In an isosceles triangle ABC,the coordinates | vedclass

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(B) Let the slope of line $AB$ be $m_1 = 2$ and the slope of line $BC$ be $m_{BC} = \frac{1 - 2}{2 - 1} = -1$.Let the slope of line $AC$ be $m_2$.The

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$y = \frac{x}{2}$

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(B) Let the slope of line $AB$ be $m_1 = 2$ and the slope of line $BC$ be $m_{BC} = \frac{1 - 2}{2 - 1} = -1$.Let the slope of line $AC$ be $m_2$.The angle $	heta$ between $AB$ and $BC$ is given by $	an 	heta = \left| \frac{m_1 - m_{BC}}{1 + m_1 m_{BC}} ight| = \left| \frac{2 - (-1)}{1 + 2(-1)} ight| = \left| \frac{3}{-1} ight| = 3$.Since $	riangle ABC$ is isosceles with $AB = AC$,the angle between $AC$ and $BC$ is also $	heta$.Thus,$	an 	heta = \left| \frac{m_{BC} - m_2}{1 + m_{BC} m_2} ight| = 3$.$\left| \frac{-1 - m_2}{1 - m_2} ight| = 3$.Case $1$: $\frac{-(1 + m_2)}{1 - m_2} = 3 \implies -1 - m_2 = 3 - 3m_2 \implies 2m_2 = 4 \implies m_2 = 2$ (This corresponds to line $AB$).Case $2$: $\frac{-(1 + m_2)}{1 - m_2} = -3 \implies 1 + m_2 = 3 - 3m_2 \implies 4m_2 = 2 \implies m_2 = \frac{1}{2}$.Since $AC$ passes through $C(2, 1)$ with slope $m_2 = \frac{1}{2}$,its equation is $y - 1 = \frac{1}{2}(x - 2) \implies y - 1 = \frac{x}{2} - 1 \implies y = \frac{x}{2}$.

In an isosceles triangle $ABC$,the coordinates of the points $B$ and $C$ on the base $BC$ are respectively $(1, 2)$ and $(2, 1)$. If the equation of the line $AB$ is $y = 2x$,then the equation of the line $AC$ is

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