To determine which point Harold used to write the linear equation in the point-slope form, let's start by analyzing the given equation and its form.
The point-slope form of a linear equation is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
In this form:
- \( m \) is the slope of the line.
- \( (x_1, y_1) \) is a specific point on the line.
The given equation is:
[tex]\[ y = 3(x - 7) \][/tex]
Let's rewrite the given equation to match the standard point-slope form.
Starting with:
[tex]\[ y = 3(x - 7) \][/tex]
We can rewrite it as:
[tex]\[ y - 0 = 3(x - 7) \][/tex]
Comparing this with the point-slope form \( y - y_1 = m(x - x_1) \), we see that:
- The slope \( m \) is 3.
- \( x_1 \) is 7.
- Since \( y - y_1 \) matches with \( y - 0 \), \( y_1 \) must be 0.
Therefore, the point \( (x_1, y_1) \) that Harold used is:
[tex]\[ (7, 0) \][/tex]
So, Harold used the point \((7, 0)\).
Among the given options, the correct one is:
[tex]\((7, 0)\)[/tex]
