Straight lines

(a)

Let (x, y) be the coordinates of B

OA // CB => slope of OA = slope of CB
3 / 4 = (y - 6) / (x - 6)
3(x - 6) = 4(y - 6)
3x - 18 = 4y - 24
3x - 4y = -6

OC // AB => slope of OC = slope of AB
6 / 6 = (y - 3) / (x - 4)
y - 3 = x - 4
y = x - 1

by solving two equations, we have x = 10, y = 9
thus, the coordinates of B are (10, 9)

(b)

OA = sqrt(4^2 + 3^2) = 5
OC = sqrt(6^2 + 6^2) = sqrt72
AC = sqrt(2^2 + 3^2) = sqrt13
cos AOC = [(5)^2 + (sqrt72)^2 - (sqrt13)^2] / 2(5)(sqrt72)
cos AOC = 84 / 10sqrt72 = 7 / (5sqrt2)
sin AOC = sqrt{1 - [7 / (5sqrt2)]^2} = 1 / (5sqrt2)
tan AOC = [1 / (5sqrt2)] / [7 / (5sqrt2)] = 1/7

a)
Method 1: simultaneous equations
let B be (x,y)
(y-3)/(x-4) = 1 => x-y=1 --(1) [AB//OC]
(y-6)/(x-6)=3/4 ==> 3x-4y=-6 --(2) [AO//BC]
solving (1) and (2) gives x=10, y=9
so coordinates of B are (10,9)

Method 2: using vector
vector OB
= vector OA+ vector OC (since OABC is //gram)
= (6i+6j)+(4i+3j)
= 10i+9j
so coordinates of B are (10,9)

b)
let angle AOC be C
........|0 0|
........|4 3| /2 = area of triangle AOC
........|6 6|
........|0 0|

hence, area of triangle AOC = 3
also, OA = 5, OC=6sqrt2 (by pyth. thm)

[(OA*OC)sinC]/2 = 3 (area of triangle ABC= (absinC)/2)
5*6sqrt2 sinC = 6
sinC = 1/(5sqrt2) = 1/sqrt50
cos²C = 1-sin²C = 1-1/50= 49/50
cos C = 7/sqrt50 (since C is acute, cosC>0)
tanC = sinC / cosC = 1/7

(or

5sqrt2 / | 1
........ /_|
the base = sqrt(50-1²) = 7
so tanC =1/7)

(a) By method of translation, when A translates to C, O will translate to B since OACB is a parallelogram.
When A translates to C, x - coordinate translates 2 units and y - coordinates translates 3 units. When O(0,0) translates x by 2 units and y by 3 units will be (2,3).
So B is (2,3).
(b)
Let mid- point of AB be M. So M is (3,3)
Let angle AOC be H and angle AO makes with the x - axis be K.
tan K = y - coordinate of A/x - coordinate of A = 4/3
tan ( H + K ) = y - coordinate of M/x - coordinate of M = 3/3 = 1.
By compound angle,
tan (H + K) = (tan H + tan K)/(1 - tan H tan K) = 1
tan H + 4/3 = 1 - (4/3) tan H
3 tan H + 4 = 3 - 4 tan H
7 tan H = 1
tan H = tangent of acute angle AOC = 1/7.





2013-01-15 10:43:46 補充：
Correction : tan K should be 3/4 ( not 4/3). So the equation should be tan H + 3/4 = 1 - (3/4) tan H. 4 tan H + 3 = 4 - 3 tan H. 7 tan H = 1, tan H = 1/7. Sorry for the mistake.