Question

If \(A(2,1)\), \(B(4,k)\), \(C(3,4)\) are vertices of triangle right angled at B and \(k\) is not an odd number, then equation of line joining orthocentre and circumcentre is

• A. \(x+y=6\)
• B. \(x-y=0\)
• C. \(3x+y=14\)
• D. \(x+3y=10\)

Hint:

In right triangles, orthocentre is the right angled vertex and circumcentre lies at midpoint of hypotenuse.

Answer 1

The Correct Option is A

Concept: In right angled triangle: \[ \text{Orthocentre = right angle vertex} \] \[ \text{Circumcentre = midpoint of hypotenuse} \] Euler line joins them.

Step 1: Use perpendicular condition.
Since angle at B is \(90^\circ\) \[ m_{AB}\cdot m_{BC}=-1 \] \[ \frac{k-1}{2}\times\frac{4-k}{-1}=-1 \] Solving: \[ (k-1)(4-k)=2 \] \[ k^2-5k+6=0 \] \[ (k-2)(k-3)=0 \] Since \(k\) not odd \[ k=2 \] Thus \[ B=(4,2) \]

Step 2: Orthocentre and circumcentre.
Orthocentre \[ H=(4,2) \] Hypotenuse AC midpoint \[ O= \left( \frac{2+3}{2}, \frac{1+4}{2} \right) = \left( \frac52,\frac52 \right) \]

Step 3: Equation through H and O.
Slope \[ =\frac{2-\frac52}{4-\frac52} = -\frac13 \] Equation \[ y-2=-\frac13(x-4) \] \[ x+3y=10 \] Hence \[ \boxed{x+3y=10} \]