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

A. (7, 3)
B. (0, 7)
C. (7, 0)
D. (3, 7)



Answer :

To determine which point Harold used to write the equation [tex]\( y = 3(x - 7) \)[/tex] in point-slope form, let's start by examining the given equation in its equivalent point-slope form:

The point-slope form of a linear equation is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]

In the given equation:
[tex]\[ y = 3(x - 7) \][/tex]

We can rewrite it to match the point-slope form:
[tex]\[ y - y_1 = 3(x - 7) \][/tex]

By comparing this with [tex]\( y - y_1 = m(x - x_1) \)[/tex], we can identify the slope [tex]\( m \)[/tex] and the coordinates of the point [tex]\( (x_1, y_1) \)[/tex]:
- The slope [tex]\( m \)[/tex] is the coefficient of [tex]\( (x - x_1) \)[/tex], which is 3.
- The term [tex]\( (x - 7) \)[/tex] indicates that [tex]\( x_1 \)[/tex] is 7.

Therefore, the equation [tex]\( y - y_1 = 3(x - 7) \)[/tex] implies that:
- [tex]\( x_1 = 7 \)[/tex]
- We need to determine [tex]\( y_1 \)[/tex] by ensuring the equation matches up.

Expanding the given equation:
[tex]\[ y = 3(x - 7) \][/tex]
[tex]\[ y = 3x - 21 \][/tex]

By comparing this with the expanded form of the point-slope equation:
[tex]\[ y - y_1 = 3x - 21 \][/tex]

We can see that:
[tex]\[ y - y_1 = 3x - 21 \][/tex]

Rewriting it, we get:
[tex]\[ y = 3x - 21 + y_1 \][/tex]

For the equation [tex]\( y = 3x - 21 \)[/tex] to match the given form, [tex]\( y_1 \)[/tex] must be zero. Thus:
[tex]\[ y_1 = 0 \][/tex]

So, Harold used the point [tex]\((7, 0)\)[/tex] to write the equation [tex]\( y = 3(x - 7) \)[/tex].

Hence, the point used by Harold is:
\[ (7, 0) \)