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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

What is the equation of the circle shown below?

Plot11

Possible Answers:

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

$\left(x - 2\right)^{2} + \left(y + 5\right)^{2} = 36$

$\left(x - 2\right)^{2} + \left(y - 5\right)^{2} = 36$

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

$\left(x + 2\right)^{2} + \left(y + 5\right)^{2} = 36$

Correct answer:

$\left(x + 2\right)^{2} + \left(y + 5\right)^{2} = 36$

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,

Plot11

we can see that the center is at $\left ( -2, -5\right )$

The next step it to find the radius. From looking at the picture, we can see that the radius is 6. With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$

We plug in the values, we get

$\left(x + 2\right)^{2} + \left(y + 5\right)^{2} = 36$

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Example Question #12 : Expressing Geometric Properties With Equations

What is the equation of the circle shown below?

Plot12

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

In order to find the equation, we must find the coordinates of the center of the circle.

If we look at the picture,
Plot12
we can see that the center is at $\left ( -2, 2\right )$ .

The next step it to find the radius. From looking at the picture, we can see that the radius is 2.

With this information, we can plug it into the general circle equation.

The general circle equation is

$\left(- a + x\right)^{2} + \left(- b + y\right)^{2} = c^{2}$

Now we substitute for $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$
We plug in the values, we get

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

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Example Question #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\textup{focus}= \left ( 1, 10\right )$

$\textup{directrix } y = 7$

Possible Answers:

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$

$\left(x - 2\right)^{2}$

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$

Correct answer:

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{6} - \frac{x_{0}}{3} + \frac{26}{3}$

So our answer is then

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

Report an Error

Example Question #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and the directrix are as follows.

$\\\text{focus}= \left ( 1, 10\right ) \\\text{directrix } y = 7$

Possible Answers:

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$

$\left(x - 2\right)^{2}$

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

Correct answer:

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{6} - \frac{x_{0}}{3} + \frac{26}{3}$

So our answer is then

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

Report an Error

Example Question #3 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 1, 10\right ) \\\text{directrix } y = 7$

Possible Answers:

$y = \frac{x^{2}}{24} - \frac{x}{12} + \frac{13}{6}$

$\left(x - 2\right)^{2}$

$y = \frac{x^{2}}{12} - \frac{x}{6} + \frac{13}{3}$

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

$\left(x - 10\right)^{2} + \left(x - 1\right)^{2}$

Correct answer:

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 1 for a 10 for b and 7 for y

$y_{0}^{2} - 14 y_{0} + 49 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} - 20 y_{0} + 101$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{6} - \frac{x_{0}}{3} + \frac{26}{3}$

So our answer is then

$y = \frac{x^{2}}{6} - \frac{x}{3} + \frac{26}{3}$

Report an Error

Example Question #4 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -10, 4\right ) \\ \text{directrix } y = -11$

Possible Answers:

$y = \frac{x^{2}}{60} + \frac{x}{3} - \frac{1}{12}$

$\left(x - 4\right)^{2} + \left(x + 10\right)^{2}$

$\left(x - 2\right)^{2}$

$y = \frac{x^{2}}{30} + \frac{2 x}{3} - \frac{1}{6}$

$y = \frac{x^{2}}{120} + \frac{x}{6} - \frac{1}{24}$

Correct answer:

$y = \frac{x^{2}}{30} + \frac{2 x}{3} - \frac{1}{6}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -10 for a 4 for b and -11 for y

$y_{0}^{2} + 22 y_{0} + 121 = x_{0}^{2} + 20 x_{0} + y_{0}^{2} - 8 y_{0} + 116$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{30} + \frac{2 x_{0}}{3} - \frac{1}{6}$

So our answer is then

$y = \frac{x^{2}}{30} + \frac{2 x}{3} - \frac{1}{6}$

Report an Error

Example Question #5 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 6, -9\right ) \\ \text{directrix } y = -5$

Possible Answers:

$y = - \frac{x^{2}}{16} + \frac{3 x}{4} - \frac{23}{4}$

$\left(x - 2\right)^{2}$

$y = - \frac{x^{2}}{8} + \frac{3 x}{2} - \frac{23}{2}$

$y = - \frac{x^{2}}{32} + \frac{3 x}{8} - \frac{23}{8}$

$\left(x - 6\right)^{2} + \left(x + 9\right)^{2}$

Correct answer:

$y = - \frac{x^{2}}{8} + \frac{3 x}{2} - \frac{23}{2}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 6 for a -9 for b and -5 for y

$y_{0}^{2} + 10 y_{0} + 25 = x_{0}^{2} - 12 x_{0} + y_{0}^{2} + 18 y_{0} + 117$

Now we can simplify, and solve for $y_{0}$

$y_{0} = - \frac{x_{0}^{2}}{8} + \frac{3 x_{0}}{2} - \frac{23}{2}$

So our answer is then

$y = - \frac{x^{2}}{8} + \frac{3 x}{2} - \frac{23}{2}$

Report an Error

Example Question #6 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 1, -6\right ) \\\text{directrix} y = -19$

Possible Answers:

$\left(x - 2\right)^{2}$

$\left(x - 1\right)^{2} + \left(x + 6\right)^{2}$

$y = \frac{x^{2}}{104} - \frac{x}{52} - \frac{81}{26}$

$y = \frac{x^{2}}{52} - \frac{x}{26} - \frac{81}{13}$

$y = \frac{x^{2}}{26} - \frac{x}{13} - \frac{162}{13}$

Correct answer:

$y = \frac{x^{2}}{26} - \frac{x}{13} - \frac{162}{13}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 1 for a -6 for b and -19 for y

$y_{0}^{2} + 38 y_{0} + 361 = x_{0}^{2} - 2 x_{0} + y_{0}^{2} + 12 y_{0} + 37$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{26} - \frac{x_{0}}{13} - \frac{162}{13}$

So our answer is then

$y = \frac{x^{2}}{26} - \frac{x}{13} - \frac{162}{13}$

Report an Error

Example Question #3 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( -10, 6\right ) \\ \text{directrix } y = 15$

Possible Answers:

$\left(x - 6\right)^{2} + \left(x + 10\right)^{2}$

$y = - \frac{x^{2}}{72} - \frac{5 x}{18} + \frac{89}{72}$

$y = - \frac{x^{2}}{36} - \frac{5 x}{9} + \frac{89}{36}$

$y = - \frac{x^{2}}{18} - \frac{10 x}{9} + \frac{89}{18}$

$\left(x - 2\right)^{2}$

Correct answer:

$y = - \frac{x^{2}}{18} - \frac{10 x}{9} + \frac{89}{18}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute -10 for a 6 for b and 15 for y

$y_{0}^{2} - 30 y_{0} + 225 = x_{0}^{2} + 20 x_{0} + y_{0}^{2} - 12 y_{0} + 136$

Now we can simplify, and solve for $y_{0}$

$y_{0} = - \frac{x_{0}^{2}}{18} - \frac{10 x_{0}}{9} + \frac{89}{18}$

So our answer is then

$y = - \frac{x^{2}}{18} - \frac{10 x}{9} + \frac{89}{18}$

Report an Error

Example Question #4 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Find the parabolic equation, where the focus and directrix are as follows.

$\\\text{focus}= \left ( 7, 5\right ) \\\text{directrix } y = -4$

Possible Answers:

$y = \frac{x^{2}}{72} - \frac{7 x}{36} + \frac{29}{36}$

$\left(x - 2\right)^{2}$

$\left(x - 7\right)^{2} + \left(x - 5\right)^{2}$

$y = \frac{x^{2}}{18} - \frac{7 x}{9} + \frac{29}{9}$

$y = \frac{x^{2}}{36} - \frac{7 x}{18} + \frac{29}{18}$

Correct answer:

$y = \frac{x^{2}}{18} - \frac{7 x}{9} + \frac{29}{9}$

Explanation:

The first step to solving this problem, it to use the equation of equal distances.

$\left|{y - y_{0}}\right| = \sqrt{\left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}}$

Let's square each side

$\left|{y - y_{0}}\right|^{2} = \left(- a + x_{0}\right)^{2} + \left(- b + y_{0}\right)^{2}$

Now we expand each binomial

$y^{2} - 2 y y_{0} + y_{0}^{2} = a^{2} - 2 a x_{0} + b^{2} - 2 b y_{0} + x_{0}^{2} + y_{0}^{2}$

Now we can substitute 7 for a 5 for b and -4 for y

$y_{0}^{2} + 8 y_{0} + 16 = x_{0}^{2} - 14 x_{0} + y_{0}^{2} - 10 y_{0} + 74$

Now we can simplify, and solve for $y_{0}$

$y_{0} = \frac{x_{0}^{2}}{18} - \frac{7 x_{0}}{9} + \frac{29}{9}$

So our answer is then

$y = \frac{x^{2}}{18} - \frac{7 x}{9} + \frac{29}{9}$

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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #11 : Derive Circle Equation: Ccss.Math.Content.Hsg Gpe.A.1

Example Question #12 : Expressing Geometric Properties With Equations

Example Question #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #3 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #4 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #5 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #6 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #3 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

Example Question #4 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2

All Common Core: High School - Geometry Resources

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