The task requires us to find the y-coordinate (ordinate) of the intersection point $A$ of two linear functions given by $f(x) = kx + b$.

Step 1: Identify the equations of the lines.

From the graph, we see two lines:

• The orange line passes through the origin $O(0, 0)$, meaning its equation is $y = k_1x$ (since $b = 0$).
• The blue line crosses the y-axis at $y = 1$, which gives its equation as $y = k_2x + 1$.

Step 2: Find the slopes.

• For the orange line, the slope can be determined from two points on the graph. It passes through $(0, 0)$ and $(1, 1)$, so the slope $k_1$ is: $k_1 = \frac{1 - 0}{1 - 0} = 1$ Thus, the equation of the orange line is $y = x$.

• For the blue line, it passes through the points $(1, 0)$ and $(0, 1)$. The slope $k_2$ is: $k_2 = \frac{0 - 1}{1 - 0} = -1$ So, the equation of the blue line is $y = -x + 1$.

Step 3: Find the intersection point.

To find the coordinates of point $A$, set the two equations equal to each other: $x = -x + 1$ Solve for $x$: $2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}$

Step 4: Find the y-coordinate of point $A$.

Substitute $x = \frac{1}{2}$ into either of the equations (I'll use $y = x$): $y = \frac{1}{2}$

Thus, the ordinate of point $A$ is $\frac{1}{2}$.

Final Answer:

The ordinate of point $A$ is $\frac{1}{2}$.

Do you have any questions or want more details?
Here are 5 related questions:

1. How do we derive the slope of a line using two points?
2. Why does the blue line's slope become negative?
3. Can linear equations with the same slope intersect?
4. How would the solution change if the lines were parallel?
5. What happens if one of the lines is vertical?

Tip: Always verify the points used to calculate slopes on a graph carefully to avoid misreading values.