A triangle has vertices `(1,1,1),(2,2,2),(1,1,y)` and its area equals to `cosec(pi/4)` sq.units. The sum of the possible values of `y` is

A triangle has vertices at (1,1,1),(2,2,2),(1,1,y) and area of the triangle is cosec (pi)/(4) square units then value of y is /are

Area of the triangle having vertices (1,2,3),(2,-1,1) and (1,2,4) in square units is

Area of the triangle ABC with vertices (1,7),(-1,2) and (3,k) is 10 square units then sum of possible values of k is

Area of the triangle ABC with vertices (1,7),(-1,2) and (3,k) is 10 square units then sum of possible values of

If the vertices of a triangle are (1,-3),(4,p) and (-9,7) and its area is 15 sq.units,find the value(s) of p.

P(x, y) moves such that the area of the triangle with vertices at P(x, y), (1, -2), (-1, 3) is equal to the area of the triangle with vertices at P(x, y), (2, -1), (3, 1). The locus of P is

A triangle has two of its vertices at (0,1) and (2,2) in the cartesian plane.Its third vertex lies on the x-axis.If the area of the triangle is 2 square units then the sum of the possible abscissae of the third vertex,is-

Let k be an integer such that the triangle with vertices (k,-3k),(5,k) and (-k,2) has area 28sq. units.Then the orthocentre of this triangle is at the point: :(1,-(3)/(4))(2)(2,(1)/(2))(3)(2,-(1)/(2))(4)(1,(3)/(4))