To find the equations of the sides $AB$ and $AC$ of the triangle, we use the slope-intercept form of a line equation:

$y - y_1 = m(x - x_1)$

where $m$ is the slope, calculated as:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Step 1: Equation of $AB$

The coordinates of $A$ are $(2, 1)$, and the coordinates of $B$ are $(-3, 5)$.

Slope of $AB$:

$m_{AB} = \frac{5 - 1}{-3 - 2} = \frac{4}{-5} = -\frac{4}{5}$

Equation of the line $AB$:

Using point $A(2, 1)$:

$y - 1 = -\frac{4}{5}(x - 2)$

Simplify:

$y - 1 = -\frac{4}{5}x + \frac{8}{5}$

$y = -\frac{4}{5}x + \frac{8}{5} + 1$

$y = -\frac{4}{5}x + \frac{13}{5}$

So, the equation of $AB$ is:

$y = -\frac{4}{5}x + \frac{13}{5}$

Step 2: Equation of $AC$

The coordinates of $A$ are $(2, 1)$, and the coordinates of $C$ are $(4, 5)$.

Slope of $AC$:

$m_{AC} = \frac{5 - 1}{4 - 2} = \frac{4}{2} = 2$

Equation of the line $AC$:

Using point $A(2, 1)$:

$y - 1 = 2(x - 2)$

Simplify:

$y - 1 = 2x - 4$

$y = 2x - 4 + 1$

$y = 2x - 3$

So, the equation of $AC$ is:

$y = 2x - 3$

Final Equations:

1. Equation of $AB$: $y = -\frac{4}{5}x + \frac{13}{5}$
2. Equation of $AC$: $y = 2x - 3$

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Here are 5 related questions you might find useful:

1. How to find the equation of the third side $BC$?
2. How to find the area of the triangle $ABC$?
3. How to verify if $A, B, C$ are collinear?
4. How to find the altitude of the triangle from $A$?
5. How to find the coordinates of the centroid of the triangle?

Tip: When solving for line equations, always ensure that the slope calculation is accurate to avoid errors in the final equation.