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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 #1 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line parallel to y = - 2 x + 88 and goes through the point $\left ( -7, \quad 10\right )$ .

Possible Answers:

y = - 2 x + 6

y = - 2 x - 4

$y = \frac{x}{2} + \frac{27}{2}$

$y = \frac{x}{2} + \frac{47}{2}$

y = - 2 x - 14

Correct answer:

y = - 2 x - 4

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is parallel, our slope () is going to be the same as the original equation.

So m = -2

Then we substitute -7 for $x_{0}$ and 10 for $y_{0}$

After plugging them in, we get.

y - 10 = - 2 x - 14

Now we solve for

y = - 2 x - 4

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Example Question #2 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line parallel to y = 6 x + 67 and goes through the point $\left ( 7, \quad 5\right )$ .

Possible Answers:

$y = - \frac{x}{6} + \frac{97}{6}$

y = 6 x - 47

y = 6 x - 27

y = 6 x - 37

$y = - \frac{x}{6} + \frac{37}{6}$

Correct answer:

y = 6 x - 37

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is parallel, our slope () is going to be the same as the original equation.

So m = 6

Then we substitute 7 for $x_{0}$ and 5 for $y_{0}$

After plugging them in, we get.

y - 5 = 6 x - 42

Now we solve for

y = 6 x - 37

Report an Error

Example Question #1 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line perpendicular to y = 14 x + 94 and goes through the point $\left ( -10, \quad 10\right )$

Possible Answers:

y = 14 x + 140

$y = - \frac{x}{28} - \frac{5}{14}$

$y = - \frac{x}{14} + \frac{65}{7}$

y = 14 x + 150

$y = - \frac{x}{14} + \frac{135}{7}$

Correct answer:

$y = - \frac{x}{14} + \frac{65}{7}$

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is perpendicular, our slope () is going to be the negative reciprocal of the original equation.

$m = - \frac{1}{14}$

Then we substitute -10 for $x_{0}$ and 10 for $y_{0}$

After plugging them in, we get.

$y - 10 = - \frac{x}{14} - \frac{5}{7}$

Now we solve for

$y = - \frac{x}{14} + \frac{65}{7}$

Report an Error

Example Question #4 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line perpendicular to y = 17 x + 91 and goes through the point $\left ( 1, \quad -2\right )$ .

Possible Answers:

y = 17 x - 19

$y = - \frac{x}{34} - \frac{407}{34}$

y = 17 x - 29

$y = - \frac{x}{17} - \frac{33}{17}$

$y = - \frac{x}{17} + \frac{137}{17}$

Correct answer:

$y = - \frac{x}{17} - \frac{33}{17}$

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is perpendicular, our slope () is going to be the negative reciprocal of the original equation.

So $m = - \frac{1}{17}$

Then we substitute 1 for $x_{0}$ and -2 for $y_{0}$

After plugging them in, we get.

$y + 2 = - \frac{x}{17} + \frac{1}{17}$

Now we solve for

$y = - \frac{x}{17} - \frac{33}{17}$

Report an Error

Example Question #5 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line perpendicular to y = - 3 x + 73 and goes through the point $\left ( -1, \quad -9\right )$ .

Possible Answers:

y = - 3 x - 22

y = - 3 x - 12

$y = \frac{x}{6} - \frac{113}{6}$

$y = \frac{x}{3} + \frac{4}{3}$

$y = \frac{x}{3} - \frac{26}{3}$

Correct answer:

$y = \frac{x}{3} - \frac{26}{3}$

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is perpendicular, our slope () is going to be the negative reciprocal of the original equation.

So $m = \frac{1}{3}$

Then we substitute -1 for $x_{0}$ and -9 for $y_{0}$

After plugging them in, we get.

$y + 9 = \frac{x}{3} + \frac{1}{3}$

Now we solve for

$y = \frac{x}{3} - \frac{26}{3}$

Report an Error

Example Question #6 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line perpendicular to y = - 9 x + 94 and goes through the point $\left ( 2, \quad -5\right )$ .

Possible Answers:

$y = \frac{x}{18} - \frac{136}{9}$

$y = \frac{x}{9} - \frac{47}{9}$

y = - 9 x + 3

y = - 9 x + 13

$y = \frac{x}{9} + \frac{43}{9}$

Correct answer:

$y = \frac{x}{9} - \frac{47}{9}$

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is perpendicular, our slope () is going to be the negative reciprocal of the original equation.

So $m = \frac{1}{9}$

Then we substitute 2 for $x_{0}$ and -5 for $y_{0}$

After plugging them in, we get.

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

Now we solve for

$y = \frac{x}{9} - \frac{47}{9}$

Report an Error

Example Question #3 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line perpendicular to y = - 5 x + 73 and goes through the point $\left ( 7, \quad -5\right )$ .

Possible Answers:

y = - 5 x + 20

$y = \frac{x}{5} + \frac{18}{5}$

$y = \frac{x}{5} - \frac{32}{5}$

y = - 5 x + 30

$y = \frac{x}{10} - \frac{157}{10}$

Correct answer:

$y = \frac{x}{5} - \frac{32}{5}$

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is perpendicular, our slope () is going to be the negative reciprocal of the original equation.

So $m = \frac{1}{5}$

Then we substitute 7 for $x_{0}$ and -5 for $y_{0}$

After plugging them in, we get.

$y + 5 = \frac{x}{5} - \frac{7}{5}$

Now we solve for

$y = \frac{x}{5} - \frac{32}{5}$

Report an Error

Example Question #4 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line parallel to y = 6 x + 39 and goes through the point $\left ( -5, \quad 6\right )$ .

Possible Answers:

y = 6 x + 46

y = 6 x + 36

$y = - \frac{x}{6} + \frac{31}{6}$

y = 6 x + 26

$y = - \frac{x}{6} + \frac{91}{6}$

Correct answer:

y = 6 x + 36

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is parallel, our slope () is going to be the same as the original equation.

So m = 6

Then we substitute -5 for $x_{0}$ and 6 for $y_{0}$

After plugging them in, we get.

y - 6 = 6 x + 30

Now we solve for

y = 6 x + 36

Report an Error

Example Question #5 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line perpendicular to y = 13 x + 53 and goes through the point $\left ( -5, \quad -4\right )$ .

Possible Answers:

$y = - \frac{x}{26} - \frac{369}{26}$

y = 13 x + 51

$y = - \frac{x}{13} - \frac{57}{13}$

$y = - \frac{x}{13} + \frac{73}{13}$

y = 13 x + 61

Correct answer:

$y = - \frac{x}{13} - \frac{57}{13}$

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$ Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is perpendicular, our slope () is going to be the negative reciprocal of the original equation.

So $m = - \frac{1}{13}$

Then we substitute -5 for $x_{0}$ and -4 for $y_{0}$

After plugging them in, we get.

$y + 4 = - \frac{x}{13} - \frac{5}{13}$

Now we solve for

$y = - \frac{x}{13} - \frac{57}{13}$

Report an Error

Example Question #10 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

In slope intercept form, find the equation of the line perpendicular to y = - 12 x + 73 and goes through the point $\left ( -5, \quad 5\right )$ .

Possible Answers:

$y = \frac{x}{24} - \frac{115}{24}$

$y = \frac{x}{12} + \frac{65}{12}$

$y = \frac{x}{12} + \frac{185}{12}$

y = - 12 x - 65

y = - 12 x - 55

Correct answer:

$y = \frac{x}{12} + \frac{65}{12}$

Explanation:

First step is to recall slope intercept form.

$y - y_{0} = m{\left (x - x_{0} \right )}$

Where is the slope, and $\left(x_0,y_0\right)$ is a point on the line.

Since we want a line that is perpendicular, our slope () is going to be the negative reciprocal of the original equation.

So $m = \frac{1}{12}$

Then we substitute -5 for $x_{0}$ and 5 for $y_{0}$

After plugging them in, we get.

$y - 5 = \frac{x}{12} + \frac{5}{12}$

Now we solve for

$y = \frac{x}{12} + \frac{65}{12}$

Report an Error

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All Common Core: High School - Geometry Resources

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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 #1 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #2 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #1 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #4 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #5 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #6 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #3 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #4 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #5 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

Example Question #10 : Prove Slope Criteria For Parallel And Perpendicular Lines: Ccss.Math.Content.Hsg Gpe.B.5

All Common Core: High School - Geometry Resources

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