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# 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

 C) $$\Large \frac{1}{2},\ 1$$

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

Similar Questions
1). 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
2). 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). $$\Large x_{1}x_{2}+y_{1}y_{2}$$
3). 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). $$\Large a\ =\ \pm \sqrt{2}b$$
4). 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). $$\Large \left(-3,\ 6\right)$$
5). 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

6). 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
7). 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). $$\Large \left( - \frac{1}{\sqrt{2}},\ \frac{7}{\sqrt{2}}\right)$$
8). Area of quadrilateral whose vertices are $$\Large \left(2,\ 3\right),\ \left(3,\ 4\right),\ \left(4,\ 5\right)\ and\ \left(5,\ 6\right)$$
9). 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). $$\Large \left(0,\ \frac{b}{2}\right)$$
10). 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