Let A(a, b),B(3, 4),and C(-6, -8) respectivel | vedclass

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(C) We know that the centroid $A$ divides the line segment joining the orthocentre $C$ and circumcentre $B$ in the ratio $2:1$.Given $C(-6, -8)$ and $

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$\frac{17 \sqrt{5}}{7}$

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(C) We know that the centroid $A$ divides the line segment joining the orthocentre $C$ and circumcentre $B$ in the ratio $2:1$.Given $C(-6, -8)$ and $B(3, 4)$,the centroid $A(a, b)$ is given by the section formula:$a = \frac{2(3) + 1(-6)}{2+1} = \frac{6-6}{3} = 0$$b = \frac{2(4) + 1(-8)}{2+1} = \frac{8-8}{3} = 0$So,$A(0, 0)$.Now,the point $P(2a+3, 7b+5)$ becomes $P(2(0)+3, 7(0)+5) = P(3, 5)$.We need to find the distance of $P(3, 5)$ from the line $L: 2x+3y-4=0$ measured parallel to the line $M: x-2y-1=0$.The slope of line $M$ is $m = \frac{1}{2}$. Thus,$	an 	heta = \frac{1}{2}$,which implies $\sin 	heta = \frac{1}{\sqrt{5}}$ and $\cos 	heta = \frac{2}{\sqrt{5}}$.The coordinates of any point on the line passing through $P(3, 5)$ with slope $m = \frac{1}{2}$ are $(3+r \cos 	heta, 5+r \sin 	heta) = (3+\frac{2r}{\sqrt{5}}, 5+\frac{r}{\sqrt{5}})$.Substituting this into the line $L: 2x+3y-4=0$:$2(3+\frac{2r}{\sqrt{5}}) + 3(5+\frac{r}{\sqrt{5}}) - 4 = 0$$6 + \frac{4r}{\sqrt{5}} + 15 + \frac{3r}{\sqrt{5}} - 4 = 0$$17 + \frac{7r}{\sqrt{5}} = 0$$\frac{7r}{\sqrt{5}} = -17$$r = -\frac{17 \sqrt{5}}{7}$Since distance is the magnitude,the required distance is $|r| = \frac{17 \sqrt{5}}{7}$.

Let $A(a, b)$,$B(3, 4)$,and $C(-6, -8)$ respectively denote the centroid,circumcentre,and orthocentre of a triangle. Then,the distance of the point $P(2a+3, 7b+5)$ from the line $2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ is

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