For the line parallel to x + 2y = 6 through (8, 3):
1. Convert x + 2y = 6 to slope-intercept form: y = -(1/2)x + 3
2. The slope of the given line is -1/2. Since parallel lines have the same slope, the line parallel to x + 2y = 6 will also have a slope of -1/2.
3. Now we have the slope (-1/2) and the point (8, 3). We can plug these values into the point-slope form: y - y₁ = m(x - x₁)
where m is the slope and (x₁, y₁) is the given point.
Substituting the values, we have: y - 3 = (-1/2)(x - 8)
Converting this equation to standard form:
y - 3 = (-1/2)x + 4
Multiply everything by 2 to eliminate the fraction:
2(y - 3) = -x + 8
2y - 6 = -x + 8
Rearrange the equation to have x and y on the same side and all the constants on the other side:
x + 2y = 14
Therefore, the equation of the line parallel to x + 2y = 6 through (8, 3) is x + 2y = 14 in standard form.
For the line perpendicular to x + 2y = 6 through (8, 3):
1. Convert x + 2y = 6 to slope-intercept form: y = -(1/2)x + 3
2. The slope of the given line is -1/2. The slope of a line perpendicular to it will be the negative reciprocal of the given slope, which is 2.
3. Now we have the slope (2) and the point (8, 3). We can plug these values into the point-slope form: y - y₁ = m(x - x₁)
where m is the slope and (x₁, y₁) is the given point.
Substituting the values, we have: y - 3 = 2(x - 8)
Converting this equation to standard form:
y - 3 = 2x - 16
2x - y = 13
Therefore, the equation of the line perpendicular to x + 2y = 6 through (8, 3) is 2x - y = 13 in standard form.
Write an equation for each line in point-slope form and then convert it to standard form. Show your work.
Write an equation of the line parallel to x + 2y = 6 through (8, 3).
Write an equation of the line perpendicular to x + 2y = 6 through (8, 3).
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