Cloud = 12
Present = 5
Reindeer = 2
5 + 4 x 24 = 101
The answer is
18 triangles.
For example:
A(2,3) , B(9,7)
AB = √(x2-x1)^2 + (y2-y1)^2
AB = √7^2 + 4^2
AB = √49 + 16
AB = √65
If the distance between the points A(p , 0)
and B(0, p) is 10, find the possible values of p.
10 = √(0-p)^2 + (p-0)^2
10 = √(-p)^2 + (p)^2
10 = √p^2 + p^2
(10)^2 = (√2p^2)^2
100 = 2p^2
√p = √50
p = ± 7.07
Given that the coordinates of the points P and Q are
(-2,6) and (9,3) respectively, find the coordinates of
the point R that lies on the y-axis such that PR = QR
P (-2,6) , R (0,a)
PR = √(0+2)^2 + (a-b)^2
PR = √4 + a^2 – 12a +36
PR = √a^2 –12a + 40
QR = Q (9,3) , R (0,a)
QR = √(0-4)^2 + (a-3)^2
QR = √81 + a^2 – 6a + 9
PR = QR
√a^2 –12a + 40 = √a^2 –6a + 90
-12a + 40 = -6a + 90
-90 + 40 = -6a + 12a
-50 = 6a
a = - 50/6
a = - 25/3
∴ R (0, -25/3)
Equation of Line
y = mx + c
Gradient
y - intercept
e.g # 1: P is the point (2,9) and Q is the point (4,6)
Find the equation of the line.
Gradient = (6 – 9) / (4 - 2) = -3/2
Y – Intercept: P(2,9)
y = mx + c
y = -3/2 x + c
9 = -3/2 (2) + c
c = 12
y = -3/2 x + 12
A line has 2 points A (9,0) B (4,8).
Find the gradient and y – intercept of the line.
M=8–0/4–9
M = 8 / -5
M = - 8/5 OR M = -1.6
Y = mx + c
4 = ¼ (2) + c
4=½+c
4 = (1 + 2c) / 2
4 * 2 = 1 + 2c
8 – 1 = 2c
C = 7/2
A line: y = -x + c passes through (1,2).
Find the y – intercept (value of c).
Y = -x + c
2 = - (1) + c
2+1=c
C=3