In a triangle `ABC, A (1,3)` and perpendicular bisector of `AB ` is `y = x` and orthocentre of triangle `ABC` is `(3, 1),` then locus of vertex `C` is

In triangle ABC, AD is the perpendicular bisector of BC (see the given figure). Show that triangle ABC is an isosceles triangle in which AB = AB .

In /_\ABC , AD is the perpendicular bisector of BC. Show that /_\ABC is an isosceles triangle in which AB=AC.

Let ABC be a triangle and A -= (1,2) ,y = x be the perpendicular bisector of AB and x-2y+1=0 be the perpendicular bisector of angle C . If the equation of BC is given by ax+by-5 = 0 then the value of a -2b is

Let ABC be a triangle and A -= (1,2) ,y = x be the perpendicular bisector of AB and x-2y+1=0 be the perpendicular bisector of angle C . If the equation of BC is given by ax+by-5 = 0 then the value of a -2b is

Let ABC be a triangle and A -= (1,2) ,y = x be the perpendicular bisector of AB and x-2y+1=0 be the perpendicular bisector of angle C . If the equation of BC is given by ax+by-5 = 0 then the value of a -2b is

In Delta ABC , AD is the perpendicular bisector of BC. Show that Delta ABC is an isoscelss tringle in which AB = AC.

In a triangle, ABC, the equation of the perpendicular bisector of AC is 3x - 2y + 8 = 0 . If the coordinates of the points A and B are (1, -1) & (3, 1) respectively, then the equation of the line BC & the centre of the circum-circle of the triangle ABC will be

In a triangle,ABC,the equation of the perpendicular bisector of AC is 3x-2y+8=0. If the coordinates of the points A and B are (1,-1)&(3,1) respectively,then the equation of the line BC & the centre of the circum-circle of the triangle ABC will be

ABC be a triangle with A(1,3),y=x,y=-2x are the equations of internal angular bisectors of /_B and /_C and area of triangle ABC is k units,then 2k is