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Aneli [31]
3 years ago
8

A figure is dilated by a scale factor of 2.5. If the origin is the center of

Mathematics
1 answer:
katen-ka-za [31]3 years ago
3 0

Answer:

C.

Step-by-step explanation:

So, if the new image coordinates are (7,9) and it is dilated from the origin there is a rule. If it is dilated from the origin you would multiply the pre-image coordinates by the scale factor so, (kx,ky) where k is the scale factor. In this case, we would divide by 2.5 so 7 divided by 2.5 is 2.8 and 9 divided by 2.5 is 3.6. The correct answer would be C. (2.8,3.6).

Hope this Helps!!! PLZ MARK BRAINLIEST!!!

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Find the locus of a point such that the sum of its distance from the point ( 0 , 2 ) and ( 0 , -2 ) is 6.
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Answer:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Step-by-step explanation:

We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.

First, we will need the distance formula, given by:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let the point on the locus be P(x, y).

So, the distance from P to (0, 2) will be:

\begin{aligned} d_1&=\sqrt{(x-0)^2+(y-2)^2}\\\\ &=\sqrt{x^2+(y-2)^2}\end{aligned}

And, the distance from P to (0, -2) will be:

\displaystyle \begin{aligned} d_2&=\sqrt{(x-0)^2+(y-(-2))^2}\\\\ &=\sqrt{x^2+(y+2)^2}\end{aligned}

So sum of the two distances must be 6. Therefore:

d_1+d_2=6

Now, by substitution:

(\sqrt{x^2+(y-2)^2})+(\sqrt{x^2+(y+2)^2})=6

Simplify. We can subtract the second term from the left:

\sqrt{x^2+(y-2)^2}=6-\sqrt{x^2+(y+2)^2}

Square both sides:

(x^2+(y-2)^2)=36-12\sqrt{x^2+(y+2)^2}+(x^2+(y+2)^2)

We can cancel the x² terms and continue squaring:

y^2-4y+4=36-12\sqrt{x^2+(y+2)^2}+y^2+4y+4

We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:

-8y=36-12\sqrt{x^2+(y+2)^2}

We can divide both sides by -4:

2y=-9+3\sqrt{x^2+(y+2)^2}

Adding 9 to both sides yields:

2y+9=3\sqrt{x^2+(y+2)^2}

And, we will square both sides one final time.

4y^2+36y+81=9(x^2+(y^2+4y+4))

Distribute:

4y^2+36y+81=9x^2+9y^2+36y+36

The 36y will cancel. So:

4y^2+81=9x^2+9y^2+36

Subtracting 4y² and 36 from both sides yields:

9x^2+5y^2=45

And dividing both sides by 45 produces:

\displaystyle \frac{x^2}{5}+\frac{y^2}{9}=1

Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.

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