In triangle ABC,the coordinates of the vertex | vedclass

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(B) Let $C = (h, k)$. Since $C$ lies on the angle bisector $7x - 4y - 1 = 0$,we have $7h - 4k - 1 = 0$,or $h = \frac{4k + 1}{7}$.Since $M$ is the midp

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$18x - y = 49$

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(B) Let $C = (h, k)$. Since $C$ lies on the angle bisector $7x - 4y - 1 = 0$,we have $7h - 4k - 1 = 0$,or $h = \frac{4k + 1}{7}$.Since $M$ is the midpoint of $AC$,$M = \left(\frac{-3 + h}{2}, \frac{1 + k}{2}ight) = \left(\frac{-3 + \frac{4k+1}{7}}{2}, \frac{1+k}{2}ight) = \left(\frac{4k - 20}{14}, \frac{1+k}{2}ight) = \left(\frac{2k - 10}{7}, \frac{1+k}{2}ight)$.Since $M$ lies on the median $BM$ $(2x + y - 3 = 0)$,we have $2\left(\frac{2k - 10}{7}ight) + \frac{1+k}{2} - 3 = 0$.Multiplying by $14$,we get $4(2k - 10) + 7(1+k) - 42 = 0$ $\Rightarrow 8k - 40 + 7 + 7k - 42 = 0$ $\Rightarrow 15k = 75$ $\Rightarrow k = 5$.Then $h = \frac{4(5) + 1}{7} = 3$. So,$C = (3, 5)$.The slope of $AC$ is $m_{AC} = \frac{5 - 1}{3 - (-3)} = \frac{4}{6} = \frac{2}{3}$.The slope of the angle bisector $CN$ is $m_{bisector} = \frac{7}{4}$.Let $m$ be the slope of $BC$. Using the angle bisector property $	an(\angle BCN) = 	an(\angle NCA)$:$\left|\frac{m - 7/4}{1 + m(7/4)}ight| = \left|\frac{7/4 - 2/3}{1 + (7/4)(2/3)}ight| = \left|\frac{21 - 8}{12 + 14}ight| = \frac{13}{26} = \frac{1}{2}$.Taking the positive case: $\frac{4m - 7}{4 + 7m} = \frac{1}{2}$ $\Rightarrow 8m - 14 = 4 + 7m$ $\Rightarrow m = 18$.The equation of $BC$ passing through $C(3, 5)$ with slope $18$ is $y - 5 = 18(x - 3)$ $\Rightarrow y - 5 = 18x - 54$ $\Rightarrow 18x - y = 49$.

In $\triangle ABC$,the coordinates of the vertex $A$ are $(-3, 1)$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $\angle C$ is $7x - 4y - 1 = 0$,then the equation of the side $BC$ is

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