If the line $2 x-3 y+5=0$ is the perpendicular bisector of the line segment joining $(1,-2)$ and $(\alpha,…
If the line $2 x-3 y+5=0$ is the perpendicular bisector of the line segment joining $(1,-2)$ and $(\alpha, \beta)$, then $\alpha+\beta=$
7
1
-1
-7
Solution
$2 x-3 y+5=0$ is perpendicular bisector then mid point of $A(1,-2)$ and $(\alpha, \beta)$ lines on the line segment AB
$\mathrm{AB}=\left(\frac{\alpha+1}{2}, \frac{\beta-2}{2}\right) \Rightarrow 2\left(\frac{\alpha+1}{2}\right)-3\left(\frac{\beta-2}{2}\right)+5=0$
$\Rightarrow 2 \alpha-3 \beta+18=0 \qquad ....\mathrm{(i)}$
$\begin{aligned}
& \overrightarrow{\mathrm{AB}} \perp 2 x-3 y+5=0 \\
& \Rightarrow\left(\frac{\beta+2}{\alpha-1}\right)\left(\frac{2}{3}\right)=-1 \Rightarrow 2 \beta+4=-3 \alpha+3 \\
& 3 \alpha+2 \beta+1=0 \qquad ....\mathrm{(ii)}
\end{aligned}$
Solving (i) and (ii), we get
$\alpha=-3, \beta=4$
So, $\alpha+\beta=1$