P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.
6.5 units
13.0 units
3.5 units
7.0 units
Correct answer is A
PQ\(^2\) = (x2 - x1)\(^2\) + (y2 - y1)\(^2\)
= 12\(^2\) + 5\(^2\)
= 144 + 25
= 169
PQ = √169 = 13
But PQ = diameter = 2r, r = PQ/2 = 6.5 units
Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.
6
2
3
8
Correct answer is A
Let the ext. angle = x
Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6
4
6
3
0
Correct answer is A
The line joining (P, 4) and (6, -2).
Gradient: \(\frac{-2 - 4}{6 - P} = \frac{-6}{6 - P}\)
The line joining (2, P) and (-1, 3)
Gradient: \(\frac{3 - P}{-1 - 2} = \frac{3 - P}{-3}\)
For perpendicular lines, the product of their gradient = -1.
\((\frac{-6}{6 - P})(\frac{3 - P}{-3}) = -1\)
\(\frac{6 - 2P}{6 - P} = -1 \implies 6 - 2P = P - 6\)
\(6 + 6 = P + 2P \implies P = \frac{12}{3} = 4\)
100°
120°
30°
60°
Correct answer is B
Cos θ° = t2 + t2 -(√3t)2
= 2t2 - 3t2
= -t2/2t2
= -1/2
Thus θ = cos-1 (-0.5) = 120°
y = (x/10) + 5
y = x + 5
√3y = - x + 5√3
√3y = x + 5√3
Correct answer is D
Cos 30 = 5/x
x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3