Question

If the equation of the line passing through the orthocentre and circumcentre of triangle \(ABC\), whose vertices are \(A(3,1)\), \(B(3,3)\), \(C(6,1)\), is \(2x+by+c=0\), then \(b+c=\)

• A. \(-6\)
• B. \(6\)
• C. \(0\)
• D. \(2\)

Hint:

For right triangles: \[ \text{Orthocentre}=\text{right angled vertex} \] and \[ \text{Circumcentre}=\text{midpoint of hypotenuse}. \] This greatly simplifies Euler line problems.

Answer 1

The Correct Option is A

Concept: The line joining the orthocentre and circumcentre of a triangle is called the Euler line. For a right triangle, the orthocentre is the vertex at the right angle and the circumcentre is the midpoint of the hypotenuse.

Step 1: Identify the nature of the triangle.
Given vertices: \[ A(3,1),\qquad B(3,3),\qquad C(6,1). \] Observe: \[ AB \text{ is vertical} \] and \[ AC \text{ is horizontal}. \] Therefore, \[ AB\perp AC. \] Hence angle \(A\) is \(90^\circ\). Thus triangle \(ABC\) is right angled at \(A\).

Step 2: Find orthocentre and circumcentre.
Since the triangle is right angled at \(A\), \[ \text{Orthocentre}=A=(3,1). \] The circumcentre is midpoint of hypotenuse \(BC\). Midpoint of \(BC\): \[ \left( \frac{3+6}{2}, \frac{3+1}{2} \right) = \left( \frac92,2 \right). \] Therefore circumcentre is \[ \left(\frac92,2\right). \]

Step 3: Find equation of line through these two points.
Slope: \[ m= \frac{2-1}{\frac92-3} = \frac1{\frac32} = \frac23. \] Using point-slope form through \((3,1)\), \[ y-1=\frac23(x-3). \] Multiplying by \(3\), \[ 3y-3=2x-6. \] Rearranging, \[ 2x-3y-3=0. \] Comparing with \[ 2x+by+c=0, \] we get \[ b=-3,\qquad c=-3. \] Hence, \[ b+c=-3-3=-6. \] Therefore, \[ \boxed{-6}. \]