A straight line 4x+y-1=0 passing through the | vedclass

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(B) The given lines are $L_1: 4x+y-1=0$ and $L_2: 3x-4y+1=0$. The slope of $L_1$ is $m_1 = -4$ and the slope of $L_2$ is $m_2 = \frac{3}{4}$.Point $A$

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$52x+89y+519=0$

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(B) The given lines are $L_1: 4x+y-1=0$ and $L_2: 3x-4y+1=0$. The slope of $L_1$ is $m_1 = -4$ and the slope of $L_2$ is $m_2 = \frac{3}{4}$.Point $A$ is $(2, -7)$. Line $AB$ is $L_1$. Point $B$ is the intersection of $L_1$ and $L_2$.Since $AB=AC$ and $A$ is common,the angle between $AB$ and $BC$ must be equal to the angle between $AC$ and $BC$. Let the slope of $AC$ be $m$. The angle $	heta$ between $AB$ and $BC$ is given by $	an 	heta = |\frac{m_1-m_2}{1+m_1m_2}| = |\frac{-4 - 3/4}{1 + (-4)(3/4)}| = |\frac{-19/4}{1-3}| = |\frac{-19/4}{-2}| = \frac{19}{8}$.Since $BC$ is the base of the isosceles triangle $ABC$,the line $BC$ makes the same angle with $AB$ and $AC$. Thus,the angle between $AC$ and $BC$ is also $	heta = 	an^{-1}(\frac{19}{8})$.Using $	an 	heta = |\frac{m-m_2}{1+mm_2}|$,we have $\frac{19}{8} = |\frac{m-3/4}{1+m(3/4)}| = |\frac{4m-3}{4+3m}|$.This gives two cases: $\frac{4m-3}{4+3m} = \frac{19}{8}$ or $\frac{4m-3}{4+3m} = -\frac{19}{8}$.Case $1$: $32m - 24 = 76 + 57m$ $\Rightarrow -25m = 100$ $\Rightarrow m = -4$ (This is line $AB$).Case $2$: $32m - 24 = -76 - 57m$ $\Rightarrow 89m = -52$ $\Rightarrow m = -\frac{52}{89}$.The equation of line $AC$ is $y - (-7) = -\frac{52}{89}(x - 2)$ $\Rightarrow 89(y+7) = -52(x-2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

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

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