To find which point Harold used when writing the equation \( y = 3(x - 7) \) in point-slope form, we need to match this equation with the standard point-slope form of a linear equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, \( m \) represents the slope of the line, and \((x_1, y_1)\) is a point on the line.
Let's rewrite the given equation \( y = 3(x - 7) \) in a form that directly shows the slope and the point:
[tex]\[ y = 3(x - 7) \][/tex]
Comparing this to the standard point-slope form, we can recognize that the slope \( m \) is 3, and the equation can also be written as:
[tex]\[ y - 0 = 3(x - 7) \][/tex]
This reveals that the point \((x_1, y_1)\) used in this form of the equation is:
[tex]\[ x_1 = 7 \][/tex]
[tex]\[ y_1 = 0 \][/tex]
Thus, the point that Harold used is \( (7, 0) \).
So, the correct answer is:
[tex]\[ (7, 0) \][/tex]
