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

Solution:

If 2x – 3y – 5 = 0 is perpendicular bisector. Then mid point of AB lies on the line mid point of AB

$\begin{array}{l} A B = (\frac{3 + α}{2}, \frac{- 4 + β}{2}) \\ \Rightarrow 2 (\frac{3 + α}{2}) - 3 (\frac{- 4 + β}{2}) - 5 = 0 \\ \Rightarrow 6 + 2 α + 12 - 3 β - 10 = 0 \\ \Rightarrow 2 α - 3 β + 8 = 0 - - (1) \\ \bar{A B} ⊥ 2 x - 3 y - 5 = 0 \\ \Rightarrow (\frac{β + 4}{α - 3}) (\frac{2}{3}) = - 1 \\ \Rightarrow 2 β + 8 = - 3 α + 9 \\ \Rightarrow 3 α + 2 β - 1 = 0 - - (2) \end{array}$

Solving (1) & (2)

$\begin{array}{l} α β 1 \\ \begin{matrix} - 3 & 8 & 2 & - 3 \\ 2 & - 1 & 3 & 2 \end{matrix} \\ \Rightarrow \frac{α}{3 - 16} = \frac{β}{24 + 2} = \frac{1}{4 + 9} \\ \Rightarrow \frac{α}{- 13} = \frac{β}{26} = \frac{1}{13} \\ ∴ α = \frac{- 13}{13} \Rightarrow - 1, β = \frac{26}{13} \Rightarrow 2 \\ ∴ α + β = - 1 + 2 = 1 \end{array}$

If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, -4) and $(α, β) t h e n α + β i s_____$

Solution:

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