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MArishka [77]
2 years ago
12

What is the reflection of the point (−2, 3) across the y-axis?

Mathematics
1 answer:
vampirchik [111]2 years ago
4 0

Answer:

  • distance to x-axis: 3 units
  • distance to y-axis: 2 units
  • the reflected point is (2, 3)

Step-by-step explanation:

The given point is (-2, 3), labeled point A in the attachment.

As you know, the distance from the x-axis is given by the y-coordinate. Here, that distance is ...

  3 units from the x-axis

__

The distance from the y-axis is given by the magnitude of the x-coordinate. Here, that distance is ...

  2 units from the y-axis

Note that the negative coordinate value means the point is located that distance to the left of the y-axis.

__

Reflecting the point across the y-axis means choosing a point that is the same distance (2 units) right of the axis instead of left of the axis. The sign of the x-coordinate value will be positive, instead of negative. (The y-coordinate remains unchanged,)

__

The reflected point is (2, 3).

_____

<em>Additional comment</em>

As we have seen, reflection across the y-axis changes the sign of the x-coordinate:

  (x, y) ⇒ (-x, y) . . . . reflection across the y-axis

Similarly, reflection across the x-axis changes the sigh of the y-coordinate:

  (x, y) ⇒ (x, -y) . . . . reflection across the x-axis

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We have these points: A(0,0) and B(6,3); its distance will be:

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We need to simplify \frac{ \sqrt{14x^3} }{ \sqrt{18x} }

First lets factor \sqrt{14x^3}

\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3}
\sqrt{14} =  \sqrt{2} \sqrt{7} by applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a} \sqrt[n]{b}
\sqrt{x^3} = x^{3/2} By applying the radical rule \sqrt[n]{x^m} = x^{m/n}

So
\sqrt{14x^3} = \sqrt{14}  \sqrt{x^3} = \sqrt{2} \sqrt{7}x^{3/2}

Now let's factor \sqrt{18x}
By applying the radical rule \sqrt[n]{ab} =  \sqrt[n]{a}  \sqrt[n]{b},
\sqrt{18x} =  \sqrt{18} \sqrt{x}
\sqrt{18} =  \sqrt{2} * 3

So \sqrt{18x} = \sqrt{2}*3 \sqrt{x}

So  \frac{ \sqrt{14x^3} }{ \sqrt{18x} } = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}  }

We know that \sqrt[n]{x} = x^{1/n} so \sqrt{x} = x^{1/2}

We now have \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}

We know that \frac{x^a}{x^b} = x^{a-b}
So \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x

We now got \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}&#10;

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} =   \frac{ \sqrt{7}x }{3}


All in All, we get \frac{ \sqrt{14x^3} }{ \sqrt{18x} } =  \frac{ \sqrt{7}x }{3}

Hope this helps! :D


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