IDNLearn.com makes it easy to get reliable answers from knowledgeable individuals. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
Certainly! Let's break down the problem into two parts:
1. Finding the equation of the line through the point (2, -5) and parallel to the given line [tex]\(3x + 4y + 5 = 0\)[/tex].
2. Finding the equation of the line through the same point (2, -5) but perpendicular to the given line.
### Part 1: Line Parallel to [tex]\(3x + 4y + 5 = 0\)[/tex]
1. Find the slope of the given line:
The given line is [tex]\(3x + 4y + 5 = 0\)[/tex]. We need it in the slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ 4y = -3x - 5 \][/tex]
[tex]\[ y = -\frac{3}{4}x - \frac{5}{4} \][/tex]
Therefore, the slope of the given line is [tex]\(-\frac{3}{4}\)[/tex].
2. Use the point-slope form for the new line that passes through the point (2, -5) and has the same slope:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\(m = -\frac{3}{4}\)[/tex], [tex]\((x_1, y_1) = (2, -5)\)[/tex].
3. Substitute the values:
[tex]\[ y - (-5) = -\frac{3}{4}(x - 2) \][/tex]
Simplifying:
[tex]\[ y + 5 = -\frac{3}{4}(x - 2) \][/tex]
[tex]\[ y + 5 = -\frac{3}{4}x + \frac{3}{2} \][/tex]
[tex]\[ y = -\frac{3}{4}x + \frac{3}{2} - 5 \][/tex]
[tex]\[ y = -\frac{3}{4}x - \frac{7}{2} \][/tex]
Converting into the standard form [tex]\(Ax + By + C = 0\)[/tex]:
[tex]\[ 3x + 4y + 14 = 0 \][/tex]
So, the equation of the line parallel to [tex]\(3x + 4y + 5 = 0\)[/tex] and passing through (2, -5) is:
[tex]\[ 3x + 4y + 14 = 0 \][/tex]
### Part 2: Line Perpendicular to [tex]\(3x + 4y + 5 = 0\)[/tex]
1. Slope of the perpendicular line is the negative reciprocal of the original slope [tex]\(-\frac{3}{4}\)[/tex]. So, if the original slope [tex]\(m = -\frac{3}{4}\)[/tex], the slope [tex]\(m_{\perp}\)[/tex] of the perpendicular line will be:
[tex]\[ m_{\perp} = \frac{4}{3} \][/tex]
2. Use the point-slope form for the new line that passes through the point (2, -5) with slope [tex]\(\frac{4}{3}\)[/tex]:
[tex]\[ y - y_1 = m_{\perp}(x - x_1) \][/tex]
Where [tex]\(m_{\perp} = \frac{4}{3}\)[/tex], [tex]\((x_1, y_1) = (2, -5)\)[/tex].
3. Substitute the values:
[tex]\[ y - (-5) = \frac{4}{3}(x - 2) \][/tex]
Simplifying:
[tex]\[ y + 5 = \frac{4}{3}(x - 2) \][/tex]
[tex]\[ y + 5 = \frac{4}{3}x - \frac{8}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{8}{3} - 5 \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{23}{3} \][/tex]
Converting into the standard form [tex]\(Ax + By + C = 0\)[/tex]:
[tex]\[ 4x - 3y - 23 = 0 \][/tex]
So, the equation of the line perpendicular to [tex]\(3x + 4y + 5 = 0\)[/tex] and passing through (2, -5) is:
[tex]\[ 4x - 3y - 23 = 0 \][/tex]
### Conclusion:
- The equation of the line through (2, -5) and parallel to [tex]\(3x + 4y + 5 = 0\)[/tex] is: [tex]\(3x + 4y + 14 = 0\)[/tex].
- The equation of the line through (2, -5) and perpendicular to [tex]\(3x + 4y + 5 = 0\)[/tex] is: [tex]\(4x - 3y - 23 = 0\)[/tex].
1. Finding the equation of the line through the point (2, -5) and parallel to the given line [tex]\(3x + 4y + 5 = 0\)[/tex].
2. Finding the equation of the line through the same point (2, -5) but perpendicular to the given line.
### Part 1: Line Parallel to [tex]\(3x + 4y + 5 = 0\)[/tex]
1. Find the slope of the given line:
The given line is [tex]\(3x + 4y + 5 = 0\)[/tex]. We need it in the slope-intercept form [tex]\(y = mx + c\)[/tex]:
[tex]\[ 4y = -3x - 5 \][/tex]
[tex]\[ y = -\frac{3}{4}x - \frac{5}{4} \][/tex]
Therefore, the slope of the given line is [tex]\(-\frac{3}{4}\)[/tex].
2. Use the point-slope form for the new line that passes through the point (2, -5) and has the same slope:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Where [tex]\(m = -\frac{3}{4}\)[/tex], [tex]\((x_1, y_1) = (2, -5)\)[/tex].
3. Substitute the values:
[tex]\[ y - (-5) = -\frac{3}{4}(x - 2) \][/tex]
Simplifying:
[tex]\[ y + 5 = -\frac{3}{4}(x - 2) \][/tex]
[tex]\[ y + 5 = -\frac{3}{4}x + \frac{3}{2} \][/tex]
[tex]\[ y = -\frac{3}{4}x + \frac{3}{2} - 5 \][/tex]
[tex]\[ y = -\frac{3}{4}x - \frac{7}{2} \][/tex]
Converting into the standard form [tex]\(Ax + By + C = 0\)[/tex]:
[tex]\[ 3x + 4y + 14 = 0 \][/tex]
So, the equation of the line parallel to [tex]\(3x + 4y + 5 = 0\)[/tex] and passing through (2, -5) is:
[tex]\[ 3x + 4y + 14 = 0 \][/tex]
### Part 2: Line Perpendicular to [tex]\(3x + 4y + 5 = 0\)[/tex]
1. Slope of the perpendicular line is the negative reciprocal of the original slope [tex]\(-\frac{3}{4}\)[/tex]. So, if the original slope [tex]\(m = -\frac{3}{4}\)[/tex], the slope [tex]\(m_{\perp}\)[/tex] of the perpendicular line will be:
[tex]\[ m_{\perp} = \frac{4}{3} \][/tex]
2. Use the point-slope form for the new line that passes through the point (2, -5) with slope [tex]\(\frac{4}{3}\)[/tex]:
[tex]\[ y - y_1 = m_{\perp}(x - x_1) \][/tex]
Where [tex]\(m_{\perp} = \frac{4}{3}\)[/tex], [tex]\((x_1, y_1) = (2, -5)\)[/tex].
3. Substitute the values:
[tex]\[ y - (-5) = \frac{4}{3}(x - 2) \][/tex]
Simplifying:
[tex]\[ y + 5 = \frac{4}{3}(x - 2) \][/tex]
[tex]\[ y + 5 = \frac{4}{3}x - \frac{8}{3} \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{8}{3} - 5 \][/tex]
[tex]\[ y = \frac{4}{3}x - \frac{23}{3} \][/tex]
Converting into the standard form [tex]\(Ax + By + C = 0\)[/tex]:
[tex]\[ 4x - 3y - 23 = 0 \][/tex]
So, the equation of the line perpendicular to [tex]\(3x + 4y + 5 = 0\)[/tex] and passing through (2, -5) is:
[tex]\[ 4x - 3y - 23 = 0 \][/tex]
### Conclusion:
- The equation of the line through (2, -5) and parallel to [tex]\(3x + 4y + 5 = 0\)[/tex] is: [tex]\(3x + 4y + 14 = 0\)[/tex].
- The equation of the line through (2, -5) and perpendicular to [tex]\(3x + 4y + 5 = 0\)[/tex] is: [tex]\(4x - 3y - 23 = 0\)[/tex].
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
