直线y=kx+b 过点P(m,n)与直线垂直的直线方程为y-n=-1/k(x-m)
y=kx+b y-n=-1/k(x-m)交点 x=(kn-bk-m)/(k^2+1) y=k(kn-bk-m)/(k^2+1) +b
P(m,n)关于这条直线的对称点P1(2(kn-bk-m)/(k^2+1)-m,2k(kn-bk-m)/(k^2+1) +2b-n)
P1(4(b-1)/5,2(1-b)/5 +2b-1)
已知一条直线解析式和一点座标,求该点关于这条直线的对称点的公式?
直线y=kx+b 过点P(m,n)与直线垂直的直线方程为y-n=-1/k(x-m)
y=kx+b y-n=-1/k(x-m)交点 x=(kn-bk-m)/(k^2+1) y=k(kn-bk-m)/(k^2+1) +b
P(m,n)关于这条直线的对称点P1(2(kn-bk-m)/(k^2+1)-m,2k(kn-bk-m)/(k^2+1) +2b-n)
P1(4(b-1)/5,2(1-b)/5 +2b-1)