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(B) The centroid $G$ of a triangle divides the line segment joining the orthocentre $H$ and circumcentre $O$ in the ratio $2:1$. Given $H = (-6, -8)$

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$16$

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(B) The centroid $G$ of a triangle divides the line segment joining the orthocentre $H$ and circumcentre $O$ in the ratio $2:1$. Given $H = (-6, -8)$ and $O = (3, 4)$,the centroid $G$ is:$G = \left( \frac{2(3) + 1(-6)}{2+1}, \frac{2(4) + 1(-8)}{2+1} ight) = (0, 0)$.The lines $2x+3y-1=0$ and $x+2y-1=0$ intersect at $(-1, 1)$.Since the orthocentre of the triangle is $(0, 0)$,the altitude from the vertex $(-1, 1)$ to the opposite side $ax+by-1=0$ passes through $(0, 0)$.The slope of the line joining $(-1, 1)$ and $(0, 0)$ is $m_1 = \frac{0-1}{0-(-1)} = -1$.The slope of the line $ax+by-1=0$ is $m_2 = -a/b$.Since the altitude is perpendicular to the side,$m_1 	imes m_2 = -1$ $\Rightarrow (-1) 	imes (-a/b) = -1$ $\Rightarrow a/b = -1$ $\Rightarrow a = -b$.The side is $ax - ay - 1 = 0$.Since the orthocentre $(0, 0)$ is the intersection of altitudes,the altitude from the intersection of $2x+3y-1=0$ and $ax-ay-1=0$ must pass through $(0, 0)$.Solving $2x+3y=1$ and $ax-ay=1$ gives the vertex $V = \left( \frac{a+3}{5a}, \frac{a-2}{5a} ight)$.The altitude from $V$ to $x+2y-1=0$ (slope $-1/2$) has slope $2$. The line passing through $(0, 0)$ with slope $2$ is $y=2x$.Substituting $V$ into $y=2x$: $\frac{a-2}{5a} = 2 \left( \frac{a+3}{5a} ight)$ $\Rightarrow a-2 = 2a+6$ $\Rightarrow a = -8$.Thus $b = 8$,and $|a-b| = |-8-8| = 16$.

If the orthocentre of the triangle formed by the lines $2x+3y-1=0$,$x+2y-1=0$,and $ax+by-1=0$ is the centroid of another triangle,whose circumcentre and orthocentre are $(3,4)$ and $(-6,-8)$ respectively,then the value of $|a-b|$ is..........

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