Let $A(1,3)$ and $B(5,1)$ be two points. If a line with slope $m$ intersects $AB$ at an angle of $45°$, then the possible values of $m$ are

Calculate the slope of line $AB$:

$m_{AB} = \dfrac{y_B - y_A}{x_B - x_A} = \dfrac{1 - 3}{5 - 1} = \dfrac{-2}{4} = -\dfrac{1}{2}$

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When two lines with slopes $m_1$ and $m_2$ intersect at angle $θ$, we use the formula:

$\tan θ = \left|\dfrac{m_2 - m_1}{1 + m_1m_2}\right|$

In our case, $m_1 = -\dfrac{1}{2}$ and $m_2 = m$. Since the angle is $45°$, $\tan(45°) = 1$.

So we have:

$1 = \left|\dfrac{m - (-\frac{1}{2})}{1 + (-\frac{1}{2})m}\right| = \left|\dfrac{m + \frac{1}{2}}{1 - \frac{m}{2}}\right|$

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Removing the absolute value gives us two cases:

Case 1: $\dfrac{m + \frac{1}{2}}{1 - \frac{m}{2}} = 1$

$m + \dfrac{1}{2} = 1 - \dfrac{m}{2}$

$m + \dfrac{m}{2} = 1 - \dfrac{1}{2}$

$\dfrac{3m}{2} = \dfrac{1}{2}$

$m = \dfrac{1}{3}$

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Case 2: $\dfrac{m + \frac{1}{2}}{1 - \frac{m}{2}} = -1$

$m + \frac{1}{2} = -1 + \frac{m}{2}$

$m - \frac{m}{2} = -1 - \frac{1}{2}$

$\frac{m}{2} = -\frac{3}{2}$

$m = -3$

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Therefore, the possible values of $m$ are $\dfrac{1}{3}$ and $-3$.