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# If two vertices of an equilateral triangle are $$\Large \left(0,\ 0\right)\ and\ \left(3,\ 3\sqrt{3}\right)$$ then the third vertex is:

 A) $$\Large \left(3,\ -3\right)$$ B) $$\Large \left(-3,\ 3\right)$$ C) $$\Large \left(-3,\ 3\sqrt{3}\right)$$ D) none of these

 C) $$\Large \left(-3,\ 3\sqrt{3}\right)$$

Angle AOM is $$\Large 30 ^{\circ}$$. Hence required point of B is $$\Large \left(-3,\ 3\sqrt{3}\right)$$ Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products

Similar Questions
1). If origin is shifted to $$\Large \left(7,\ -4\right)$$ then point $$\Large \left(4,\ 5\right)$$ shifted to
 A). $$\Large \left(-3,\ 9\right)$$ B). $$\Large \left(3,\ 9\right)$$ C). $$\Large \left(11,\ 1\right)$$ D). none of these
2). The feet of the perpendicular drawn from P to the sides of a triangle ABC are collinear, then P is:
 A). circumcentre of triangle ABC B). lies on the circumcircle of triangle ABC C). excentre of triangle ABC D). none of these
3). The points $$\Large \left(k,\ 2-2k\right)$$, $$\Large \left(-k+1,\ 2k\right)$$, $$\Large \left(-4-k,\ 6-2k\right)$$ are collinear then k is equal to:
 A). 2, 3 B). 1, 0 C). $$\Large \frac{1}{2},\ 1$$ D). 1, 2
4). Let AB is divided internally and externally at P and Q in the same ratio. Then AP, AB, AQ are in
 A). AP B). GP C). HP D). none of these
5). If O be the origin and if $$\Large P_{1} \left(x_{1},\ y_{1}\right)\ and\ P_{2} \left(x_{2},\ y_{2}\right)$$ be two points, then $$\Large \left(OP_{1} \parallel \ OP_{2} \right) \cos \left( \angle P_{1}\ OP_{2}\right)$$ is equal to:
 A). $$\Large x_{1}y_{2}+x_{2}y_{1}$$ B). $$\Large \left(x^{2}_{1}+x^{2}_{2}+y^{2}_{2}\right)$$ C). $$\Large \left(x_{1}-x_{2}\right)^{2}+ \left(y_{1}-y_{2}\right)^{2}$$ D). $$\Large x_{1}x_{2}+y_{1}y_{2}$$

6). The median BE and AD of triangle with vertices $$\Large A \left(0,\ b\right), B \left(0,\ 0\right)\ and\ C \left(a,\ 0\right)$$ are perpendicular to each other if;
 A). $$\Large a=\frac{b}{2}$$ B). $$\Large b=\frac{a}{2}$$ C). ab = 1 D). $$\Large a\ =\ \pm \sqrt{2}b$$
 A). $$\Large \left(3,\ 6\right)$$ B). $$\Large \left(1,\ 2\right)$$ C). $$\Large \left(4,\ 8\right)$$ D). $$\Large \left(-3,\ 6\right)$$
8). If $$\Large A \left(x_{1},\ y_{1}\right), B \left(x_{2},\ y_{2}\right)\ and\ C \left(x_{3},\ y_{3}\right)$$ are the vertices of a triangle, then the excentre with respect to B is:
 A). $$\Large \frac{ \left(ax_{1}-bx_{2}+cx_{3}\right) }{a-b+c},\ \frac{ay_{1}-by_{2}+cy_{3}}{a-b+c}$$ B). $$\Large \frac{ \left(ax_{1}+bx_{2}-cx_{3}\right) }{a+b-c},\ \frac{ay_{1}+by_{2}-cy_{3}}{a-b-c}$$ C). $$\Large \frac{ \left(ax_{1}-bx_{2}-cx_{3}\right) }{a-b-c},\ \frac{ay_{1}-by_{2}-cy_{3}}{a-b-c}$$ D). none of these
9). The four distinct points $$\Large \left(0,\ 0 \right),\ \left(2,\ 0\right),\ \left(0,\ -2\right)\ and\ \left(k,\ -2\right)$$ are conocyclic, if k is equal to:
10). The co-ordinate axis rotated through an angle $$\Large 135 ^{\circ}$$. If the co-ordinates of a points P in the new system are known to be $$\Large \left(4,\ -3\right)$$, then the co-ordinates of P in the original systems are:
 A). $$\Large \left(\frac{1}{\sqrt{2}},\ \frac{7}{\sqrt{2}}\right)$$ B). $$\Large \left(\frac{1}{\sqrt{2}},\ -\frac{7}{\sqrt{2}}\right)$$ C). $$\Large \left( - \frac{1}{\sqrt{2}},\ -\frac{7}{\sqrt{2}}\right)$$ D). $$\Large \left( - \frac{1}{\sqrt{2}},\ \frac{7}{\sqrt{2}}\right)$$