What is the point-slope form of a line with slope [tex]\frac{4}{5}[/tex] that contains the point [tex](-2,1)[/tex]?

A. [tex]y + 1 = \frac{4}{5}(x + 2)[/tex]
B. [tex]y - 1 = \frac{4}{5}(x + 2)[/tex]
C. [tex]y + 1 = \frac{4}{5}(x - 2)[/tex]
D. [tex]y - 1 = \frac{4}{5}(x - 2)[/tex]



Answer :

To find the point-slope form of the equation of a line given a slope and a point, we use the following formula:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is a point on the line.

Given:
- Slope [tex]\( m = \frac{4}{5} \)[/tex]
- Point [tex]\((x_1, y_1) = (-2, 1)\)[/tex]

Using the point-slope formula:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]

Substitute [tex]\( m \)[/tex] and [tex]\((x_1, y_1)\)[/tex] into the formula:
[tex]\[ y - 1 = \frac{4}{5}(x - (-2)) \][/tex]

Simplify the terms inside the parenthesis:
[tex]\[ y - 1 = \frac{4}{5}(x + 2) \][/tex]

Thus, the point-slope form of the line is:
[tex]\[ y - 1 = \frac{4}{5}(x + 2) \][/tex]

Now, let's compare this equation with the given options:

A. [tex]\( y + 1 = \frac{4}{5}(x + 2) \)[/tex]

B. [tex]\( y - 1 = \frac{4}{5}(x + 2) \)[/tex]

C. [tex]\( y + 1 = \frac{4}{5}(x - 2) \)[/tex]

D. [tex]\( y - 1 = \frac{4}{5}(x - 2) \)[/tex]

The equation matches option B.

Therefore, the correct option is:
[tex]\[ \boxed{y - 1 = \frac{4}{5}(x + 2)} \][/tex]