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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

 A) AP

Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products

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1). 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). $$\Large x_{1}x_{2}+y_{1}y_{2}$$
2). 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). $$\Large a\ =\ \pm \sqrt{2}b$$
3). ABC is a triangle with vertices A = (-1, 4), B = (6, -2) and C = (-2, 4). D, E and F are points which divide each AB, BC and CA respectively in the ratio 3 : 1 internally. Then centroid of the triangle DEF is:
 A). $$\Large \left(3,\ 6\right)$$ B). $$\Large \left(1,\ 2\right)$$ C). $$\Large \left(4,\ 8\right)$$ D). $$\Large \left(-3,\ 6\right)$$
4). 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
5). 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:
 A). -2 B). 2 C). 1 D). 0

6). 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). $$\Large \left( - \frac{1}{\sqrt{2}},\ \frac{7}{\sqrt{2}}\right)$$
7). Area of quadrilateral whose vertices are $$\Large \left(2,\ 3\right),\ \left(3,\ 4\right),\ \left(4,\ 5\right)\ and\ \left(5,\ 6\right)$$
8). Co-ordinates of the foot of the perpendicular drawn from $$\Large \left(0,\ 0\right)$$ to the line joining $$\Large \left(a \cos \alpha ,\ a \sin \alpha\right)\ and\ \left(a \cos \beta ,\ a \sin \beta \right)$$ are:
 A). $$\Large \left(\frac{a}{2},\ \frac{b}{2}\right)$$ B). $$\Large \left(\frac{a}{2} \left(\cos \alpha + \cos \beta \right), \frac{a}{2} \left(\sin \alpha + \sin \beta \right) \right)$$ C). $$\Large \left(\cos\frac{ \alpha + \beta }{2},\ \sin\frac{ \alpha + \beta }{2}\right)$$ D). $$\Large \left(0,\ \frac{b}{2}\right)$$
9). An equilateral triangle has each side equal to a the co-ordinates of its vertices are $$\Large \left(x_{1},\ y_{1}\right),\ \left(x_{2},\ y_{2}\right)\ and\ \left(x_{3},\ y_{3}\right)$$, then the square of determinant $$\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}$$ equals
 A). $$\Large 3a^{4}$$ B). $$\Large \frac{3}{4}a^{4}$$ C). $$\Large 4a^{4}$$ D). none of these
10). The incentre of triangle with vertices $$\Large \left(1,\ \sqrt{3}\right),\ \left(0,\ 0\right)\ and\ \left(2,\ 0\right)$$ is:
 A). $$\Large \left(1,\ \frac{\sqrt{3}}{2}\right)$$ B). $$\Large \left(\frac{2}{3},\ \frac{1}{\sqrt{3}}\right)$$ C). $$\Large \left(\frac{2}{3},\ \frac{\sqrt{3}}{2}\right)$$ D). $$\Large \left(1,\ \frac{1}{\sqrt{3}} \right)$$