If a point p is at a distance of 5 units on one of the lines y - √3 | x | = 2 from their intersection point, then the foot of perpendicular drawn from p on the bisector of the angle between lines is
(It depends on that the point p lies on which line)
If the gradient of lines passing through A (3,2) is 3/4,the points on the lines at a distance 5 units from A are.
(-7,5) (-1,-1)
(7,5) (-1,-1)
(1,1) (5,-5)
(-7,-5) (1,1)
The opposite vertices of a square are (1,2) and (3,8) , then equation of diagonal passing through (1,2) is
3x - y - 1 = 0
3x - y + 1 = 0
y + 3x - 1 = 0
y + 3x + 1 = 0
The ortho centre of the triangle formed by vertices (0,0) , (8,0) and (4,6) is
(3,4)
(4,3)
(-3,4)
(4, 8/3)
Line L is perpendicular to line 5x - y = 1.The area of triangle formed by coordinates axis and line is 5 sq . units.Then its equation is
x + y = √2
x + 5y = ± 5 √2
x + 5y = -5
x + 5y = - √2
The ortho-center of triangle whose vertices are (0,0), (2,-1) and (1,3) is
A line moves in such a way that the sum of reciprocals of intersection on two mutually perpendicular lines by the line remains constant than the line passes through
Fixed Point
A variable point
Origin
None of these
The equation of a line passing through (3,-4) and perpendicular to line 3x + 4y = 5 is
4x + 3y = 24
y - x = x + 3
3y - 4x = 24
y + 4 = 4/3 (x - 3)
The foot of perpendicular drawn from origin to the line joining points (a cos a, a sin a) and (a cos β , a sin β) is
The ortho-center of the triangle formed by the lines x = 3, y =4 and 3x + 4y = 6 is
(0,0)
(3,0)
(0,4)