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--- 2d_graphics [2012/08/02 23:13] – javapimp
+++ 2d_graphics [2023/08/18 18:15] (current) – external edit 127.0.0.1
@@ Line 31: / Line 31: @@
 Intersection: `(2, 3)`
-|<html>
+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="ASCIIsvg" src="http://www.fourthwoods.com/js/d.svgd.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); '/>
+<html>
-</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:
@@ Line 50: / Line 51: @@
 `y = -2x + 7`
-|<html>
-<embed class="ASCIIsvg" src="http://www.fourthwoods.com/js/d.svgd.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);'/>
+<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);'/>
-</html>|
 ===== Intersection of two lines =====
@@ Line 67: / Line 68: @@
 amath
-/2 x + 2 = -2x + 7\\
+/2 x + 2 = -2x + 7
+Now solve for x:
 /2 x + 2x + 2 = 7\\
 /2 x + 2 = 7\\
@@ Line 85: / Line 89: @@
 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 =====