L_1 equiv 2x+y-3=0 and L_2 equiv ax+by+c=0 ar | vedclass

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(A) The slopes of the lines are $m_1 = -2$,$m_2 = -\frac{a}{b}$,and $m_3 = -\frac{1}{2}$.Since $L_1$ and $L_2$ are equal sides,the angle between $L_1$

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$\frac{121}{2}$

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(A) The slopes of the lines are $m_1 = -2$,$m_2 = -\frac{a}{b}$,and $m_3 = -\frac{1}{2}$.Since $L_1$ and $L_2$ are equal sides,the angle between $L_1$ and $L_3$ must equal the angle between $L_2$ and $L_3$.Using the formula $	an 	heta = |\frac{m_1-m_2}{1+m_1m_2}|$,we have $|\frac{-2 - (-1/2)}{1 + (-2)(-1/2)}| = |\frac{-a/b - (-1/2)}{1 + (-a/b)(-1/2)}|$.$|\frac{-3/2}{2}| = |\frac{-2a+b}{2b+a}| \Rightarrow \frac{3}{4} = |\frac{2a-b}{a+2b}|$.Case $1$: $\frac{3}{4} = \frac{2a-b}{a+2b}$ $\Rightarrow 3a+6b = 8a-4b$ $\Rightarrow 5a = 10b$ $\Rightarrow a=2b$.Since $(5,1)$ lies on $L_2$,$5a+b+c=0$ $\Rightarrow 10b+b+c=0$ $\Rightarrow c=-11b$.Then $\frac{b^2}{|ac|} = \frac{b^2}{|(2b)(-11b)|} = \frac{b^2}{22b^2} = \frac{1}{22}$.Case $2$: $-\frac{3}{4} = \frac{2a-b}{a+2b}$ $\Rightarrow -3a-6b = 8a-4b$ $\Rightarrow 11a = -2b$ $\Rightarrow a=-\frac{2b}{11}$.Since $(5,1)$ lies on $L_2$,$5(-\frac{2b}{11})+b+c=0 \Rightarrow c = \frac{10b}{11}-b = -\frac{b}{11}$.Then $\frac{b^2}{|ac|} = \frac{b^2}{|(-\frac{2b}{11})(-\frac{b}{11})|} = \frac{b^2}{2b^2/121} = \frac{121}{2}$.

$L_1 \equiv 2x+y-3=0$ and $L_2 \equiv ax+by+c=0$ are two equal sides of an isosceles triangle. If $L_3 \equiv x+2y+1=0$ is the third side of this triangle and $(5,1)$ is a point on $L_2=0$,then $\frac{b^2}{|ac|}=$

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