2d_graphics
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| 2d_graphics [2012/08/01 00:19] – javapimp | 2d_graphics [2023/08/18 18:15] (current) – external edit 127.0.0.1 | ||
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| ===== Perpendicular Lines ===== | ===== Perpendicular Lines ===== | ||
| - | The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line. For example, given the line: `y = mx + b`, the slope of a line perpendicular is: `m_1 = -1/m` | + | The slope of a line perpendicular to a given line is the //negative reciprocal// of the slope of the given line. For example, given the line: `y = mx + b`, the slope of a line perpendicular is: `m_1 = -1/m` |
| Perpendicular lines will intersect at a single point. The point of intersection is where the `x` and `y` values satisfy both equations. | Perpendicular lines will intersect at a single point. The point of intersection is where the `x` and `y` values satisfy both equations. | ||
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| Intersection: | Intersection: | ||
| - | \begin{graph} width=400; height=300; xmin=-8.3; xmax=8.3; xscl=1; plot(x-2); plot(2x-1) \end{graph} | + | a``graph width=320; height=220; xmin=-8.3; xmax=8.3; ymin=-3.3; ymax=8.3; xscl=1; yscl=1; plot(1/2*x + 2); enda``graph |
| + | < | ||
| + | <embed class=" | ||
| + | </ | ||
| We can find the slope of the second line by taking the negative reciprocal of the slop of the first: | We can find the slope of the second line by taking the negative reciprocal of the slop of the first: | ||
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| Next we need to find the Y-intercept. In this case, we know the slope of the line and a point that satisfies the line. All we need to do is plug in those values and solve for b: | Next we need to find the Y-intercept. In this case, we know the slope of the line and a point that satisfies the line. All we need to do is plug in those values and solve for b: | ||
| + | |||
| amath | amath | ||
| y = mx + b\\ | y = mx + b\\ | ||
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| So, the equation of the perpendicular line is: | So, the equation of the perpendicular line is: | ||
| `y = -2x + 7` | `y = -2x + 7` | ||
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| + | |||
| + | <embed class=" | ||
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| ===== Intersection of two lines ===== | ===== Intersection of two lines ===== | ||
| - | Two lines intersect when the point `(x, y)` satisfies both equations. To determine the point of intersections | + | Two lines intersect when the point `(x, y)` satisfies both equations. To determine the point of intersection |
| - | amath | + | `y_1 = 1/2 x_1 + 2`\\ |
| - | y_1 = 1/2 x_1 + 2\\ | + | `y_2 = -2x_2 + 7` |
| - | y_2 = -2x_2 + 7 | + | |
| - | endamath | + | |
| Given that `y_1 = y_2` we can write: | Given that `y_1 = y_2` we can write: | ||
| Line 62: | Line 68: | ||
| amath | amath | ||
| - | 1/2 x + 2 = -2x + 7\\ | + | 1/2 x + 2 = -2x + 7 |
| + | |||
| + | Now solve for x: | ||
| 1/2 x + 2x + 2 = 7\\ | 1/2 x + 2x + 2 = 7\\ | ||
| 5/2 x + 2 = 7\\ | 5/2 x + 2 = 7\\ | ||
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| The point of intersection of the two lines is `(2, 3)` | The point of intersection of the two lines is `(2, 3)` | ||
| + | |||
| + | ===== Find point on a line ===== | ||
| + | Given a point on a line `(x, y)` a slope `m` and a distance `d` along the line from the given point, what is the `(x_1, y_1)` of the new point? | ||
| + | |||
| + | `c = cos(theta) = 1/sqrt(1 + m^2)`\\ | ||
| + | `s = sin(theta) = m/sqrt(1 + m^2)`\\ | ||
| + | `x_1 = x + d * c`\\ | ||
| + | `y_1 = y + d * s` | ||
| ===== Pythagorean theorem ===== | ===== Pythagorean theorem ===== | ||
2d_graphics.1343780375.txt.gz · Last modified: 2023/08/18 18:15 (external edit)