Geometry - 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 and goes through the point $\left ( 7, \quad 5\right )$ .

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

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

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.

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

After plugging them in, we get.

Now we solve for

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$y = - \frac{x}{10} + \frac{87}{10}$

$y = - \frac{x}{10} + \frac{187}{10}$

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.

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

After plugging them in, we get.

Now we solve for

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

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

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

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.

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

After plugging them in, we get.

Now we solve for

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

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

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

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

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

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

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

$y = - \frac{x}{11} - \frac{70}{11}$

$y = - \frac{x}{11} + \frac{40}{11}$

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.

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

After plugging them in, we get.

Now we solve for

Prove Slope Criteria for Parallel and Perpendicular Lines: CCSS.Math.Content.HSG-GPE.B.5

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