The lines `(a+b)x + (a-b) y-2ab=0, (a-b)x+(a+b)y-2ab = 0 and x+y=0` form an isosceles triangle whose vertical angle is

Show that the lines (a+b)x+(a-b)y-2ab=0,(a-b)x+(a+b)y-2ab=0andx+y=0 form an isosceles triangle whose vertical angle is 2tan^(-1)((a)/(b)) .

Two sides of a triangle are the lines (a+b)x+(a-b)y-2ab=0 and (a-b)x+(a+b)y-2ab=0 . If the triangle is isosceles and the third side passes through point (b-a,a-b) then the equation of third side can be :

Triangle formed by variable lines (a+b)x+(a-b)y-2ab=0 and (a-b)x+(a+b)y-2ab=0 and x+y=0 is (where a, b in R )

The straight lines 3x+y-4=0,x+3y-4=0 and x+y=0 form a triangle which is : a) isosceles b) right-angled c) equilateral d) scalene

If the straight lines 2x+3y-1=0,x+2y-1=0,and ax+by-1=0 form a triangle with the origin as orthocentre, then (a , b) is given by

A moving line intersects the lines x + y = 0 and x-y= 0 at the points A, B respectively such that the area of the triangle with vertices (0, 0), A & B has a constant area C. The locus of the mid-point AB is given by the equation

x+y=7 and a x^2+2h x y+a y^2=0,(a!=0) , are three real distinct lines forming a triangle. Then the triangle is isosceles (b) scalene equilateral (d) right angled

Let A,B,C be angles of triangles with vertex A -= (4,-1) and internal angular bisectors of angles B and C be x - 1 = 0 and x - y - 1 = 0 respectively. If A,B,C are angles of triangle at vertices A,B,C respectively then cot ((B)/(2))cot .((C)/(2)) =

Prove that (a b+x y)(a x+b y)>4a b x y(a , b ,x ,y >0)dot