Let A equiv (3, 2) and B equiv (5, 1). An equ | vedclass

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(D) The midpoint $M$ of $AB$ is $\left( \frac{3+5}{2}, \frac{2+1}{2} \right) = (4, 3/2)$.The slope of $AB$ is $m_{AB} = \frac{1-2}{5-3} = -1/2$.The sl

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$\left( 4 + \frac{1}{6}\sqrt{3}, \frac{3}{2} + \frac{1}{3}\sqrt{3} \right)$

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(D) The midpoint $M$ of $AB$ is $\left( \frac{3+5}{2}, \frac{2+1}{2} \right) = (4, 3/2)$.The slope of $AB$ is $m_{AB} = \frac{1-2}{5-3} = -1/2$.The slope of the altitude $PM$ is $m_{PM} = -1/m_{AB} = 2$.The length of $AB$ is $a = \sqrt{(5-3)^2 + (1-2)^2} = \sqrt{4+1} = \sqrt{5}$.The height of the equilateral triangle is $h = \frac{a\sqrt{3}}{2} = \frac{\sqrt{15}}{2}$.The unit vector perpendicular to $AB$ is $\vec{n} = \frac{(1, 2)}{\sqrt{5}}$.Since $P$ is remote from the origin,we move from $M$ in the direction of $\vec{n}$ away from the origin.$P = M + h \cdot \frac{(1, 2)}{\sqrt{5}} = (4, 3/2) + \frac{\sqrt{15}}{2} \cdot \frac{(1, 2)}{\sqrt{5}} = (4, 3/2) + \frac{\sqrt{3}}{2}(1, 2) = (4 + \frac{\sqrt{3}}{2}, 3/2 + \sqrt{3})$.In an equilateral triangle,the orthocentre is the same as the centroid $G = \frac{A+B+P}{3}$.$G = \left( \frac{3+5+4+\frac{\sqrt{3}}{2}}{3}, \frac{2+1+\frac{3}{2}+\sqrt{3}}{3} \right) = \left( \frac{12+\frac{\sqrt{3}}{2}}{3}, \frac{\frac{9}{2}+\sqrt{3}}{3} \right) = \left( 4 + \frac{\sqrt{3}}{6}, \frac{3}{2} + \frac{\sqrt{3}}{3} \right)$.

Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. An equilateral triangle $ABP$ is constructed on the side of $AB$ remote from the origin. The orthocentre of triangle $ABP$ is:

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