Answer :

To find the equation of the line that passes through the point [tex]\((6, 2)\)[/tex] and has a slope of [tex]\(\frac{1}{3}\)[/tex], we use the point-slope form of the equation of a line. The point-slope form can be written as:

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

where:
- [tex]\( (x_1, y_1) \)[/tex] is the given point the line passes through,
- [tex]\( m \)[/tex] is the slope of the line.

Given:
- [tex]\((x_1, y_1) = (6, 2)\)[/tex]
- [tex]\( m = \frac{1}{3} \)[/tex]

Substitute these values into the point-slope form:

[tex]\[ y - 2 = \frac{1}{3}(x - 6) \][/tex]

Next, we simplify this equation to get it into the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.

First, distribute [tex]\(\frac{1}{3}\)[/tex] on the right side:

[tex]\[ y - 2 = \frac{1}{3}x - \frac{1}{3} \cdot 6 \][/tex]
[tex]\[ y - 2 = \frac{1}{3}x - 2 \][/tex]

Now, isolate [tex]\( y \)[/tex] by adding 2 to both sides of the equation:

[tex]\[ y = \frac{1}{3}x - 2 + 2 \][/tex]
[tex]\[ y = \frac{1}{3}x \][/tex]

Since there is no constant term on the right side after simplifying, the y-intercept [tex]\( b \)[/tex] is 0.

So, the equation of the line in slope-intercept form is:

[tex]\[ y = \frac{1}{3}x \][/tex]

Therefore, the equation of the line that passes through the point [tex]\((6, 2)\)[/tex] and has a slope of [tex]\(\frac{1}{3}\)[/tex] is:

[tex]\[ y = \frac{1}{3}x \][/tex]

Here, the slope [tex]\( m \)[/tex] is [tex]\( \frac{1}{3} \)[/tex] and the y-intercept [tex]\( b \)[/tex] is 0.