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(B) The centroid $C$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 +

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$(-9, -6)$

Vedclass · Answer

(B) The centroid $C$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.For vertices $(3, -1)$,$(1, 3)$,and $(2, 4)$,the centroid is $C = \left(\frac{3 + 1 + 2}{3}, \frac{-1 + 3 + 4}{3}\right) = (2, 2)$.To find the intersection point $P$ of $x + 3y - 1 = 0$ and $3x - y + 1 = 0$,we solve the system:$x + 3y = 1$ and $3x - y = -1$.Multiplying the second equation by $3$: $9x - 3y = -3$.Adding the two equations: $(x + 3y) + (9x - 3y) = 1 - 3$ $\Rightarrow 10x = -2$ $\Rightarrow x = -\frac{1}{5}$.Substituting $x = -\frac{1}{5}$ into $x + 3y = 1$: $-\frac{1}{5} + 3y = 1$ $\Rightarrow 3y = \frac{6}{5}$ $\Rightarrow y = \frac{2}{5}$.So,$P = \left(-\frac{1}{5}, \frac{2}{5}\right)$.The slope $m$ of the line passing through $C(2, 2)$ and $P\left(-\frac{1}{5}, \frac{2}{5}\right)$ is $m = \frac{2 - 2/5}{2 - (-1/5)} = \frac{8/5}{11/5} = \frac{8}{11}$.The equation of the line is $y - 2 = \frac{8}{11}(x - 2)$ $\Rightarrow 11y - 22 = 8x - 16$ $\Rightarrow 8x - 11y + 6 = 0$.Checking the options,for $(-9, -6)$: $8(-9) - 11(-6) + 6 = -72 + 66 + 6 = 0$.Thus,the line passes through $(-9, -6)$.

Let $C$ be the centroid of the triangle with vertices $(3, -1)$,$(1, 3)$,and $(2, 4)$. Let $P$ be the point of intersection of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0$. Then the line passing through the points $C$ and $P$ also passes through the point

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