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(C) The coordinates of point $A$ using the section formula are $\left( \frac{3k - 5}{k + 1}, \frac{5k + 1}{k + 1} \right)$.Given $B = (1, 5)$ and $C =

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$7, 31/9$

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(C) The coordinates of point $A$ using the section formula are $\left( \frac{3k - 5}{k + 1}, \frac{5k + 1}{k + 1} ight)$.Given $B = (1, 5)$ and $C = (7, -2)$,the area of $	riangle ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 2$.Substituting the coordinates: $\frac{1}{2} \left| \frac{3k - 5}{k + 1}(5 - (-2)) + 1(-2 - \frac{5k + 1}{k + 1}) + 7(\frac{5k + 1}{k + 1} - 5) ight| = 2$.$\frac{1}{2} \left| \frac{7(3k - 5) - (2k + 2 + 5k + 1) + 7(5k + 1 - 5k - 5)}{k + 1} ight| = 2$.$\frac{1}{2} \left| \frac{21k - 35 - 7k - 3 - 28}{k + 1} ight| = 2$.$|14k - 66| = 4|k + 1|$.Case $1$: $14k - 66 = 4k + 4$ $\Rightarrow 10k = 70$ $\Rightarrow k = 7$.Case $2$: $14k - 66 = -4k - 4$ $\Rightarrow 18k = 62$ $\Rightarrow k = 31/9$.

The point $A$ divides the join of the points $(-5, 1)$ and $(3, 5)$ in the ratio $k : 1$. The coordinates of the points $B$ and $C$ are $(1, 5)$ and $(7, -2)$ respectively. If the area of the triangle $ABC$ is $2$ square units,then $k =$

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