弄個題目來算好了
L : 3x + 2y + 10 = 0
線外一點 Q(1, 2)
已知線上一點 K,使QK 垂直 L 於 K 點
求 K 點座標 = ?
3x+2y = -10
3(x-1)+2(y-2) = -10-3-4 = -17
(3 , 2).(x-1 , y-2) = -17
(3 , 2).QK = -17
設 QK = m(3 , 2)
(3 , 2).m(3 , 2) = -17
m = -17/13
QK = (-51/13 , -34/13)
K 點座標 = (1-51/13 , 2-34/13) = (-38/13 , -8/13)
K 代入 L : 3x + 2y + 10 = -114/13 - 16/13 + 10 = 0
沒錯 |