A line L_1 passing through A(3,4) and having | vedclass

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(A) The equation of line $L_1$ passing through $A(3,4)$ with slope $m_1 = 1$ is:$y - 4 = 1(x - 3) \implies x - y + 1 = 0$.Let the slope of line $AC$ b

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$7x - y - 17 = 0$

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(A) The equation of line $L_1$ passing through $A(3,4)$ with slope $m_1 = 1$ is:$y - 4 = 1(x - 3) \implies x - y + 1 = 0$.Let the slope of line $AC$ be $m$. The line $BC$ has the equation $2x - y + 4 = 0$,so its slope is $m_{BC} = 2$.Since $AB = AC$,$	riangle ABC$ is an isosceles triangle with $A$ as the vertex. Thus,the base angles are equal: $\angle B = \angle C$.The angle between $AC$ (slope $m$) and $BC$ (slope $2$) is equal to the angle between $AB$ (slope $1$) and $BC$ (slope $2$).Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,we have:$\left| \frac{m - 2}{1 + 2m} ight| = \left| \frac{1 - 2}{1 + (1)(2)} ight| = \left| \frac{-1}{3} ight| = \frac{1}{3}$.This gives two cases:$1) \frac{m - 2}{1 + 2m} = \frac{1}{3} \implies 3m - 6 = 1 + 2m \implies m = 7$.$2) \frac{m - 2}{1 + 2m} = -\frac{1}{3} \implies 3m - 6 = -1 - 2m \implies 5m = 5 \implies m = 1$.Since $m = 1$ corresponds to line $L_1$ (which is $AB$),the slope of $AC$ must be $m = 7$.The equation of $AC$ passing through $A(3,4)$ with slope $7$ is:$y - 4 = 7(x - 3) \implies y - 4 = 7x - 21 \implies 7x - y - 17 = 0$.

$A$ line $L_1$ passing through $A(3,4)$ and having slope $1$ cuts another line $L_2$ passing through $C$ at $B$,such that $AB = AC$. If the equation of line $BC$ is $2x - y + 4 = 0$,then the equation of $AC$ is

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