The vertices of a triangle are (1, 2), (h, -3) and (-4, k). Find the value of `sqrt({(h+k)^(2)+(h+3k)^(2)})`. If the centroid of the triangle be at point (5, -1).

The vertices of a triangle are (1,2),(h-3) and (-4,k) . If the centroid of the triangle is at the point (5,-1) then find value of sqrt((h+k)^(2) + (h+3k)^(2))

The vertices of a triangle are (1,2), (h-3), and (-4,k). If the centroid of triangle is at the points (5,-1) then find the value of sqrt((h+k)^(2)+(h+3k)^(2))

The mid point of the sides of a triangle are (1,5,-1) (0,4,-2) and (2,3,4) find its vertices also find the centroid of the triangle

If the area of the triangle with vertices (2,5), (7, k) and (3,1) is 10, then find the value of k

Area of the triangle with vertices (k,0),(1,1) and (0,3) is 5 sq. units. Find the values of k.

Find the value of k, if the area of the triangle formed by the points (k,0),(3,4) and (5,-2) is 10.

Find the value of k, if the area of the triangle formed by the points (k,0),(3,4) and (5,-2) is 10.

If the centroid of the triangle whose vertices are (2, 4), (3, k) and (4, 2) is (k, 3) then k =

What type of triangle is DeltaABC , with the coordinates of the vertices are A(-1, 0) B(1, 0) and C(0, sqrt(3)) . Calculate its Area. OR Find the values of k so that the area of the triangle with vertices (1, -1) (-4, 2k) and (-k, -5) is 24 sq. units.