点到直线距离公式证明

点到直线距离公式的推导如下：对于点P（x0,y0)作PQ垂直直线Ax+By+C=0于Q作PM平行Y轴,交直线于M；作PN平行X轴,交直线于N设M（x1,y1)x1=x0,y1=(-Ax0+C)/B.PM=|y0-y1|=|y0+(Ax0+C)/B|=|(Ax0+By0+C)/B|同理,设N(x2,y2).y...请问Ax+By+C=0于QB的含义借用房东的追问解答一下“乐乐刷钻”的疑问：S△PMN=PM×PN/2，同时S△=MN×PQ/2所以PM×PN=MN×PQ，PQ=PM×PN/MN代入PM、PN表达形式：PQ=|(Ax0+By0+C)/B|×|(Ax0+By0+C)/A|/√[|(Ax0+By0+C)/B|²+|(Ax0+By0+C)/A|²]=[(Ax0+By0+C)²/(AB)]/√[(Ax0+By0+C)²/B²+(Ax0+By0+C)²/A²]=[(Ax0+By0+C)²/(AB)]/√[A²(Ax0+By0+C)²/(A²B²)+B²(Ax0+By0+C)²/(A²B²)]=[(Ax0+By0+C)²/(AB)]/√[(A²+B²)(Ax0+By0+C)²/(A²B²)]=[(Ax0+By0+C)²/(AB)]/[(Ax0+By0+C)√(A²+B²)/(AB)]=[(Ax0+By0+C)²/(AB)]×AB/(Ax0+By0+C)√(A²+B²)=|Ax0+By0+C|/√(A²+B²)这样如果再看不明白我也没办法了