If the line 2x - 3y + 5 = 0 is the perpendicu | vedclass

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(B) Let the points be $A(1, -2)$ and $B(\alpha, \beta)$. The midpoint of $AB$ is $M = \left(\frac{1+\alpha}{2}, \frac{-2+\beta}{2}\right)$.Since $M$ l

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(B) Let the points be $A(1, -2)$ and $B(\alpha, \beta)$. The midpoint of $AB$ is $M = \left(\frac{1+\alpha}{2}, \frac{-2+\beta}{2}ight)$.Since $M$ lies on the line $2x - 3y + 5 = 0$,we have $2\left(\frac{1+\alpha}{2}ight) - 3\left(\frac{-2+\beta}{2}ight) + 5 = 0$,which simplifies to $1 + \alpha + 3 - 1.5\beta + 5 = 0$,or $2\alpha - 3\beta = -18$.The slope of $AB$ is $m_{AB} = \frac{\beta - (-2)}{\alpha - 1} = \frac{\beta + 2}{\alpha - 1}$.The slope of the given line $2x - 3y + 5 = 0$ is $m_L = \frac{2}{3}$.Since the line is the perpendicular bisector,$m_{AB} 	imes m_L = -1$,so $\left(\frac{\beta + 2}{\alpha - 1}ight) 	imes \left(\frac{2}{3}ight) = -1$.This gives $2\beta + 4 = -3\alpha + 3$,or $3\alpha + 2\beta = -1$.Solving the system of equations $2\alpha - 3\beta = -18$ and $3\alpha + 2\beta = -1$:Multiply the first by $2$ and the second by $3$: $4\alpha - 6\beta = -36$ and $9\alpha + 6\beta = -3$.Adding them gives $13\alpha = -39$,so $\alpha = -3$.Substituting $\alpha = -3$ into $3\alpha + 2\beta = -1$ gives $3(-3) + 2\beta = -1$,so $2\beta = 8$,$\beta = 4$.Thus,$\alpha + \beta = -3 + 4 = 1$.

If the line $2x - 3y + 5 = 0$ is the perpendicular bisector of the line segment joining $(1, -2)$ and $(\alpha, \beta)$,then $\alpha + \beta =$

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