Question
Find the general equation of the line that passes through the point P(1;-4) and is parallel to the line L:3x+y-8=0
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Answer to a math question Find the general equation of the line that passes through the point P(1;-4) and is parallel to the line L:3x+y-8=0
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Given line \( L: 3x + y - 8 = 0 \)
1. The slope-intercept form of the line \( L \) is found by solving for \( y \):
y = -3x + 8
2. From the slope-intercept form, the slope \( m \) of the line \( L \) is:
m = -3
3. Since parallel lines have the same slope, the new line passing through point \( P(1, -4) \) also has a slope of \( -3 \). Using the point-slope form of a line equation:
y - y_1 = m(x - x_1)
where \( (x_1, y_1) = (1, -4) \) and \( m = -3 \),
y + 4 = -3(x - 1)
4. Solve for \( y \) to get it in slope-intercept form:
y + 4 = -3x + 3
y = -3x + 3 - 4
y = -3x - 1
5. Convert back to the standard form of the equation:
3x + y + 1 = 0
So, the general equation is:
3x + y + 1 = 0
1. The slope-intercept form of the line \( L \) is found by solving for \( y \):
2. From the slope-intercept form, the slope \( m \) of the line \( L \) is:
3. Since parallel lines have the same slope, the new line passing through point \( P(1, -4) \) also has a slope of \( -3 \). Using the point-slope form of a line equation:
where \( (x_1, y_1) = (1, -4) \) and \( m = -3 \),
4. Solve for \( y \) to get it in slope-intercept form:
5. Convert back to the standard form of the equation:
So, the general equation is:
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