The equation of the plane passing through the intersection of the planes x + 2y + 3z + 4 = 0 and 4x + 3y + 2z + 1 = 0 and the origin is :
3x + 2y + z + 1 = 0
3x + 2y + z = 0
2x + 3y + z = 0
x + y + z = 0
The equation of the plane passing through (2, 3, 4) and parallel to the plane 5x - 6y + 7z = 3 is :
5x - 6y + 7z + 20 = 0
5x - 6y + 7z - 20 = 0
- 5x + 6y - 7z + 3 = 0
5x + 6y + 7z + 3 = 0
The inclination of the straight line passing through the point (- 3, 6) and the mid point of the line joining the points ( 4, - 5) and (- 2, 9) is:
A point moves such that the area of the triangle formed by it with the points (1, 5) and (3, - 7) is 21 sq unit Then locus of the point is :
6x + y - 32 = 0
6x - y + 32 = 0
x + 6y - 32 = 0
6x - y - 32 = 0
The line = 1 cuts the x-axis at P. The equation of the line through P perpendicular to the given line is :
x + y = ab
x + y = a + b
ax + by = a2
bx + ay = b2
The value of . for which the lines 3x + 4y = 5, 2x + 3y = 4and x + 4y = 6 meet at a point, is :
2
1
4
3
Three vertices of a parallelogram taken in order are (- 1, - 6), (2, - 5) and (7, 2). The fourth vertex is :
(1, 4)
(1, 1)
(4, 4)
(4, 1)
The orthocentre of the triangle whose vertices are (5, - 2), - 1, 2) and (1, 4), is :
A.
Let A(5, - 2), B (- 1, 2)andC (1, 4)arevertices of a triangle ABC.
Slope of line BC =
slope of line perpendicular to BC = - 1
So equation of altitude through A is
(y + 2) = - 1(x - 5)
...(i)
and slope of CA =
slope of a line perpendicular to CA =
so equation of altitude through C is
(y - 4) =
.