To solve this, we need to find the missing $y$-coordinate of the point $(33, y)$. The diagram shows a line segment that moves in two steps:

1. From $(5, 5)$ to $(12, 13)$.
2. From $(12, 13)$ to $(33, y)$.

Step 1: Calculate the slope of the line

The slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

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

For the first segment from $(5, 5)$ to $(12, 13)$: $m = \frac{13 - 5}{12 - 5} = \frac{8}{7}$

The slope is $\frac{8}{7}$. Since this is a straight line, the same slope applies to the second segment.

Step 2: Use the slope to find the missing $y$-coordinate

We now use the point-slope formula of a line to find $y$ at $x = 33$: $m = \frac{y - y_1}{x - x_1}$

Let $(x_1, y_1) = (12, 13)$, and substitute $m = \frac{8}{7}$ and $x = 33$: $\frac{y - 13}{33 - 12} = \frac{8}{7}$

Simplify: $\frac{y - 13}{21} = \frac{8}{7}$

Cross-multiply: $7(y - 13) = 21 \cdot 8$

$7(y - 13) = 168$

$y - 13 = \frac{168}{7} = 24$

$y = 13 + 24 = 37$

Final Answer:

The missing coordinate is: $(33, 37)$

Would you like a breakdown of the calculation or further details?

Related Questions:

1. How do you calculate the slope of a line segment between two points?
2. What is the equation of the line passing through $(5, 5)$ and $(12, 13)$?
3. How do you determine if three points lie on the same straight line?
4. Can you generalize this approach for 3D coordinates?
5. How is the point-slope form of a line derived?

Tip:

When working with slopes, always double-check your subtraction order to avoid sign errors.