Answer:
Point used by Harold was:
(7, 0)
Step-by-step explanation:
Given that
Equation of linear function used by Harold:
[tex]y = 3(x - 7)[/tex]
We know that linear equation in point slope form can be represented as:
[tex]y - y_1 = m(x - x_1)[/tex]
Where [tex](x_1,y_1)[/tex] are the coordinates of a given point.
[tex]m[/tex] is the slope of line.
Formula for Slope, m is given as:
[tex]m = \dfrac{y_2-y_1}{x_2-x_1}[/tex]
Where [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] are the two points on the line.
If slope and a point with coordinates [tex](x_1,y_1)[/tex] is know, the equation of a line can be represented in linear form as:
[tex]y - y_1 = m(x - x_1)[/tex] ....... (1)
Now, the given equation is:
[tex]y = 3(x - 7)[/tex]
Re-writing the equation with a slight modification:
[tex]y-0 = 3(x - 7)[/tex]
Now, comparing the above equation with equation (1):
We get that:
[tex]x_1=7\\y_1=0[/tex]
So, the point used by Harold is (7, 0).
