A triangle ABC is given where vertex A is (1,1) and the orthocenter is (2,4). Also sides AB and AC are members of the family of lines ax+by+c=0 where a,b,c are in A.P. Find the coordinates of the circumcentre. Also find the angles of the triangle justify whether the trianlge is acute angled triangle of obtuse.

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

1. The vertex B is
(A) (2, 1)
(B) (1, -2)
(C) (-1, 2)
(D) None of these

2. Triangle ABC is a/an
(A) obtuse angled triangle
(B) right angled triangle
(C) acute angled triangle
(D) equilateral triangle

3. The coordinate of the circumcentre is?

Solution:
Since ax + by + c = 0 is a family of lines which satisfy AB and Ac, so A(1, 1) must lie on it. It gives us
a(1) + b(1) + c = 0
a + b + c = 0
Now a, b and c are in A.P. Let us assume that the common difference is d. So,
a = a
b = a + d
c = a + 2d

Putting these values into a + b + c = 0 , we get;
a = -d
So b = 0 and c = -a
Now let us get back to ax + by + c = 0
ax + (0)y -a = 0
ax = a
Therefore x = 1
If we look the the options, we can clearly say it is (B) (1, -2)

Now Again for coordinate of C lets say (h,k).

So C is (-17,4)
Now we have A (1,1,), B(2,-1), and C(-17,4)
SO we can find circum center and angles by using direct formula.
Regards