If two vertices of a triangle are (–2, 3) and (5, –1), orthocentre lies at the origin and centroid on the line x + y = 7, then the third vertex lies at
None of these
Let O(0, 0) be the orthocentre; A(h, k) the third vertex; B(–2, 3) and C(5,–1) the other two vertices.
Then the slope of the line through A and O is k/h, while the line throughB and C has the slope (–1 –3)/(5 + 2) = –4/7. By the property of the orthocentre, these two lines must be perpendicular, so we have
Which is not satisfied by the points given in options.
Let P = (–1, 0), Q = (0, 0) and R = be three points. Then the equation of the bisector of the angle PQR is
A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. Then the point O divides the segment PQ in the ratio.
Let A0, A1 A2 A3 A4 A5 be a regular hexagon described in a circle of unit radius. Then the product of the length of the line segments A0 A1, A0 A2and A0 A4 is
The diagonals of a parallelogram PQRS are long the lines
x + 3y = 4 and 6x – 2y = 7, then PQRS must be a
The orthocenter of the triangle formed by the lines xy = 0 and x + y = 1 is
The straight lines x + y = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is
If sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is
If the circumcentre of a triangle lies at the origin and centroid is the middle point of the line joining the points (a2 + 1, a2 + 1) and (2a, –2a), then the orthocenter lies on the line.
If a, b, c are unequal and different from 1 such that the points are collinear, then
The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p andq are the intercepts of the line L on the new axes, then