Let the coordinates of the two points A and B | vedclass

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(C) The points are $A(1, 2)$ and $B(7, 5)$. The point of trisection nearer to $B$ divides $AB$ in the ratio $2:1$. Using the section formula,the coord

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$3x - y - 11 = 0$

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(C) The points are $A(1, 2)$ and $B(7, 5)$. The point of trisection nearer to $B$ divides $AB$ in the ratio $2:1$. Using the section formula,the coordinates are $P = \left( \frac{2(7) + 1(1)}{2+1}, \frac{2(5) + 1(2)}{2+1} ight) = \left( \frac{15}{3}, \frac{12}{3} ight) = (5, 4)$. The slope of line $AB$ is $m = \frac{5-2}{7-1} = \frac{3}{6} = \frac{1}{2}$. Let the new slope be $m'$. Since the line is rotated by $45^{\circ}$ anti-clockwise,$m' = 	an(	heta + 45^{\circ}) = \frac{	an 	heta + 	an 45^{\circ}}{1 - 	an 	heta 	an 45^{\circ}}$,where $	an 	heta = \frac{1}{2}$. $m' = \frac{1/2 + 1}{1 - (1/2)(1)} = \frac{3/2}{1/2} = 3$. The equation of the new line passing through $(5, 4)$ with slope $3$ is $y - 4 = 3(x - 5)$. $y - 4 = 3x - 15 \Rightarrow 3x - y - 11 = 0$.

Let the coordinates of the two points $A$ and $B$ be $(1, 2)$ and $(7, 5)$ respectively. The line $AB$ is rotated through $45^{\circ}$ in the anti-clockwise direction about the point of trisection of $AB$ which is nearer to $B$. The equation of the line in the new position is:

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