`y = mx + b`
`m = (Delta y)/(Delta x) = (y_2 - y_1)/(x_2 - x_1)`
amath
y = mx + b
y = m(0) + b
y = 0 + b
y = b
endamath
`y = m(x - x_1) + y_1`
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
3 = -2 * 2 + b
3 = -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);'/>
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 1/2 x + 2 = -2x + 7
Now solve for x:
1/2 x + 2x + 2 = 7
5/2 x + 2 = 7
5/2 x = 5
5x = 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)`
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`
`c^2 = a^2 + b^2` or `c = sqrt(a^2 + b^2)`
