| Parallel and Perpendicular Slopes |
| GOAL: I can derive the equation of a line parallel to and perpendicular to a given line and through a specific point |
| Learning Standards |
Common Core Standard: G-GPE |
| Resources |
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RED BOOK Reference: Pages: 292 ─ 297
Practice on page 295-296,
#5 - 39 odds. |
 BLACK BOOK Reference:
Pages: 327 ─ 332
Practice on page 331,
#7 - 27 odds. |
| Lesson Vocabulary |
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| Parallel Lines |
Lines that never intersect and have the same slope. | | Perpendicular Lines | Lines that intersect at a 90° angle and have opposite, reciprocal slopes. |
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Strategies and Content Practice |
Begin
(prepare) |
What do the slopes of parallel lines have in common?
What can be said about the slopes of perpendicalar lines? |
| Engage |
Parallel line have the exact same slope.
A typical parallel slope problem involves using a given line equation (y = mx + b) and a given point (x, y) and finding the equation of a second line that is parallel to the given line and passes through the given point. Here is a example.
Given the line equation y = ½x + (-5) and the point (4, 3) find the equation of a line parallel to the given line that passes through the point (4, 3).
Start with the original line equation y = ½x + (-5). Since we are looking for a parallel line and we know parallel lines have the same slope, we need to keep the slope (½) from the original line equation and use it in the new line equation. The "generic" new line equation starts out as
y = ½x + b, where be is the intercept point that we need to solve for.
Before we can solve for b, we need to substitute the given (desired) point into the generic new equation. Thus we get 3 = ½(4) + b, this leads to 3 = 2 + b, and solving for b we get b = 1.
Now we use the b = 1 and substitute it into our new equation to get y = ½x + 1.
So the equation of the line that passes through the point (4, 3) and is parallel to y = ½x + (-5) is
y = ½x + 1.
A typical perpendicular slope problem involves using a given line equation (y = mx + b) and a given point (x, y) and finding the equation of a second line that is parallel to the given line and passes through the given point. We can use the information in the previous example except we can solve fro the perpendicular line instead of the parallel line.
Given the line equation y = ½x + (-5) and the point (4, 3) find the equation of a line perpendicular to the given line that passes through the point (4, 3).
Start with the original line equation y = ½x + (-5). Since perpendicular lines have opposite reciprocal slopes, we need to take the original slope (½), flip it upside-down, and then take its opposite. So the reciprocal slope of (½) becomes (2/1) or simply 2. Then, since the (2) is positive, its opposite would be negative. So the slope for the perpendicular line is (-2). The "generic" new line equation starts out as
y = -2x + b, where be is the intercept point that we need to solve for.
Before we can solve for b, we need to substitute the given (desired) point into the generic new equation. Thus we get 3 = -2(4) + b, this leads to 3 = -8 + b, and solving for b we get b = 11.
Now we use the b = 11 and substitute it into our new equation to get y = -2x + 11.
So the equation of the line that passes through the point (4, 3) and is perpendicular to y = ½x + (-5) is
y = -2x + 11.
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Assess
(formative) |
After example problems and videos, try to solve the Practice Worksheets above. The answers to the practices are on the last page so you can check how well you are doing.
Please see me for more help if you are having difficulty with the practices. |
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