To find the equation of a line that is perpendicular to the given line [tex]\(2x + y = -5\)[/tex] and passes through the point [tex]\((-1, -2)\)[/tex], follow these steps:
1. Determine the slope of the given line [tex]\(2x + y = -5\)[/tex]:
- Rewrite the equation in slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.
- [tex]\(2x + y = -5\)[/tex] can be rewritten as [tex]\(y = -2x - 5\)[/tex].
- Therefore, the slope [tex]\(m\)[/tex] of the given line is [tex]\(-2\)[/tex].
2. Find the slope of the line perpendicular to the given line:
- The slope of a line perpendicular to another is the negative reciprocal of the slope of the original line.
- The negative reciprocal of [tex]\(-2\)[/tex] is [tex]\(\frac{1}{2}\)[/tex].
3. Form the equation of the line in slope-intercept form [tex]\(y = mx + b\)[/tex]:
- Use the point-slope form of the line equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
- Substitute the slope ([tex]\(\frac{1}{2}\)[/tex]) and the given point [tex]\((-1, -2)\)[/tex]:
[tex]\[ y - (-2) = \frac{1}{2}(x - (-1)) \][/tex]
[tex]\[ y + 2 = \frac{1}{2}(x + 1) \][/tex]
4. Solve for the [tex]\(y\)[/tex]-intercept [tex]\(b\)[/tex]:
- Distribute and solve for [tex]\(y\)[/tex]:
[tex]\[ y + 2 = \frac{1}{2}x + \frac{1}{2} \][/tex]
[tex]\[ y = \frac{1}{2}x + \frac{1}{2} - 2 \][/tex]
[tex]\[ y = \frac{1}{2}x - \frac{3}{2} \][/tex]
Therefore, the equation of the line perpendicular to [tex]\(2x + y = -5\)[/tex] that passes through the point [tex]\((-1, -2)\)[/tex] is [tex]\(y = \frac{1}{2}x - \frac{3}{2}\)[/tex].
Among the given options, the correct equation is:
[tex]\[ \boxed{y = \frac{1}{2}x - \frac{3}{2}} \][/tex]
