Given the line $y-4=\frac{1}{2}(x-1)$:
The equation of the line is given in the point-slope form: \(y - 4 = \frac{1}{2}(x - 1)\). The point-slope form of a linear equation is generally expressed as \(y - y_0 = m(x - x_0)\), where \(m\) represents the slope of the line and \((x_0, y_0)\) is a specific point lying on that line. Identifying these parameters allows us to understand the geometric properties of the line directly from its equation.
By comparing the given equation \(y - 4 = \frac{1}{2}(x - 1)\) with the standard form \(y - y_0 = m(x - x_0)\), we can extract the required values. The coefficient of the parenthesis \((x - 1)\) is the slope \(m\). Thus, by inspection, we find that \(m = \frac{1}{2}\). Similarly, by matching the constants \(y_0\) and \(x_0\) from the terms \((y - y_0)\) and \((x - x_0)\), we see that \(y_0 = 4\) and \(x_0 = 1\). This identifies \((1, 4)\) as a point on the line.
In summary, the slope of the line is \(m = \frac{1}{2}\) and a point on the line is \((1, 4)\). These values fully characterize the line in the Cartesian plane. The slope is \(\boxed{m = \frac{1}{2}}\) and the point is \(\boxed{(1, 4)}\).
Understanding the point-slope form is essential for modeling linear relationships in real-world scenarios. For example, if you know the starting temperature of a chemical reaction is \(4^\circ C\) at time \(1\) minute, and the temperature increases at a constant rate of \(0.5^\circ C\) per minute, you can represent the temperature \(y\) at time \(x\) as \(y - 4 = 0.5(x - 1)\). Mastering this form helps in quickly determining the rate of change and the initial conditions of various dynamic systems.