Consider the triangle ABC having vertex A(1,1) and its orthocentre is (2,4). Also side AB & BC are members of the family of line, `ax + by + c = 0` where a,b,c are in A.P.
`DeltaABC` is a:

Consider the triangle ABC having vertex A(1,1) and its orthocentre is (2,4). Also side AB & BC are members of the family of line, ax + by + c = 0 where a,b,c are in A.P. The vertex C is :

Consider the triangle ABC having vertex A(1,1) and its orthocentre is (2,4). Also side AB & BC are members of the family of line, ax + by + c = 0 where a,b,c are in A.P. The vertex C is :

A triangle ABC is given where vertex A is (1,1) and the orthocenter is (2,4). Also sides AB and BC are members of the family of lines ax+by+c=0 where a,b,c are in A.P . Based on above written,solve the following questions.The vertex B is

If in Delta ABC the sides a,b,c are in A.P. Then

If in Delta ABC , sides a, b, c are in A.P. then

If in Delta ABC , sides a, b, c are in A.P. then

In triangle ABC are in A.P and b:c= sqrt(3): sqrt(2) , then

The sides a,b,c of Delta ABC are in G.P. , where log a - log 2b, log 2b -log 3c, log 3c - log a , are in A.P., then Delta ABC is :

A triangle ABC with vertices A(-1,0),B(-2,3/4) & C(-3,-7/6) has its orthocentre H, then the orthocentre of triangle BCH will be

Equation of line which is equally inclined to the axis and passes through a common points of family of lines 4acx + y(ab + bc + ca- abc) +abc = 0 (where a, b, c> 0 are in H.P.) is