A(1, -2), B(-2, 3), C(-1, -3) are the vertice | vedclass

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(C) The slope of $BC$ is $\frac{-3 - 3}{-1 - (-2)} = \frac{-6}{1} = -6$.The slope of the altitude $L_1$ from $A$ to $BC$ is the negative reciprocal of

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$13$

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(C) The slope of $BC$ is $\frac{-3 - 3}{-1 - (-2)} = \frac{-6}{1} = -6$.The slope of the altitude $L_1$ from $A$ to $BC$ is the negative reciprocal of the slope of $BC$, which is $\frac{1}{6}$.The equation of $L_1$ passing through $A(1, -2)$ is $y - (-2) = \frac{1}{6}(x - 1) \Rightarrow y + 2 = \frac{x}{6} - \frac{1}{6} \Rightarrow y = \frac{x}{6} - \frac{13}{6}$ ... $(i)$The midpoint of $AB$ is $\left(\frac{1 - 2}{2}, \frac{-2 + 3}{2}\right) = \left(-\frac{1}{2}, \frac{1}{2}\right)$.The slope of $AB$ is $\frac{3 - (-2)}{-2 - 1} = \frac{5}{-3} = -\frac{5}{3}$.The slope of the perpendicular bisector $L_2$ is $\frac{3}{5}$.The equation of $L_2$ is $y - \frac{1}{2} = \frac{3}{5}(x + \frac{1}{2}) \Rightarrow y = \frac{3x}{5} + \frac{3}{10} + \frac{5}{10} \Rightarrow y = \frac{3x}{5} + \frac{4}{5}$ ... (ii)Equating $(i)$ and (ii) to find the intersection point $(l, m)$:$\frac{l}{6} - \frac{13}{6} = \frac{3l}{5} + \frac{4}{5} \Rightarrow \frac{l}{6} - \frac{3l}{5} = \frac{4}{5} + \frac{13}{6} \Rightarrow \frac{5l - 18l}{30} = \frac{24 + 65}{30} \Rightarrow -13l = 89 \Rightarrow 13l = -89$.Since $13l = -89$, we have $l = -\frac{89}{13}$.Substituting $l$ into $(i)$: $m = \frac{-89/13}{6} - \frac{13}{6} = \frac{-89}{78} - \frac{169}{78} = \frac{-258}{78} = -\frac{43}{13}$.Then $26m - 3 = 26(-\frac{43}{13}) - 3 = 2(-43) - 3 = -86 - 3 = -89$.Since $13l = -89$, we have $26m - 3 = 13l$.

$A(1, -2), B(-2, 3), C(-1, -3)$ are the vertices of a triangle $ABC$. $L_1$ is the perpendicular drawn from $A$ to $BC$ and $L_2$ is the perpendicular bisector of $AB$. If $(l, m)$ is the point of intersection of $L_1$ and $L_2$, then $26m - 3 =$ (in $l$)

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