A . Maths

(a) 因為L1和L2為平行線 , 所以線段上每一點的距離均一樣。
從L2 , 當x = 1 , 3(1) -4y +13 = 0
y = 4
兩線之間距離為 | [ 3(1) - 4(4) -2 ] / 開方(3^2 + 4^2) |
= | ( -15 ) / 5 |
= 3sq. units

(bi) 從P點作一直線L3與L2互相垂直 , 該線段長度為3sq. units
L3與L2相交於R點.
QR長度則可以用Pyth. Thm. 得出 , 即 √[ (√13)^2 - (3)^2 ] = 2
tanθ 即等於 PR / QR 即是3/2
(bii) L2 的斜率為3/4 , tanθ = | ( 3/4 + m ) / ( 1 - (3/4)m) |
3/2 = | (3/4 + m ) / ( 1 - (3/4)m) |
(3/4 + m) / (1 - (3/4)m) = 3/2 or -3/2
m = 6/17 or 18

2008-07-08 15:34:48 補充：
唔好意思..有更正..
(bii) L2 的斜率為3/4 , tanθ = | ( 3/4 - m ) / ( 1 + (3/4)m) |
3/2 = | (3/4 - m ) / ( 1 + (3/4)m) |
(3/4 - m) / (1 + (3/4)m) = 3/2 or -3/2
m = -6/17 or -18

調轉左正負號的..SORRY..=[

2008-07-08 15:37:02 補充：
回應第一位回答者.."
L 的斜率不可能只有一個 , 你可以試作一圖 , 有兩個可能性的。
而且 , 你犯了和我同一個的錯誤 , 就是把正負號也調轉了.."

你只有得到一個ANS..主要原因是你忘記了絕對值..=]

Please refer to: http://hk.knowledge.yahoo.com/question/question?qid=7007112404355

(a) distance = | 13 - ( -2) | / √( 3*3 4*4) = 15/5 = 3 units.


(b) cos θ = 3 / √13, so sin θ = √( 13 - 3*3 ) / √13 = 2 / √13

( i) tan θ = sin θ / cos θ = 2/3.

(ii) m1 = slope of L1 = 3/4 , m2 = slope of L ,

therefore, tan θ = ( m1 + m2 )/( 1 - m1*m2)

i.e. 2/3 = ( 3/4 m2 ) / ( 1 - 3/4*m2)

i.e. m2 = - 1/18
So, slope of L is -1/18.