In an isosceles triangle ABC , coordinates of points B and C on base BC are respectively (1, 2) and

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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

$y = \frac{1}{2}(x - 1)$

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

$y = x - 1$

$2y = x + 3$

Solution

(b) $\angle ABC = \tan \theta = \frac{{\frac{1}{2} – 1}}{{1 + \frac{1}{2}}} = – \frac{1}{3}$ (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,$y – 1 = \frac{1}{2}(x – 2)$ ==> $y = \frac{x}{2}$.

Standard 11

Mathematics

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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

Solution

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