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--- 2d_graphics [2012/07/31 22:10] – [Y-intercept] javapimp
+++ 2d_graphics [2023/08/18 18:15] (current) – external edit 127.0.0.1
@@ Line 20: / Line 20: @@
 ===== Point-Slope form =====
 `y = m(x - x_1) + y_1`
+===== 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`
+Perpendicular lines will intersect at a single point. The point of intersection is where the `x` and `y` values satisfy both equations.
+Given a point on the line, find the equation of the line perpendicular to that line intersecting at that point:
+Line: `y = 1/2 x + 2`\\
+Intersection: `(2, 3)`
+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
+<html>
+<embed class="ASCIIsvg" script=' 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); '/>
+</html>
+We can find the slope of the second line by taking the negative reciprocal of the slop of the first:
+`m_1 = -2/1 = -2`
+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
+y = mx + b\\
+= -2 * 2 + b\\
+= -4 + b\\
+b = 7\\
+endamath
+So, the equation of the perpendicular line is:
+`y = -2x + 7`
+<embed class="ASCIIsvg" src="http://www.fourthwoods.com/js/d.svg" wmode="transparent" script=' 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); plot(-2x + 7);'/>
+===== Intersection of two lines =====
+Two lines intersect when the point `(x, y)` satisfies both equations. To determine the point of intersection of two lines:
+`y_1 = 1/2 x_1 + 2`\\
+`y_2 = -2x_2 + 7`
+Given that `y_1 = y_2` we can write:
+`1/2 x_1 + 2 = -2x_2 + 7`
+Since `x_1 = x_2` we can write:
+amath
+/2 x + 2 = -2x + 7
+Now solve for x:
+/2 x + 2x + 2 = 7\\
+/2 x + 2 = 7\\
+/2 x = 5\\
+x = 10 \\
+x = 2
+endamath
+Now that we have an `x` value we can use one of the equations to find the `y`.
+amath
+y = 1/2 x + 2\\
+y = 1/2 * 2 + 2\\
+y = 1 + 2\\
+y = 3
+endamath
+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 =====