To determine if the line passing through points $A(-3, -2)$ and $B(2, 1)$ also passes through point $C(5, 3)$, we can follow these steps:

Step 1: Find the slope of the line passing through $A$ and $B$

The slope $m$ of a line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

Using points $A(-3, -2)$ and $B(2, 1)$:

$m = \frac{1 - (-2)}{2 - (-3)} = \frac{1 + 2}{2 + 3} = \frac{3}{5}$

So, the slope of the line is $\frac{3}{5}$.

Step 2: Equation of the line passing through $A$ and $B$

We now use the point-slope form of the equation of a line:

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

Using point $A(-3, -2)$ and slope $\frac{3}{5}$, the equation becomes:

$y - (-2) = \frac{3}{5}(x - (-3))$ $y + 2 = \frac{3}{5}(x + 3)$ $y + 2 = \frac{3}{5}x + \frac{9}{5}$ $y = \frac{3}{5}x + \frac{9}{5} - \frac{10}{5}$ $y = \frac{3}{5}x - \frac{1}{5}$

So, the equation of the line is:

$y = \frac{3}{5}x - \frac{1}{5}$

Step 3: Check if point $C(5, 3)$ satisfies the line equation

Substitute $x = 5$ and $y = 3$ into the equation of the line:

$y = \frac{3}{5}(5) - \frac{1}{5}$ $y = 3 - \frac{1}{5} = \frac{15}{5} - \frac{1}{5} = \frac{14}{5}$

This gives $y = \frac{14}{5}$, which is approximately 2.8, not 3. Hence, point $C(5, 3)$ does not lie on the line.

Conclusion:

The line passing through points $A(-3, -2)$ and $B(2, 1)$ does not pass through point $C(5, 3)$.

Would you like more details on any part of this?

Here are some related questions:

1. How do we find the equation of a line using different forms (e.g., slope-intercept)?
2. Can we use a different method to verify if a point lies on a line?
3. What are parallel and perpendicular lines in relation to slopes?
4. How do we find the distance from a point to a line?
5. What is the significance of slope in geometry?

Tip: The slope of a line represents its steepness; lines with the same slope are parallel!