Select the correct answer from each drop-down menu.
1 answer:
Answer:
A non-equilateral rhombus.
Step-by-step explanation:
We can solve this graphically.
We start with square:
ABCD
with:
A = (11, - 7)
B = (9, - 4)
C = (11, - 1)
D = (13, - 4)
Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:
AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13
BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13
CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13
DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13
And we change C by C' = (11, 1)
In the image you can see the 5 points and the figure that they make:
The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.
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Answer:
(- 2, - 1 ), (2, 1 )
Step-by-step explanation:
Given the 2 equations
x - 2y = 0 → (1)
x² - y² = 3 → (2)
Rearrange (1) expressing x in terms of y by adding 2y to both sides
x = 2y → (3)
Substitute x = 2y into (2)
(2y)² - y² = 3, that is
4y² - y² = 3
3y² = 3 ( divide both sides by 3 )
y² = 1 ( take the square root of both sides )
y = ± = ± 1
Substitute these values into (3) for corresponding values of x
y = - 1 ⇒ x = 2(- 1) = - 2 ⇒ (- 2, - 1 )
y = 1 ⇒ x = 2(1) = 2 ⇒ (2, 1 )