The equation of the line that passes through the point (8, -4) and is parallel to the line x - 4y = 12 is x - 4y = - 8.
Two lines with slopes m₁ and m₂ are parallel when m₁ = m₂ and perpendicular when m₁*m₂ = -1.
In the question, we are asked for the equation of the line that passes through the point (8, -4) and is parallel to the line x - 4y = 12.
The given line can be shown as:
x - 4y = 12,
or, - 4y = - x + 12,
or, y = x/4 - 12/4,
or, y = x/4 - 3.
Comparing it to the slope-intercept form of a line, y = mx + b, we get the slope, m = 1/4.
The line parallel to this will also have the slope, m = 1/4.
The required line passes through the point (8, -4).
Using the one-point formula, (y - y₁) = m(x - x₁), we can write the required equation as:
y - (-4) = (1/4)(x - 8),
or, y + 4 = x/4 - 2,
or, y = x/4 + 2,
or, x/4 - y = -2,
or, x - 4y = - 8.
Thus, the equation of the line that passes through the point (8, -4) and is parallel to the line x - 4y = 12 is x - 4y = - 8.
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