Write the equation of a line in slope-intercept form that is parallel to y = 2x and goes through the point (1, 9).
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Answer:
2x +12y = -63 . . . in standard form
y = -1/6x -21/4 . . . in slope-intercept form
Step-by-step explanation:
It is useful to find the midpoint of the segment. That is the average of the end points:
M = ((-1, -2) +(-2, -8))/2 = ((-1-2)/2, (-2-8)/2) = (-3/2, -5)
It is also useful to find the changes in coordinates from B to A:
Δ = A-B = (-1-(-2), -2-(-8)) = (1, 6)
From here, there are a couple of ways you can write the equation of the perpendicular line.
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One way is to use the Δ values to compute the slope of the segment. The perpendicular line will have a slope that is the negative reciprocal of that.
Δy/Δx = 6/1 = 6
m = -1/6 . . . . . slope of the perpendicular line
Now we have a point and a slope for the desired line, so we can use a point-slope form of the equation for a line:
y = m(x -h) +k
y = (-1/6)(x -(-3/2)) +(-5)
y = (-1/6)x -21/4 . . . . . . . . eliminate parentheses; point-slope form
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Another way to write the perpendicular line is to use the Δ values directly as coefficients in the standard form equation:
Δx(x -h) +Δy(y -k) = 0
1(x -(-3/2)) + 6(y -(-5)) = 0 . . . substitute values
x +6y +31.5 = 0 . . . . . . . . . . .collect terms
2x +12y = -63 . . . . . . . . . . . . multiply by 2, put in standard form
