7x+y-24=0 and x+7y-24=0 represent the equal s | vedclass

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(B) Let the lines be $L_1: 7x+y-24=0$ and $L_2: x+7y-24=0$. The slopes are $m_1 = -7$ and $m_2 = -\frac{1}{7}$.Since the lines are equal sides of an i

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$x+y=0$

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(B) Let the lines be $L_1: 7x+y-24=0$ and $L_2: x+7y-24=0$. The slopes are $m_1 = -7$ and $m_2 = -\frac{1}{7}$.Since the lines are equal sides of an isosceles triangle,the third side makes equal angles with $L_1$ and $L_2$.Let the slope of the third side be $m$. Then $\left| \frac{m - (-7)}{1 + m(-7)} \right| = \left| \frac{m - (-1/7)}{1 + m(-1/7)} \right|$.$\left| \frac{m+7}{1-7m} \right| = \left| \frac{7m+1}{7-m} \right|$.Case $1$: $\frac{m+7}{1-7m} = \frac{7m+1}{7-m}$ $\Rightarrow (m+7)(7-m) = (7m+1)(1-7m)$ $\Rightarrow 49-m^2 = 1-49m^2$ $\Rightarrow 48m^2 = -48$ (No real solution).Case $2$: $\frac{m+7}{1-7m} = -\frac{7m+1}{7-m}$ $\Rightarrow (m+7)(7-m) = -(7m+1)(1-7m)$ $\Rightarrow 49-m^2 = -(1-49m^2-7m+7m) = 49m^2-1$ $\Rightarrow 50m^2 = 50$ $\Rightarrow m = \pm 1$.For $m = -1$,the line passing through $(-1, 1)$ is $y-1 = -1(x+1)$ $\Rightarrow y-1 = -x-1$ $\Rightarrow x+y=0$.For $m = 1$,the line passing through $(-1, 1)$ is $y-1 = 1(x+1)$ $\Rightarrow y-1 = x+1$ $\Rightarrow x-y+2=0$.

$7x+y-24=0$ and $x+7y-24=0$ represent the equal sides of an isosceles triangle. If the third side passes through $(-1, 1)$,then a possible equation for the third side is

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