光线自直线L1:2X-3Y+4=0上的点P(4,4)发出,沿直线L1经直线L:X+5y-11=0反射,求反射光线所在的直
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P关于的对称点为P'(a, b)

x + 5y - 11 = 0, y = -x/5 + 11/5, 斜率k = -1/5

PP'斜率k' = 5, 方程: y - 4 = 5(x - 4)

与y = -x/5 + 11/5联立得交点A(7/2, 3/2)

A为PP'的中点:

7/2 = (a + 4)/2, a = 3

3/2 = (b + 4)/2, b = 1

P'(3, 1)

联立L1, L得交点B(1, 2)

P'B即为L2 (不明白再问):

(y - 1)/(2 - 1) = (x - 3)/1 - 3)

x + 2y - 5 = 0