数学公式：点到直线的距离

\[l的点斜式为\\ y = -\frac{A}{B}x - \frac{C}{B}\\ x = -\frac{B}{A}y - \frac{C}{A}\\ 求垂直斜率，通过斜率相乘得-1求得。k = \frac{B}{A}\\ 则 m 的点斜式方程为\\ y = \frac{B}{A}(x-a) + b\\ 一般式为\\ Ay = B(x-a)+Ab\\ Ay = Bx-Ba+Ab\\ -Bx=-Ay-Ba+Ab\\ Bx=Ay+Ba-Ab\\ x的式子为\\ x=\frac{A}{B}y+a - \frac{Ab}{B} \]

\[-\frac{A}{B}x-\frac{C}{B} = \frac{B}{A}x-\frac{Ba}{A}+b\\ -\frac{AAB}{B}x-\frac{ABC}{B} = \frac{ABB}{A}x-\frac{ABBa}{A}+ABb \\ -A^2x-AC = B^2x-B^2a+ABb\\ -B^2x - A^2x = -B^2a+ABb +AC\\ A^2x + B^2x = B^2a-ABb - AC\\ (A^2+B^2)x = B^2a-ABb - AC\\ x = \frac{B^2a-ABb - AC}{A^2+B^2} \]

\[-\frac{B}{A}y - \frac{C}{A} = \frac{A}{B}y+a - \frac{Ab}{B}\\ -\frac{ABB}{A}y - \frac{ABC}{A} = \frac{ABA}{B}y+ABa - \frac{ABAb}{B}\\ -B^2y - BC = A^2y + ABa - A^2b \\ -BC-ABa+A^2b = (A^2+B^2)y\\ y=\frac{A^2b-ABa-BC}{A^2+B^2} \]

\[(\frac{B^2a-ABb - AC}{A^2+B^2},\frac{A^2b-ABa-BC}{A^2+B^2}) \]

\[\begin{align} |PQ| &= \sqrt{(x-a)^2+(y-b)^2}\\ &=\sqrt{(\frac{B^2a-ABb - AC}{A^2+B^2}-a)^2+(\frac{A^b-ABa-BC}{A^2+B^2}-b)^2}\\ &=\sqrt{(\frac{B^2a-ABb - AC-a(A^2+B^2)}{A^2+B^2})^2+(\frac{A^2b-ABa-BC-b(A^2+B^2)}{A^2+B^2})^2}\\ &=\sqrt{(\frac{B^2a-ABb - AC-A^2a-B^2a}{A^2+B^2})^2+(\frac{A^2b-ABa-BC-A^2b-B^2b}{A^2+B^2})^2}\\ &=\sqrt{(\frac{-ABb - AC-A^2a}{A^2+B^2})^2+(\frac{-ABa-BC-B^2b}{A^2+B^2})^2}\\ &=\sqrt{(\frac{A(-Bb - C-Aa)}{A^2+B^2})^2+(\frac{B(-Aa-C-Bb)}{A^2+B^2})^2}\\ &=\sqrt{(\frac{A^2(-Bb - C-Aa)^2}{(A^2+B^2)^2})+(\frac{B^2(-Aa-C-Bb)^2}{(A^2+B^2)^2})}\\ &=\sqrt{(\frac{A^2(-Bb - C-Aa)^2+B^2(-Aa-C-Bb)^2}{(A^2+B^2)^2})}\\ &=\sqrt{(\frac{A^2(-Aa-Bb-C)^2+B^2(-Aa-Bb-C)^2}{(A^2+B^2)^2})}\\ &=\sqrt{(\frac{(A^2+B^2)(-Aa-Bb-C)^2)}{(A^2+B^2)^2})}\\ &=\sqrt{(\frac{(-Aa-Bb-C)^2}{A^2+B^2})}\\ &=\frac{\left| -(-Aa-Bb-C) \right|}{\sqrt{A^2+B^2}}\\ &=\frac{\left| Aa+Bb+C \right|}{\sqrt{A^2+B^2}} \end{align} \]