If the equation of the median through vertex | vedclass

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(B) The median through vertex $A$ passes through the midpoint $M$ of the side $BC$. Midpoint $M$ of $BC$ is given by $(\frac{2-4}{2}, \frac{1+5}{2}) =

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$2$

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(B) The median through vertex $A$ passes through the midpoint $M$ of the side $BC$. Midpoint $M$ of $BC$ is given by $(\frac{2-4}{2}, \frac{1+5}{2}) = (-1, 3)$. The median passes through $A(3, k)$ and $M(-1, 3)$. The slope of the median $AM$ is $m = \frac{3-k}{-1-3} = \frac{3-k}{-4}$. The equation of the line $AM$ is $y - 3 = \frac{3-k}{-4}(x + 1)$. $-4y + 12 = (3-k)x + (3-k)$. $(3-k)x + 4y = 9+k$. Comparing this with the given equation $x + 4y = p$,we equate the coefficients of $x$ and $y$. Since the coefficient of $y$ is $4$ in both,we have $3-k = 1$,which gives $k = 2$. Thus,$k = 2$.

If the equation of the median through vertex $A(3, k)$ of $\triangle ABC$ with vertices $B(2, 1)$ and $C(-4, 5)$ is $x + 4y = p$,then $k = ?$ where $p$ and $k$ are constants.

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