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The equation of a linear function in point-slope form is [tex]$y - y_1 = m \left(x - x_1\right)$[/tex]. Harold correctly wrote the equation [tex]$y = 3(x - 7)$[/tex] using a point and the slope. Which point did Harold use?

A. [tex][tex]$(7,3)$[/tex][/tex]
B. [tex]$(0,7)$[/tex]
C. [tex]$(7,0)$[/tex]
D. [tex][tex]$(3,7)$[/tex][/tex]



Answer :

GinnyAnswer
To determine which point Harold used to write the equation [tex]\( y = 3(x - 7) \)[/tex] using the point-slope form, we start by recalling the point-slope form of a linear equation:

[tex]\[ y - y_1 = m(x - x_1) \][/tex]

In this form:
- [tex]\( m \)[/tex] is the slope of the line.
- [tex]\( (x_1, y_1) \)[/tex] is a specific point on the line.

Given the equation [tex]\( y = 3(x - 7) \)[/tex], let's compare this with the point-slope form:

[tex]\[ y - y_1 = m(x - x_1) \][/tex]

From the equation, we can observe:
- The slope [tex]\( m \)[/tex] is [tex]\( 3 \)[/tex].
- The term [tex]\( x - 7 \)[/tex] implies that [tex]\( x_1 \)[/tex] is [tex]\( 7 \)[/tex].

Now, since [tex]\( y \)[/tex] on the left-hand side is not subtracted by anything, it implies that [tex]\( y_1 \)[/tex] is [tex]\( 0 \)[/tex] (because [tex]\( y - 0 \)[/tex] is just [tex]\( y \)[/tex]).

Thus, the point [tex]\( (x_1, y_1) \)[/tex] that Harold used is:

[tex]\[ (7, 0) \][/tex]

Therefore, the correct point is:

[tex]\[ (7, 0) \][/tex]

### Conclusion:
Harold used the point [tex]\((7, 0)\)[/tex]. Hence, the correct answer is:

[tex]\[ \boxed{(7,0)} \][/tex]

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