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Alexeev081 [22]

3 years ago

9

triangle ABC has the vertices A, (-6,2) B (-9,1), C (1,-6). What would be the vertices of ABC' after a reflection across the x-a

xis

1 answer:

brilliants [131]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the x- axis

a point (x, y) → (x, - y), hence

A(- 6, 2) → A'(- 6, - 2)

B(- 9, 1) → B'(- 9, - 1)

C(1, - 6) → C'(1, 6)

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Solving for <em>Angles</em>

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Solving for <em>Edges</em>

$\displaystyle b^2 + a^2 - 2ba\:cos∠C = c^2 \\ c^2 + a^2 - 2ca\:cos∠B = b^2 \\ c^2 + b^2 - 2cb\:cos∠A = a^2$

You would use this law under <em>two</em> conditions:

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_____________________________________________

Now, JUST IN CASE, you would use the Law of Sines under <em>three</em> conditions:

Two angles and one edge defined, while trying to solve for the <em>second edge</em>
One angle and two edges defined, while trying to solve for the <em>second angle</em>
ALL three angles defined [<em>of which does not occur very often, but it all refers back to the first bullet</em>]

* I HIGHLY suggest you keep note of all of this significant information. You will need it going into the future.

I am delighted to assist you at any time.

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