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Sidana [21]
3 years ago
6

Triangle ABC is labeled with points A(5, 2), B(1, 2) and C(3, 6) on a coordinate plane. Find the coordinates of A' B' C' after a

reflection over the x-axis.
Mathematics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

After reflection over the x-axis, we have the coordinates as follows;

A’ (5,-2)

B’ ( 1,-2)

C’ (3,-6)

Step-by-step explanation:

Here, we want to find the coordinates A’ B’ and C’ after a reflection over the x-axis

By reflecting over the x-axis, the y-coordinate is bound to change in sign

So if we have a Point (x,y) and we reflect over the x-axis, the image of the point after reflection would turn to (x,-y)

We simply go on to negate the value of the y-coordinate

Mathematically if we apply these to the given points, what we get are the following;

A’ (5,-2)

B’ ( 1,-2)

C’ (3,-6)

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Answer:

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Using <em>factorization</em> method:

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Now, Making factors of 2y^{4} + 40y^{3} + 200y^{2}

Taking 2y^{2} common:

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Using <em>factorization</em> method:

\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10)        ............ (2)

The underlined parts show the Highest Common Factor(HCF).

i.e. <em>HCF</em> is 2y^{2} (y+ 10).

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HCF \times LCM = p \times q

Using equations <em>(1)</em> and <em>(2)</em>: \Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times  4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)

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