The quadrilateral WXYZ has vertices W(3,-5) X(1,-3) Y(-1,-5) and Z(1,-7)

1 answer:

5 0

The preimage WXYZ is to undergo a rotation of 90° in the clockwise direction.
A 90° clockwise rotation leads to the following change in the coordinates;
(x,y) -> after rotation (y,-x)
this rule applies for all the vertices in the preimage
W - (3,-5) --> W' (-5,-3)
X - (1,-3) ---> X' (-3,-1)
Y - (-1,-5)---> Y' (-5,1)
Z - (1,-7) ---> Z' (-7,-1)
therefore the new coordinates are W' (-5,-3), X' (-3,-1), Y' (-5,1) and Z' (-7,-1)

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In each of the following ,make y the subject and hence find the value of y when a=2, b=3 ,and c=4.

zvonat [6]

Part i

$\dfrac{2a-7}{6y} = \dfrac{3b-5}{2c}$

$y = \dfrac{ 2c(2a-7)}{6(3b-5)} = \dfrac{c(2a-7)}{9b-15}$

Substituting,

$y = \dfrac{4(2(2)-7)}{9(3)-15} = \dfrac{ -12}{12} = -1$

Answer: -1

Part ii

I'm not sure that one's typed in correctly but I'll solve it as written.

$3-34y + 2a = \dfrac{3b-5}{2c + 1}$

$34y = 3+2a-\dfrac{3b-5}{2c + 1}$

$y = \frac 1 {34}\left(3+2a-\dfrac{3b-5}{2c + 1} \right)$

We're not asked to simplify it so I wont. Substituting,

$y = \frac 1 {34}\left(3+2(2)-\dfrac{3(3)-5}{2(4) + 1} \right) = \frac 1 {34}(7-4/9) = \dfrac{59}{306}$

Answer: 59/306

3 0

3 years ago

iren2701 [21]

Answer:

The distance between points A and B is 13

Step-by-step explanation:

We need to find distance between points A and B

Looking at the graph,

Point A : (2,6) and Point B: (-3,-6)

We need to find the distance between these points.

The formula used is: $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\$

We have Points A : (2,6) and B: (-3,-6)

So, $x_1=2, y_1=6, x_2=-3, y_2=-6$

Putting values in formula and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\Distance=\sqrt{(-3-2)^2+(-6-6)^2} \\Distance=\sqrt{(-5)^2+(-12)^2} \\Distance=\sqrt{25+144} \\Distance =\sqrt{169}\\Distance = 13$