点P（m-n,-m）到直线（x/m）+（y/n）=1的距离为?

Question

点P（m-n,-m）到直线（x/m）+（y/n）=1的距离为?

Answer 1

由于:x/m+y/n=1 则：nx+my-mn=0 由点到直线的距离公式得：距离d =|n(m-n)+m(-m)-mn|/√[n^2+m^2] =|(mn-n^2)-m^2-mn|/√[n^2+m^2] =|-n^2-m^2|/√[n^2+m^2] 由于：m^2>=0,n^2>=0 则：|-n^2-m^2| =m^2+n^2 ={√[n^2+m^2]}^2 故：d=√[n^2+m^2]