- Identify the point \((10, 3)\) from the original point-slope equation \(y - 3 = \frac{1}{5}(x - 10)\).
- Use the point \((10, 3)\) and the new slope \(m = 2\) in the point-slope form \(y - y_1 = m(x - x_1)\).
- Substitute the values to obtain the new equation \(y - 3 = 2(x - 10)\).
- The final answer is \(\boxed{y-3=2(x-10)}\).
Explanation
- Analyzing the Original Equation
The initial equation representing the line is \(y - 3 = \frac{1}{5}(x - 10)\). The point-slope form of a line is defined as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. By comparing the given equation to the point-slope form, we can identify that the line passes through the point \((10, 3)\) and has a slope of \(m_1 = \frac{1}{5}\).
- Formulating the New Equation
The teacher modifies the line such that it maintains the same point \((10, 3)\) but adopts a new slope of \(m_2 = 2\). To find the equation of this new line, we use the point-slope form \(y - y_1 = m(x - x_1)\) again, substituting the point \((x_1, y_1) = (10, 3)\) and the new slope \(m = 2\). This yields the equation \(y - 3 = 2(x - 10)\).
- Selecting the Correct Equation
Comparing our derived equation \(y - 3 = 2(x - 10)\) to the provided options, we see that it matches the first choice. Therefore, the correct equation for the new line is \(y - 3 = 2(x - 10)\).
Examples
This problem demonstrates the importance of the point-slope form in linear algebra. In real-life scenarios, this concept is crucial for modeling trends or paths where you know a starting point and need to adjust the rate of change. For example, if you are designing a ramp for accessibility, you might start with a specific point where the ramp meets the ground and adjust the slope to meet safety standards. Knowing how to manipulate the equation to reflect these changes helps engineers and designers update their models efficiently.