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Rectangular and Cartesian products
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
Correct Answer:
B) lies on the circumcircle of triangle ABC
Part of solved Rectangular and Cartesian products questions and answers :
>> Elementary Mathematics
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1). 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
-- View Answer
2). 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
-- View Answer
3). 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). \( \Large x_{1}x_{2}+y_{1}y_{2} \)
-- View Answer
4). 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). \( \Large a\ =\ \pm \sqrt{2}b \)
-- View Answer
5). 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). \( \Large \left(-3,\ 6\right) \)
-- View Answer
6). 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
-- View Answer
7). 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
-- View Answer
8). 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). \( \Large \left( - \frac{1}{\sqrt{2}},\ \frac{7}{\sqrt{2}}\right) \)
-- View Answer
9). Area of quadrilateral whose vertices are \( \Large \left(2,\ 3\right),\ \left(3,\ 4\right),\ \left(4,\ 5\right)\ and\ \left(5,\ 6\right) \)
A). 0
B). 4
C). 6
D). none of these
-- View Answer
10). 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). \( \Large \left(0,\ \frac{b}{2}\right) \)
-- View Answer