To determine the equation of a line that passes through the point [tex]\((-6, 2)\)[/tex] and has a slope of 1, we will use the point-slope form of the equation of a line. The point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line, and [tex]\(m\)[/tex] is the slope.
Given:
- The point [tex]\((x_1, y_1) = (-6, 2)\)[/tex]
- The slope [tex]\(m = 1\)[/tex]
Substitute these values into the point-slope form equation:
[tex]\[ y - 2 = 1(x + 6) \][/tex]
Now, we will simplify this equation to put it into the slope-intercept form [tex]\(y = mx + b\)[/tex]:
First, distribute the slope [tex]\(1\)[/tex] on the right side of the equation:
[tex]\[ y - 2 = x + 6 \][/tex]
Next, we need to isolate [tex]\(y\)[/tex]. To do this, add 2 to both sides of the equation:
[tex]\[ y = x + 6 + 2 \][/tex]
Combine the constant terms on the right side:
[tex]\[ y = x + 8 \][/tex]
Therefore, the equation of the line that passes through the point [tex]\((-6, 2)\)[/tex] and has a slope of 1 is:
[tex]\[ y = x + 8 \][/tex]
