The acute angle between the line 4x - 2y + 13 | vedclass

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(B) The slope of the line $4x - 2y + 13 = 0$ is $m_1 = -\frac{4}{-2} = 2$.The line making equal intercepts with the coordinate axes has the form $\fra

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$	an^{-1}(3)$

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(B) The slope of the line $4x - 2y + 13 = 0$ is $m_1 = -\frac{4}{-2} = 2$.The line making equal intercepts with the coordinate axes has the form $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$. The slope of this line is $m_2 = -1$.The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values: $	an 	heta = \left| \frac{2 - (-1)}{1 + (2)(-1)} ight| = \left| \frac{3}{1 - 2} ight| = \left| \frac{3}{-1} ight| = 3$.Therefore,$	heta = 	an^{-1}(3)$.

The acute angle between the line $4x - 2y + 13 = 0$ and the line which makes equal intercepts with the coordinate axes is

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