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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].
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