Question
Line $\frac{x}{a} + \frac{y}{b} = 1$ cuts the coordinate axes at $A(a,0)$ and $B(0,b)$ and the line $\frac{x}{a^{'}} + \frac{y}{b^{'}} = - 1$ at $A^{'}\left( - a^{'},0 \right)$ and $B^{'}\left( 0, - b^{'} \right)$. If the points $A,B,A^{'},B^{'}$ are concyclic then the orthocentre of the triangle $ABA^{'}$ is:
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Solution
Sol. [Key Concept: Concyclic intercept points fix $a'b' = ab$; the orthocentre of $\bigtriangleup \ ABA'$ then has two possible positions.]
$H$ is the image of $D$
$$\Rightarrow \ H \equiv \left( 0,b^{'} \right)$$
Applying power of ' $O$ '
$$aa^{'} = bb^{'} \Rightarrow b^{'} = \frac{aa^{'}}{b}$$
✘ (a) not the orthocentre for the concyclic configuration.
✔ (b) $(0,b')$ is a valid orthocentre.
✔ (c) $(0,b)$ is the other valid orthocentre.
✘ (d) does not satisfy the orthocentre condition.
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