If 2x + 3y + 4 = 0 is the perpendicular bisec | vedclass

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(A) The line $2x + 3y + 4 = 0$ is the perpendicular bisector of $AB$. First,the midpoint $M$ of $AB$ lies on the line: $M = (\frac{1+\alpha}{2}, \frac

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$-81$

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(A) The line $2x + 3y + 4 = 0$ is the perpendicular bisector of $AB$. First,the midpoint $M$ of $AB$ lies on the line: $M = (\frac{1+\alpha}{2}, \frac{2+\beta}{2})$ $2(\frac{1+\alpha}{2}) + 3(\frac{2+\beta}{2}) + 4 = 0$ $1 + \alpha + 3 + \frac{3\beta}{2} + 4 = 0$ $2\alpha + 3\beta + 16 = 0$ ... $(i)$ Second,the slope of $AB$ is perpendicular to the slope of the given line $(-2/3)$: $\frac{\beta - 2}{\alpha - 1} 	imes (-\frac{2}{3}) = -1$ $2(\beta - 2) = 3(\alpha - 1)$ $2\beta - 4 = 3\alpha - 3$ $3\alpha - 2\beta + 1 = 0$ ... $(ii)$ Solving $(i)$ and $(ii)$: Multiply $(i)$ by $2$ and $(ii)$ by $3$: $4\alpha + 6\beta = -32$ $9\alpha - 6\beta = -3$ Adding them: $13\alpha = -35 \Rightarrow \alpha = -\frac{35}{13}$ Substituting $\alpha$ in $(ii)$: $3(-\frac{35}{13}) - 2\beta + 1 = 0 -\frac{105}{13} + 1 = 2\beta -\frac{92}{13} = 2\beta$ $\Rightarrow \beta = -\frac{46}{13}$ Finally,$13\alpha + 13\beta = 13(-\frac{35}{13}) + 13(-\frac{46}{13}) = -35 - 46 = -81$.

If $2x + 3y + 4 = 0$ is the perpendicular bisector of the line segment joining the points $A(1, 2)$ and $B(\alpha, \beta)$,then the value of $13\alpha + 13\beta$ equals $......$

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