求点到直线的距离公式是一个很有魅力 的数学问题,它吸引广大师生为之苦苦思索, 得到很多证法.现介绍一种证法,供大家参考. 已知定点为P(x0,y0),定直线为Ax+By +C=0,求证点P到定直线的距离为 证明 设Q(x,y)为定直线上任意一点, 则d为|PQ|的最小值. ∵ C=-Ax-By, ∴Ax0+By0+C =-Ax-By+Ax0+By0 =A(x0-x)+B(y0-y). 再由柯西不等式: Seeking a straight line to the distance formula is a very attractive mathematical problem, it attracted the majority of teachers and students to think hard, get a lot of card law. Now introduce a card law for your reference. Known fixed-point P (x0 , y0), and set the straight line as Ax + By + C = 0, and verify that the distance from point P to a straight line is the minimum value of | PQ | to prove that Q (x, y) C = -Ax-By, ∴Ax0 + By0 + C = -Ax-By + Ax0 + By0 = A (x0-x) + B (y0-y)