Question
Let $ABC$ be a triangle. Let $A$ be the point $(1,2),y = x$ is the perpendicular bisector of $AB$ and $x - 2y + 1 = 0$ is the angle bisector of angle $C$. If the equation of $BC$ is given by $ax + by - 5 = 0$, then the value of $a + b$ is :
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Solution
Sol. [Key Concept: $A$ lies on the perpendicular bisector of $AB$ and on the angle bisector of $\angle A$; reflect to locate $B,C$ and form line $BC$.]
Image of $A$ say $A^{'}$ w.r.t. $x - 2y + 1 = 0$ lies on $BC$.
$$\frac{x - 1}{1} = \frac{y - 2}{- 2} = - 2\frac{(1 - 4 + 1)}{1 + 2^{2}} = \frac{4}{5}A^{'} \equiv \left( \frac{9}{5},\frac{2}{5} \right) $$Equation of $BC$ joining $A^{'}\left( \frac{9}{5},\frac{2}{5} \right)$ and $B(2,1)$ is
$${y - 1 = \frac{1 - \frac{2}{5}}{2 - \frac{9}{5}}(x - 2) = \frac{3}{1}(x - 2) }{3x - y - 5 = 0 \Rightarrow a + b = 3 - 1 = 2}$$
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