The equation of the line joining the centroid | vedclass

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Question

(B) Let the vertices be $A(-2, 3), B(2, -1)$,and $C(4, 0)$.First,find the orthocentre $H$. The slope of $BC = \frac{0 - (-1)}{4 - 2} = \frac{1}{2}$. T

Vedclass · Accepted Answer

$11x-y-14=0$

Vedclass · Answer

(B) Let the vertices be $A(-2, 3), B(2, -1)$,and $C(4, 0)$.First,find the orthocentre $H$. The slope of $BC = \frac{0 - (-1)}{4 - 2} = \frac{1}{2}$. The altitude from $A$ is perpendicular to $BC$,so its slope is $-2$. The equation is $y - 3 = -2(x + 2) \Rightarrow 2x + y + 1 = 0$.The slope of $AC = \frac{0 - 3}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}$. The altitude from $B$ is perpendicular to $AC$,so its slope is $2$. The equation is $y - (-1) = 2(x - 2)$ $\Rightarrow y + 1 = 2x - 4$ $\Rightarrow 2x - y - 5 = 0$.Solving $2x + y + 1 = 0$ and $2x - y - 5 = 0$ by adding them: $4x - 4 = 0 \Rightarrow x = 1$. Substituting $x=1$ into $2x + y + 1 = 0$ gives $2(1) + y + 1 = 0 \Rightarrow y = -3$. Thus,the orthocentre $H = (1, -3)$.Next,find the centroid $G = \left(\frac{-2+2+4}{3}, \frac{3-1+0}{3}\right) = \left(\frac{4}{3}, \frac{2}{3}\right)$.The equation of the line joining $H(1, -3)$ and $G\left(\frac{4}{3}, \frac{2}{3}\right)$ is given by $y - y_1 = m(x - x_1)$,where $m = \frac{\frac{2}{3} - (-3)}{\frac{4}{3} - 1} = \frac{\frac{11}{3}}{\frac{1}{3}} = 11$.So,$y - (-3) = 11(x - 1)$ $\Rightarrow y + 3 = 11x - 11$ $\Rightarrow 11x - y - 14 = 0$.

The equation of the line joining the centroid with the orthocentre of the triangle formed by the points $(-2, 3), (2, -1), (4, 0)$ is

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