Let ABC is a triangle whose vertices are `A(1, –1), B(0, 2), C(x', y')` and area of `triangleABC` is 5 and C(x', y') lie on `3x + y – 4lambda = 0`, then

If the points A(x_1,y_1), B(x_2,y_2) and C(x_3,y_3) are collinear, then area of triangle ABC is:

Let two points be A(1,-1) and B(0,2) .If a point P(h,k) be such that the area of triangle PAB is 5 sq units and it lies on the line 3x+y-4 lambda =0 ,then the value of lambda is

Let two points be A(1,-1) and B(0, 2) . If a point P(x', y') be such that the area of DeltaPAB = 5 sq. units and it lies on the line, 3x + y - 4lambda =0 , then a value of lambda is:

Let two points be A(1,-1) and B(0,2) . If a point P(x',y') be such that the area of DeltaPAB = 5sq.unit and it lies on the line 3x+y-4lambda=0 then value of lambda is

The vertices of a triangle are A (3, 7), B(3, 4) and C(5, 4). The equation of the bisector of the angle ABC is a) y = x+1 b) y = x-1 c) y= 3x - 5 d) y = x

Let two points be A(1,-1) and B(0,2) .If a point P(h,k) be such that the area of Delta PAB=5 sq units and it lies on the line 3x+y-4 lambda=0 then a value of lambda is

Let A(4, 2), B (6, 5) and C (1, 4) be the vertices of triangleABC . :- If (x_1,y_1) , B (x_2,y_2) and C (x_3,y_3) the vertices of triangleABC , find the coordinates of the centroid of the triangle.

Find the vertices of the triangle whose sides are y+2x=3, 4y+x=5 and 5y+3x=0

Find the vertices of the triangle whose sides are y+2x=3, 4y+x=5 and 5y+3x=0

Find the value of k, if x-2y+k=0 is a median of the triangle ABC whose vertices are A(-1,3), B(0,4) and C(-5,2) .