2D Graphics

Formula for a line

$y = m x + b$

Slope of a line

$m = \frac{Δ y}{Δ x} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Y-intercept

Y-intercept - where the line intersects with the line defined by the Y-axis
The line intersects the Y-axis when $x$ is $0$ .

$y =$ mx $+ b$
$y = m (0) + b$
$y = 0 + b$
$y = b$

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: $y = m x + b$ , the slope of a line perpendicular is: $m_{1} = - \frac{1}{m}$

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: $y = \frac{1}{2} x + 2$
Intersection: $(2, 3)$

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We can find the slope of the second line by taking the negative reciprocal of the slop of the first: $m_{1} = - \frac{2}{1} = - 2$

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:

$y =$ mx $+ b$
$3 = - 2 \cdot 2 + b$
$3 = - 4 + b$
$b = 7$

So, the equation of the perpendicular line is: $y = - 2 x + 7$

Intersection of two lines

Two lines intersect when the point $(x, y)$ satisfies both equations. To determine the point of intersection of two lines:

$y_{1} = \frac{1}{2} x_{1} + 2$
$y_{2} = - 2 x_{2} + 7$

Given that $y_{1} = y_{2}$ we can write:

$\frac{1}{2} x_{1} + 2 = - 2 x_{2} + 7$

Since $x_{1} = x_{2}$ we can write:

$\frac{1}{2} x + 2 = - 2 x + 7$

Now solve for $x :$

$\frac{1}{2} x + 2 x + 2 = 7$
$\frac{5}{2} x + 2 = 7$
$\frac{5}{2} x = 5$
$5 x = 10$
x = 2

Now that we have an $x$ value we can use one of the equations to find the $y$ .

$y = \frac{1}{2} x + 2$
$y = \frac{1}{2} \cdot 2 + 2$
$y = 1 + 2$
y = 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?

$c = cos (θ) = \frac{1}{\sqrt{1 + m^{2}}}$
$s = sin (θ) = \frac{m}{\sqrt{1 + m^{2}}}$
$x_{1} = x + d \cdot c$
$y_{1} = y + d \cdot s$

Pythagorean theorem

$c^{2} = a^{2} + b^{2}$ or $c = \sqrt{a^{2} + b^{2}}$