F3 Maths Problems

1)

Since AC is perpendicular to CB,
(Slope of AC)×(Slope of CB) = -1
[(3-1)/(2-c)]×[(5-1)/(8-c)] = -1
2/(2-c)×4/(8-c) = -1
8 = -(2-c)(8-c)
8 = -c^2 + 10c - 16
c^2 - 10c + 24 = 0
(c-6)(c-4) = 0
c = 4 or 6

2)
(a)
Mean = 2.5
⇒ (1×5+2x+3×7+4y)/(5+x+7+y) = 2.5
(26+2x+4y)/(12+x+y) = 5/2
2(26+2x+4y) = 5(12+x+y)
52+4x+8y = 60+5x+5y
52-60+8y-5y = 5x-4x
3y-8 = x

(b) If a family is randomly selected, the probability that the family has 2 family members is 0.2. Find the value of N.

P(the family has 2 members) = x/(5+x+7+y) = (3y-8)/(5+3y-8+7+y) = (3y-8)/(4+4y) = 0.2
⇒ 5(3y-8)=4+4y
15y - 40 = 4 + 4y
11y = 44
y = 4
N = 5 + x + 7 + y = 12 + (3y-8) + y = 4+4y = 4+4×4 = 20

∴ The value of N is 20.

3)

(a)

Let the coordinates of B be (x,0). (as B is on the x-axis)
AB = 10
sqrt[(9-x)²+(8-0)²] = 10
(9-x)² + 8² = 10²
81-18x+x² = 100 - 64
x² - 18x + 45 = 0
(x-15)(x-3) = 0
x = 3 or 15
∴ The possible coordinates of B: (3,0) or (15,0)

(b) If the slope of L is positive, find the equation of L.

slope of L = slope of AB = (0-8)/(x-9) > 0
-8/(x-9) > 0
x-9 < 0
x < 9
Therefore the coordinates of B can only be (3,0) in this case.

The equation of L: (0-8)/(3-9) = (y-8)/(x-9)
-8/-6 = (y-8)/(x-9)
8(x-9) = 6(y-8)
8x - 72 =6y - 48
∴ The equation of L: 8x - 6y - 24 = 0