(1,1) only
(1,1) and (1,2)
(0,0) and (1,1)
(√2,√2) only
Correct answer is A
x2 + y2 = (√2)2
x2 + y2 = 2
but y = x
Thus; x2 + x2 = 2
2x2 = 2
x2 = + or - 1
But x2 + y2 = 2
12 + y2 = 2
1 + y2 = 2
y2 = 2 - 1
y2 = 1
y = + or - 1
Thus point (x,y) = (1,1) only.
8
6
4
3
Correct answer is B
2x + x = 180°, => 3x = 180°, and thus x = 60°
Each exterior angle = 60° but size of ext. angle = 360°/n
Therefore 60° = 360°/n
n = 360°/60° = 6 sides
if P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is
A straight line
a circle
the bisector of angle PXQ
the perpendicular bisector of PQ
Correct answer is D
No explanation has been provided for this answer.
-4/3
-3/4
3/4
4/3
Correct answer is A
Grad of 3y = 4x - 1
y = 4x/3 - 1/3
Grad = 4/3
Grad of Ky = x + 3
y = x/k + 3/4
Grad = 1/k
Since the two lines are perpendicular,
1/k = -3/4
-3k = 4
k = -4/3
An equilateral triangle of side √3cm is inscribed in a circle. Find the radius of the circle.
2/3 cm
2 cm
1 cm
3 cm
Correct answer is C
Since the inscribed triangle is equilateral, therefore the angles at all the points = 60°
Using the formula for inscribed circle,
2R = \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)
where R = radius of the circle; a, b and c are the sides of the triangle.
⇒ 2R = \(\frac{\sqrt{3}}{\sin 60}\)
2R = \(\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}\)
2R = 2
R = 1cm