If 2x – 3y – 5= 0 is the perpendicular bisector of the line segment joining (3,−4) and (α,β). then find α+β.

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If 2x – 3y – 5= 0 is the perpendicular bisector $of the line segment joining (3, - 4) and (α, β) .$ $then find α + β .$

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Detailed Solution

$If (α, β) is the image of the point (3, - 4) w.r.t$ st.line 2x – 3y – 5 = 0 then
$\frac{h - x_{1}}{a} = \frac{k - y_{1}}{b} = \frac{- 2 ({ax}_{1} + {by}_{1} + c)}{a^{2} + b^{2}}$
$\begin{array}{l} \frac{α - 3}{2} = \frac{β + 4}{- 3} = \frac{- 2 (2 .3 - 3 (- 4) - 5)}{(2)^{2} + (3)^{2}} \\ \Rightarrow \frac{α - 3}{2} = \frac{β + 4}{- 3} = \frac{- 2 (6 + 12 - 5)}{13} \\ ∴ α = - 1, β = 2 \\ ∴ α + β = 1 \end{array}$