A line 4x + y = 1 passes through the point A( | vedclass

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(A) The slopes of $AB$ and $BC$ are $m_1 = -4$ and $m_2 = \frac{3}{4}$ respectively. Let $\alpha$ be the angle between $AB$ and $BC$. Then,$	an \alph

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$52x + 89y + 519 = 0$

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(A) The slopes of $AB$ and $BC$ are $m_1 = -4$ and $m_2 = \frac{3}{4}$ respectively. Let $\alpha$ be the angle between $AB$ and $BC$. Then,$	an \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{-4 - \frac{3}{4}}{1 + (-4)(\frac{3}{4})} ight| = \left| \frac{-\frac{19}{4}}{1 - 3} ight| = \left| \frac{-\frac{19}{4}}{-2} ight| = \frac{19}{8}$.Since $AB = AC$,the triangle $ABC$ is isosceles with $\angle ABC = \angle ACB = \alpha$. Thus,the line $AC$ also makes an angle $\alpha$ with the line $BC$. Let $m$ be the slope of line $AC$. The equation of line $AC$ passing through $A(2, -7)$ is $y + 7 = m(x - 2)$.The angle between $AC$ and $BC$ is $\alpha$,so $	an \alpha = \left| \frac{m - \frac{3}{4}}{1 + m(\frac{3}{4})} ight| = \frac{19}{8}$.$\frac{4m - 3}{4 + 3m} = \pm \frac{19}{8}$.Case $1$: $8(4m - 3) = 19(4 + 3m)$ $\Rightarrow 32m - 24 = 76 + 57m$ $\Rightarrow -25m = 100$ $\Rightarrow m = -4$ (This is the slope of line $AB$).Case $2$: $8(4m - 3) = -19(4 + 3m)$ $\Rightarrow 32m - 24 = -76 - 57m$ $\Rightarrow 89m = -52$ $\Rightarrow m = -\frac{52}{89}$.Using $m = -\frac{52}{89}$ in $y + 7 = m(x - 2)$,we get $89(y + 7) = -52(x - 2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

$A$ line $4x + y = 1$ passes through the point $A(2, -7)$ and meets the line $BC$,whose equation is $3x - 4y + 1 = 0$,at the point $B$. The equation of the line $AC$ such that $AB = AC$ is

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