Projection point of point on line

Linear equation ：
$$ax+by+c=0$$
spot ：
$$(x_0,y_0)$$
First find the normal equation as ：
$$bx-ay+(-b{x}_0+a{y}_0)=0$$
By finding the intersection of the line and the normal, that is, the projection point of the point on the line , namely ：
$$(\frac{b(-b{x}_0+a{y}_0)+ac}{-a^2-b^2},\frac{bc-a(-b{x}_0+a{y}_0)}{-a^2-b^2})$$

reference ：
The intersection of two straight lines ：https://blog.csdn.net/RobotLife/article/details/114884084