A line 4x+y=1 passes through the point A(2,-7 | vedclass

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(A) The slope of line $AB$ $(4x+y=1)$ is $m_1 = -4$. The slope of line $BC$ $(3x-4y+1=0)$ is $m_2 = \frac{3}{4}$.Let $\alpha$ be the angle between $AB

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$52x+89y+519=0$

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(A) The slope of line $AB$ $(4x+y=1)$ is $m_1 = -4$. The slope of line $BC$ $(3x-4y+1=0)$ is $m_2 = \frac{3}{4}$.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 $\angle ABC = \angle ACB = \alpha$.Let the slope of line $AC$ be $m$. Since $AC$ passes through $A(2,-7)$,its equation is $y+7 = m(x-2)$.The angle between $AC$ and $BC$ is also $\alpha$,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 $AB$).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 $y+7 = m(x-2)$:$89(y+7) = -52(x-2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

$A$ 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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