In an isosceles triangle ABC, the coordinates | vedclass

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(b) $\angle ABC = 	an 	heta = \frac{{\frac{1}{2} - 1}}{{1 + \frac{1}{2}}} = - \frac{1}{3}$ (Here ${m_1} = \frac{1}{2},\,{m_2} = 1)$
$AB = AC$;&nbsp

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $\angle ABC = 	an 	heta = \frac{{\frac{1}{2} - 1}}{{1 + \frac{1}{2}}} = - \frac{1}{3}$ (Here ${m_1} = \frac{1}{2},\,{m_2} = 1)$
$AB = AC$;  $\angle ABC = \angle ACB$

Hence $ - \frac{1}{3} = \frac{{m - 1}}{{1 + m}}$ ==> $m = \frac{1}{2}$

(Here m is the gradient of line $AC$)

 Equation of line AC is,$y - 1 = \frac{1}{2}(x - 2)$ ==> $y = \frac{x}{2}$.

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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