Prove Slope Criteria for Parallel and Perpendicular Lines: CCSS.Math.Content.HSG-GPE.B.5 - Common Core: High School - Geometry

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Question

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

Answer

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

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

After plugging them in, we get.

Now we solve for

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Question

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

Answer

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

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

After plugging them in, we get.

Now we solve for

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Question

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

Answer

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

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

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

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

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Question

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

Answer

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

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

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Question

In slope intercept form, find the equation of the line parallel to and goes through the point $\left ( -4, \quad -6\right )$ .

Answer

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

Then we substitute -4 for $x_{0}$ and -6 for $y_{0}$

After plugging them in, we get.

Now we solve for

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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

After plugging them in, we get.

Now we solve for

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Question

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

Answer

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

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

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Question

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

Answer

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

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

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Question

In slope intercept form, find the equation of the line parallel to and goes through the point $\left ( 7, \quad 8\right )$ .

Answer

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

Then we substitute 7 for $x_{0}$ and 8 for $y_{0}$

After plugging them in, we get.

Now we solve for

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