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Proof of Formula for Distance Between Line and Point in 2D
Find the least distance of the line {jatex options:inline}ax+by=c{/jatex} from the point {jatex options:inline}(x_0,y_0){/jatex}
Rearrange the line as {jatex options:inline}y=- \frac{a}{b}x+ \frac{c}{b}{/jatex}.
The gradient of this line is {jatex options:inline}- \frac{a}{b}{/jatex}.
The gradient of the perpendicular line is {jatex options:inline}\frac{b}{a}{/jatex}.
The perpendicular line passes through the point {jatex options:inline}(x_0,y_0){/jatex}.
{jatex options:inline}y_0=\frac{b}{a}x_0+c \rightarrow c=y_0- \frac{b}{a}x_0{/jatex}
The equation of the perpendicular line is {jatex options:inline}y=\frac{b}{a}x+y_0- \frac{b}{a}x_0{/jatex} Now find the point of intersection of these two lines by solving the simultaneous equations {jatex options:inline}ax+by=c, \; y=\frac{b}{a}x+y_0- \frac{b}{a}x_0{/jatex}, equivalent to
{jatex options:inline}ax+by=c{/jatex} (1)
{jatex options:inline}bx-ay=bx_0-ay_0{/jatex} (2)
Add {jatex options:inline}a{/jatex} times (1) to {jatex options:inline}b{/jatex} times (2) to give {jatex options:inline}a^2x+b^2x=ac-aby_0-a^2x_0 \rightarrow x= \frac{ac-aby_0-a^2x_0}{a^2+b^2}{/jatex}.
Subtract {jatex options:inline}a{/jatex} times (2) from {jatex options:inline}b{/jatex} times (1) to give
{jatex options:inline}a^2y+b^2y= bc-abx_0+a^2y_0 \rightarrow y= \frac{bc-abx_0+a^2y_0}{a^2+b^2}{/jatex}.
The distance {jatex options:inline}d{/jatex} is then the distance between the points {jatex options:inline}(0,0){/jatex} and {jatex options:inline}(\frac{ac-aby_0-a^2x}{a^2+b^2}, \frac{bc-abx_0+a^2y_0}{a^2+b^2}){/jatex}
{jatex options:inline}\begin{equation} \begin{aligned} d &=\sqrt{(\frac{ac-aby_0+b^2x_0}{a^2+b^2}-x_0)^2+(\frac{bc-abx_0+a^2y_0}{a^2+b^2}-y_0)^2} \\ &=\sqrt{(\frac{ac-aby_0-a^2x_0}{a^2+b^2})^2+(\frac{bc-abx_0-b^2y_0}{a^2+b^2})^2} \\ &= \sqrt{\frac{(a^2+b^2)(c-by_0-ax_0)^2}{a^2+b^2)^2}} \\ &= \frac{c-ax_0-by_0}{\sqrt{a^2+b^2}}\end{aligned} \end{equation}{/jatex}
Solution of Linear Congruence
If {jatex options:inline}gcd(a,n)=1{/jatex} then {jatex options:inline}a^{\phi (n)} \equiv 1 \; (mod \; n){/jatex} where {jatex options:inline}\phi (n){/jatex} is the number of integers between 1 and {jatex options:inline}n{/jatex} relatively prime to {jatex options:inline}n{/jatex}.
Using this fact we can solve the congruence {jatex options:inline}ax \equiv b \; (mod \; n){/jatex}.
{jatex options:inline}a^{\phi (n) -1}ax =a^{\phi (n)}x \equiv 1 x= a^{\phi (n) -1}b \; (mod \; n){/jatex}.
For example, the linear congruence {jatex options:inline}5x \equiv 11 \; (mod \; 12){/jatex} has solution {jatex options:inline}x \equiv 5^{\phi (12)-1} \times 11 \equiv 5^{3} \times 11 \; (mod \; 12) \equiv 5 \times 11 \; (mod \; 12) \equiv 7 \; (mod \; 13){/jatex}.