The orthocentre of the triangle formed by the lines x + y = 1 and 2y2 - xy - 6x2 = 0
43, 43
23, 23
23, - 23
43, - 43
The incentre of the triangle formed by the straight lines y = 3x, y = - 3x and y = 3 is and y = 3 is
(0, 2)
(1, 2)
(2, 0)
(2, 1)
The points on the straight line 3x - 4y + 1 = 0 which are at a distance of 5 units from the point (3, 2) are
- 2, - 74 - 3, - 52
4, 114 - 1, - 1
1, 122, 54
7, 5, - 1, - 1
D.
d We have, line 3x - 4y - 1 = 0 and point Ax1, y1which is at 5 units distance from (3, 2) equations of line PA are
x1 - 3cosθ = y1 - 2sinθ = ± 5⇒ x1 = 3 ± 5cosθy1 = 2 ± 5sinθSince, x1, y1 lies on the 3x - 4y - 1 = 0∴ 33 ± 5cosθ - 42 ± 5sinθ - 1 = 0⇒ 9 ± 15cosθ - 8 + 20sinθ - 1= 0⇒15cosθ - 20sinθ = 0⇒ 3cosθ = ± 4sinθ⇒tanθ =± 34cosθ = ± 45sinθ = ± 35∴ x1 = 3 ± 455 = 7, - 1 y1 = 2 ± 535 = 5, - 1∴ Coordinates are 7, 5 and - 1, - 1