數學問題!!直線方程

a)(i)
OB的斜度是 -2
因為AP垂直於OB
AP的斜度為 (-1) / (-2) = 1/2

AP的方程式是
(y - 0) / [x - (-5)] = 1/2
2y = x + 5

將OB的方程式 y = -2x
代入AP的方程式就可得出兩直線的交點P
2(-2x) = x + 5
-4x = x + 5
-5x = 5
x = -1

y = (-2)(-1)
y = 2

所以P (-1, 2)


Q是AP線與y軸的交點, 將x = 0 代入,
會得出 2y = 5
y = 5/2

所以Q (0, 5/2)


a)(ii)
OP = PB
因此PB = 2OP
PB : OP = 2 : 1

設B的座標是 (m, n)
m = 2 (-1) = -2
n = 2 (2) = 4

所以B (-2, 4)


b)
B (-2, 4), 而 Q (0, 5/2)
BQ的方程式為
(y - 4) / [x - (-2)] = (5/2 - 4) / [0 - (-2)]
(y - 4) / (x + 2) = (-3/2) / 2
(y - 4) / (x + 2) = (-3) / 4
4y - 16 = -3x - 6
4y = -3x + 10

R 為BQ與x軸的交點, 將y = 0 代入,
會得出 4(0) = -3x + 10
3x = 10
x = 10/3

所以 R (10/3, 0)

BQ的長度是
√{[(-2) - 0]^2 + (4 - 5/2)^2}
= √(4 + 9/4)
= √(25/4)
= 5/2

QR的長度是
√[(0 - 10/3)^2 + (5/2 - 0)^2]
= √(100/9 + 25/4)
= √(625/36)
= 25/6

BQ / QR = (5/2) / (25/6)
BQ / QR = 3/5
BQ : QR = 3 : 5