To find the equation of a line in slope-intercept form [tex]\( y = mx + b \)[/tex] that passes through a specific point and is parallel to a given line, follow these steps:
1. Identify the slope of the given line:
- The given line is [tex]\( y = 2x + 4 \)[/tex].
- The slope-intercept form of a line is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope.
- For the given line, the slope [tex]\( m \)[/tex] is 2.
2. Use the fact that parallel lines have the same slope:
- Since the new line is parallel to the given line, it must have the same slope.
- Therefore, the slope [tex]\( m \)[/tex] of our new line is also 2.
3. Use the point-slope form of the equation of a line:
- The point-slope form of a line is [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.
- We are given the point [tex]\( (3, -2) \)[/tex].
- Thus, [tex]\( x_1 = 3 \)[/tex], [tex]\( y_1 = -2 \)[/tex], and the slope [tex]\( m = 2 \)[/tex].
4. Substitute the slope and the point into the point-slope form:
- Plugging in the values, we get:
[tex]\[
y - (-2) = 2(x - 3)
\][/tex]
- Simplify the equation:
[tex]\[
y + 2 = 2(x - 3)
\][/tex]
5. Convert the equation to slope-intercept form [tex]\( y = mx + b \)[/tex]:
- Distribute the slope [tex]\( 2 \)[/tex] on the right-hand side:
[tex]\[
y + 2 = 2x - 6
\][/tex]
- Isolate [tex]\( y \)[/tex] by subtracting 2 from both sides:
[tex]\[
y = 2x - 8
\][/tex]
So, the equation of the line in slope-intercept form that passes through the point [tex]\( (3, -2) \)[/tex] and is parallel to the line [tex]\( y = 2x + 4 \)[/tex] is:
[tex]\[
y = 2x - 8
\][/tex]
