Prove Simple Geometric Theorems by using Coordinates: CCSS.Math.Content.HSG-GPE.B.4 - Common Core: High School - Geometry

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Question

True or False:

The point lies on a circle with a center of and a radius .

Answer

In order to solve this problem, we need to write out the general equation of a circle.

$\left(- h + x\right)^{2} + \left(- k + y\right)^{2} = r^{2}$

Now let's substitute for , , and .

$\left(x - 8\right)^{2} + \left(y + 15\right)^{2} = 4$

Now we can plug in the point, and see if it lies on the circle.

The point does not lie on the circle, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the ellipse

$\frac{1}{15}$ $\left(x - 2\right)^{2} + \frac{1}{11} \left(y + 1\right)^{2} = 1$

Answer

In order to solve this problem, we need to plug in the point, and see if both sides of the equation are equal.

$\frac{1}{15}$ $\left(x - 2\right)^{2} + \frac{1}{11} \left(y + 1\right)^{2} = 1$

$\frac{1}{15}$ $\left(2 - 2\right)^{2} + \frac{1}{11} \left(2 + 1\right)^{2} = 1$

$\frac{1}{15}$ $\cdot 0 + \frac{1}{11} \cdot 9= 1$

$\frac{9}{11} = 1$

The point does not lie on the ellipse, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the ellipse

$\frac{1}{9} \left(x - 4\right)^{2} + \frac{1}{12} \left(y + 1\right)^{2} = 1$

Answer

In order to solve this problem, we need to plug in the point, and see if both sides of the equation are equal.

$\frac{1}{9} \left(x - 4\right)^{2} + \frac{1}{12} \left(y + 1\right)^{2} = 1$

$\frac{1}{9} \left(3 - 4\right)^{2} + \frac{1}{12} \left(2 + 1\right)^{2} = 1$

$\frac{1}{9} \left(-1\right)^{2} + \frac{1}{12} \left(3\right)^{2} = 1$

$\frac{1}{9} + \frac{1}{12} \cdot 9 = 1$

$\frac{1}{9} + \frac{9}{12} = 1$

$\frac{31}{36} = 1$

The point does not lie on the ellipse, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the line

Answer

In order to solve this problem, we need to plug in the point and see if both sides of the equation are equal or not.

$\8 = 4 (10) + 7 \8=40+7$

The point does not lie in the line, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on a circle with a center of and a radius .

Answer

In order to solve this problem, we need to write out the general equation of a circle.

$\left(- h + x\right)^{2} + \left(- k + y\right)^{2} = r^{2}$

Now let's substitute for , , and .

$\left(x + 9\right)^{2} + \left(y - 10\right)^{2} = 9$

Now we can plug in the point, and see if it lies on the circle.

The point does not lie on the circle, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the line

Answer

In order to solve this problem, we need to plug in the point and see if both sides of the equation are equal or not.

$\24 = 9 (4) + 5 \24=36+5$

The point does not lie in the line, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the ellipse

$\frac{1}{6} \left(x - 1\right)^{2} + \frac{1}{10} \left(y + 1\right)^{2} = 1$

Answer

In order to solve this problem, we need to plug in the point, and see if both sides of the equation are equal.

$\frac{1}{6} \left(x - 1\right)^{2} + \frac{1}{10} \left(y + 1\right)^{2} = 1$

$\frac{1}{6} \left(9 - 1\right)^{2} + \frac{1}{10} \left(10 + 1\right)^{2} = 1$

$\frac{1}{6} \left(8\right)^{2} + \frac{1}{10} \left(11\right)^{2} = 1$

$\frac{1}{6} (64) + \frac{1}{10} (121) = 1$

$\frac{683}{30} = 1$

The point does not lie on the ellipse, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on a circle with a center of and a radius .

Answer

In order to solve this problem, we need to write out the general equation of a circle.

$\left(- h + x\right)^{2} + \left(- k + y\right)^{2} = r^{2}$

Now let's substitute for , , and .

$\left(x - 9\right)^{2} + \left(y - 19\right)^{2} = 1$

Now we can plug in the point, and see if it lies on the circle.

The point does not lie on the circle, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the ellipse

$\frac{1}{6} \left(x - 1\right)^{2} + \frac{1}{14} \left(y + 3\right)^{2} = 1$

Answer

In order to solve this problem, we need to plug in the point, and see if both sides of the equation are equal.

$\frac{1}{6} \left(x - 1\right)^{2} + \frac{1}{14} \left(y + 3\right)^{2} = 1$

$\frac{1}{6} \left(1 - 1\right)^{2} + \frac{1}{14} \left(7 + 3\right)^{2} = 1$

$\frac{1}{14} \left(100\right) = 1$

$\frac{50}{7} = 1$

The point does not lie on the ellipse, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the line

Answer

In order to solve this problem, we need to plug in the point and see if both sides of the equation are equal or not.

$\34 = 3 (7) + 8 \34=21+8$

The point does not lie in the line, because each side is not equal to each other. So our answer is False.

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Question

True or False:

The point lies on the line

Answer

In order to solve this problem, we need to plug in the point and see if both sides of the equation are equal or not.

The point does not lie in the line, because each side is not equal to each other. So our answer is False.

Compare your answer with the correct one above

Question

True or False:

The point lies on the ellipse

$\frac{1}{5} \left(x - 6\right)^{2} + \frac{1}{6} \left(y - 6\right)^{2} = 1$

Answer

In order to solve this problem, we need to plug in the point, and see if both sides of the equation are equal.

$\frac{1}{5} \left(x - 6\right)^{2} + \frac{1}{6} \left(y - 6\right)^{2} = 1$

$\frac{1}{5} \left(7 - 6\right)^{2} + \frac{1}{6} \left(4 - 6\right)^{2} = 1$

$\frac{1}{5} \left(1\right)^{2} + \frac{1}{6} \left(-2\right)^{2} = 1$

$\frac{13}{15} = 1$

The point does not lie on the ellipse, because each side is not equal to each other. So our answer is False.

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