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5 Questions around this concept.
Two equal sides of an isosceles triangle have the equations 7x – y + 3 = 0 and x + y = 3 and its third side passes through the point (-2, –1). Then the equation of the third side.
Line Equally Inclined with two lines
If the two lines with slope $m_1$ and $m_2$ are equally inclined to a line having slope $m$, then
$
\left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right)
$
Two lines with slope $m_1$ and $m_2$ intersect at point $A$
As from the fig
$
\begin{aligned}
& \angle \mathrm{PAQ}=\angle \mathrm{QAR}=\angle \theta \\
& \tan (\angle \mathrm{PAQ})=\frac{\mathrm{m}_1-\mathrm{m}}{1+\mathrm{m}_1 \mathrm{~m}}=\tan \theta \\
& \text { and } \\
& \tan (\angle \mathrm{QAR})=\frac{\mathrm{m}-\mathrm{m}_2}{1+\mathrm{mm}_2}=\tan \theta
\end{aligned}
$
Hence,
$
\left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right)
$
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