一次函数y=kx+b(k=/=0)的图像是一条直线，它更一般的形式为Ax+Bx+c=0(A.B.C为常数且A.B不同时为0）。设点P(Xo,Yo),直线L:Ax+By+c...
一次函数y=kx+b(k=/=0)的图像是一条直线，它更一般的形式为Ax+Bx+c=0(A.B.C为常数且A.B不同时为0）。设点P(Xo,Yo),直线L:Ax+By+c=0(A.B不同时为0），则P到L的距离d=|AXo+BYo+c|/根号(A^2+B^2)谁能解释为什么啊？
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点M到直线的距离,即过点M向已知直线作垂线，设垂足为N，则垂线段MN的长即是所求的点到直线的距离。但如何求此线段的长呢？同学们给出了不同的解决方法。方法一：求出过点M且与已知直线aX+bY+c=0（a、b均不为零）垂直的直线方程，而后联立方程组，求出垂足N点的坐标，然后利用两点间的距离公式求出点到直线的距离。方法二：过点M分别作垂直于两坐标轴的直线，且交已知直线分别于C、D两点，三角形MCD为直角三角形，点到直线的距离即是直角三角形MCD斜边上的高。而C、D两点的坐标较易求解，利用平行于坐标轴的两点间的距离公式，可得到两直角边MC、MD的长度，再利用勾股定理求出斜边的长，最后利用等面积法求出点到直线的距离。
参考资料：
http://www.thjy.edu.cn/fuchuner/UpFiles/200612301038209430.ppt
本回答被提问者和网友采纳展开全部
先求过P点平行L的直线解析式L1：Ax+By-AX0-BY0=0L与L1在x轴上的截距m=|AXo+BYo+c|/Ad=m*sina，a为L与x轴的夹角，tana=-A/B sina=A/根号(A^2+B^2)所以d=|AXo+BYo+c|/根号(A^2+B^2)B=0的情况单独讨论下就行
展开全部设直线为：Ax+Bx+C=0，...(1)点为M（x0,y0）则过点M作原直线的垂线，方程为Bx-Ay=Bx0-Ay0...........(2)(1)(2)两式联立得交点纵坐标Y=((A^2)*y0-ABx0-BC)/(A^2+B^2)所以交点与点M纵向距离为|Y-y0|=B|AXo+BYo+c|/(A^2+B^2)再除以垂线与x轴夹角正弦B/根号(A^2+B^2)即得距离公式。展开全部你可以利用图形去做啊.利用数行结合的思想啊,点到直线的距离实际上还是在利用一个直角三角形,勾古定理啊.在这又不好画图.自己好好想想了.
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