# 2條MATHS LQ MOCK

Rating

12a) No.

Because P(A∪B) = 7/10 < 1.

b) P(A∩B)

= P(A) * P(B|A)

= (1 - P(A')) * P(B|A)

= (1 - 4/5) * 1/4

= 1/20

c)P(A) + P(B) - P(A∩B) = P(A∪B)

1/5 + P(B) - 1/20 = 7/10

P(B) = 11/20

d) P(A) * P(B) = 1/5 * 11/20 = 11/100 ≠ P(A∩B) = 1/20 therefore

A and B are not independent.

e)A and B are not mutually exclusive since P(A∩B) = 1/20 > 0.

14i)The center of the circle x² + y² = 1 is O(0,0).Let (x , y) be any point on PC ,

The slope of OC * The slope of PC = - 1 since OC ⊥ PC (tangent ⊥ radius)

y1 / x1 * (y1 - y) / (x1 - x) = - 1

yy1 - y1² = x1² - xx1

xx1 + yy1 = x1² + y1²

xx1 + yy1 = 1 which is the equation of PC.

ii) Similarly , the equation of PD is xx2 + yy2 = 1.

iii) Sub. x = 3 and y = 4 into the equation of PC and PD as (3,4) is their meeting point : 3x1 + 4y1 = 1

{

3x2 + 4y2 = 13x1 + 4y1 = 3x2 + 4y2

4(y1 - y2) = 3(x2 - x1)

(y2 - y1) / (x2 - x1) = - 3/4 which is the slope of CD.On the other hand , (x1 , y1) and (x2 , y2) are two solution sets of x² + y² = 1 ... (1)

{

3x + 4y = 1 ... (2)By (2) , y = (1 - 3x)/4 , sub. into (1) :x² + (1 - 3x)² / 16 = 1

7x² - 6x - 15 = 0

The x-coordinate of the mid point of CD is (x1 + x2)/2 = (6/7)/2 = 6/14,

then the y-coordinate of the mid point of CD is (y1 + y2)/2

= (( 1 - 3x1)/4 + (1 - 3x2)/4 ) / 2

= ( 2 - 3(x1 + x2) ) / 8

= ( 2 - 3(6/7) ) / 8

= - 1/14By point slope form , the equation of CD is

(y + 1/14) / (x - 6/14) = - 3/4

(14y + 1) / (14x - 6) = - 3/4

56y + 4 = 18 - 42x

3x + 4y - 1 = 0