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astra-53 [7]
4 years ago
5

Given triangle ABC with coordinates A(−2, 7), B(−2, 4), and C(−4, 3), and its image A′B′C′ with A′(2, 3), B′(−1, 3), and C′(−2,

1), find the line of reflection.
The line of reflection is at y =
Mathematics
1 answer:
Setler [38]4 years ago
6 0

Answer:

y = x + 5

Step-by-step explanation:

given an obtuse scalene triangle (∆ABC) with vertices A(-2,7), B(-2,4), and C(-4,3) (representing the preimage of the reflection). And congruent triangle (∆A'B'C') with vertices A'(2,3), B'(-1,3), and C'(-2,1) (representing the image of the reflection / triangle after the reflection).

The easiest way to look at this is by taking the inverse of each point on the preimage (swapping y and x coordinates) and the adding 5 to the y and subtracting 5 from the x to get the image.

You can test this, and you will see that this will match with vertices of the image.

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Check the picture below.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
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