Find the equation of a line in cartesian form passes through the vector
A line passes through the point (2, -1, 3) and is Perpendicular to the line find its equation
The value of p and q so that the points (p, q,1), (-1, 4, -2) and (0, 2, -1) are collinear.
-2, -2
2, 2
2, -2
0, 2
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 equation of bisector of acute angle between lines 3x - 4y + 7 = 0 and 12x + 5y - 2 = 0 is.
21x + 77y - 101 = 0
11x - 3y + 9 = 0
31x + 77y +101 = 0
11x - 3x - 9 = 0
The equation of a line passing through the point(-1, 2, 3) and having direction ratios proportional to -4, 5, 6 is
None of these
Find the vector equation for the line passing through the point (-1, 0, 2) and (3, 4, 6) is
Two vertices of a triangle are (4,-3) and (-2,5) and its ortho centre is (1,2) , then its third vertex is
(-33,-26)
(33,26)
(26,33)
(-33,26)
The cartesian equation of the line is . The vector equation of the line is
The co ordinates of the point where the line through A(5, 1, 6) and B (3, 4, 1) cross the yz - plane
(17/2, 0, -13/2)
(0, 0, -13/2)
(17/2, 0, 1)
(0, 17/2, -13/2)
