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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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