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 $M$ of $AB$ is $\left(\frac{\alpha + 1}{2}, \frac{\beta - 2}{2}\right)$.Since $M

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(B) Let the points be $A(1, -2)$ and $B(\alpha, \beta)$. The midpoint $M$ of $AB$ is $\left(\frac{\alpha + 1}{2}, \frac{\beta - 2}{2}ight)$.Since $M$ lies on the line $2x - 3y + 5 = 0$,we have $2\left(\frac{\alpha + 1}{2}ight) - 3\left(\frac{\beta - 2}{2}ight) + 5 = 0$,which simplifies to $2\alpha - 3\beta + 18 = 0$ $(i)$.The slope of line $AB$ is $m_1 = \frac{\beta + 2}{\alpha - 1}$. The slope of the given line $2x - 3y + 5 = 0$ is $m_2 = \frac{2}{3}$.Since $AB$ is perpendicular to the given line,$m_1 	imes m_2 = -1$,so $\left(\frac{\beta + 2}{\alpha - 1}ight) 	imes \frac{2}{3} = -1$,which gives $2\beta + 4 = -3\alpha + 3$,or $3\alpha + 2\beta + 1 = 0$ $(ii)$.Solving equations $(i)$ and $(ii)$ simultaneously: Multiply $(i)$ by $2$ and $(ii)$ by $3$: $4\alpha - 6\beta + 36 = 0$ and $9\alpha + 6\beta + 3 = 0$.Adding these gives $13\alpha + 39 = 0$,so $\alpha = -3$. Substituting $\alpha = -3$ into $(ii)$,we get $3(-3) + 2\beta + 1 = 0$,so $2\beta = 8$,which means $\beta = 4$.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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