new quadrillateral are $A'(x,y) = (0, 2)$ , $B'(x,y) = (-1, -2)$ , $C'(x,y) = \left(-3,-2\right)$ and $D'(x,y) = (-4, -1)$ , respectively. $\blacksquare$

b) A translation along a vector is defined by following operation:

$P'(x,y) = P(x,y) +T(x,y)$ (2)

Where $T(x,y)$ is the transformation vector.

If we know that $T(x,y) = (-5,-1)$ , $A(x,y) = (0,6)$ , $B(x,y) = (-3,-6)$ , $C(x,y) = (-9, -6)$ and $D(x,y) = (-12, -3)$ ,

$A'(x,y) = (0,6) + (-5, -1)$

$A'(x,y) = (-5, 5)$

$B'(x,y) = (-3, -6) + (-5, -1)$

$B'(x,y) = (-8,-7)$

$C'(x,y) = (-9, -6) + (-5, -1)$

$C'(x,y) = (-13, -7)$

$D'(x,y) = (-12,-3)+(-5,-1)$

$D'(x,y) = (-17, -4)$

The points of the new quadrillateral are $A'(x,y) = (-5, 5)$ , $B'(x,y) = (-8,-7)$ , $C'(x,y) = (-13, -7)$ and $D'(x,y) = (-17, -4)$ , respectively. $\blacksquare$

To learn more on transformation rules, we kindly invite to check this verified question: brainly.com/question/4801277

7 0

2 years ago

Somebody pls help me!!!

Setler [38]

<h3>Answer: 60 units</h3>

=====================================================

Explanation:

The similar red marks on the angles tell us those angles are congruent. So they are the same measure.

The sides opposite those congruent angles are the same length. This means AC = BC. This triangle is isosceles.

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Solve for x

AC = BC

5x-8 = 2x+7 .... substitution

5x-8-2x = 7 ... subtract 2x from both sides

5x-2x = 7+8 ..... add 8 to both sides

3x = 15

x = 15/3 ... divide both sides by 3

x = 5

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Use that x value to compute each side length

AB = 3x+11 = 3*5+11 = 15+11 = 26

BC = 2x+7 = 2*5+7 = 10+7 = 17

CA = 5x-8 = 5*5-8 = 25-8 = 17

Note how BC and CA are the same length (17 units). This helps confirm we have the right x value.

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The last step is to add up all the side lengths to get the perimeter

perimeter of triangle = sum of all the sides

perimeter of triangle = side1 + side2 + side3

perimeter of triangle ABC = AB + BC + CA

perimeter of triangle ABC = 26 + 17 + 17

perimeter of triangle ABC = 60 units

4 0

2 years ago

Read 2 more answers

I’m actually so dumb!!!<br> How???<br> Answer???

Pani-rosa [81]

Standard form: 2x - 16 = 0

Factorization: 2(x - 8) = 0

Solutions: x = 16/2 = 8

Hope this helps!

6 0

3 years ago

Read 2 more answers

Write the slope intercept form of a line that passes through (3, -2) and is 1 to y = x + 3.