In a triangle ABC , if the equation of sides AB,BC and CA are `2x- y + 4 = 0 , x - 2y - 1 = 0 and x + 3y - 3 = 0 ` respectively ,
The equation of external bisector of angle B is

In a triangle ABC , if the equation of sides AB,BC and CA are 2x- y + 4 = 0 , x - 2y - 1 = 0 and x + 3y - 3 = 0 respectively ,Tangent of internal angle A is equal to

In a triangle ABC , if the equation of sides AB,BC and CA are 2x- y + 4 = 0 , x - 2y - 1 = 0 and x + 3y - 3 = 0 respectively ,Tangent of internal angle A is equal to

In a triangle AbC , if the equation of sides AB,BC and CA are 2x- y + 4 = 0 , x - 2y - 1 = 0 and x + 3y - 3 = 0 respectively , Tangent of internal angle A is equal to

In a triangle ABC , if the equation of sides Ab,BC and CA are 2x- y + 4 = 0 , x - 2y - 1 = 0 and x + 3y - 3 = 0 respectively , The image of point b w.r.t the side cA is

The equations of the sides AB, BC and CA of a triangle ABC are x-2=0, y+1=0 and x + 2y - 4=0 respectively. What is the equation of the altitude through B on AC?

The equations of the sides AB,BC,CA of a triangle ABC are 2x-y-3=0,6x+y-21=0 and 2x+y-5=0 respectively.The external bisector of angle A passes through the point

In Delta ABC equation of the right bisectors of the sides Ab and AC are x+y = 0 and x - y = 0 respectively .If A = (5,7) then equation of side BC is

Let ABC be a triangle with equations of the sides AB, BC and CA respectively x - 2 = 0,y- 5 = 0 and 5x + 2y - 10 = 0. Then the orthocentre of the triangle lies on the line

In triangle ABC, the equation of the right bisectors of the sides AB and AC are x+y=0 and y-x=0. respectively. If A-=(5,7) the find the equation of side BC.

The vertex A of a triangle ABC is the point (-2,3) . Whereas the line perpendicular to the sides AB and AC are x - y - 4 = 0 and 2x - y - 5 = 0 respectively. The right bisectors of sides meet at P (3/2,5/2) . Then the equation of the median of side BC is