Find an equation for the line passing through the point -2,5 that is parallel to the line x+y=2, then graph both lines. First I put the equation in point slope y=mxtb to get y=-1+2. The slope being -1 over 1, And the points being 0,2. Next I used the point intercept form y-y1=m(x-x1) to get y=-1+3. The slope being -1 over 1 and the points being 0,3. Is this correct?
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When we find the equation of lines parallel and perpendicular to a given line, we have to remember, the slopes of parallel lines are equal and he slopes of perpendicular lines are "opposite-reciprocals."
You started off, right; the equation of the line in slope-intercept is y = -1x + 2; so the slope is -1 (-1/1).
For parallel, we do use y-y1 = m(x-x1) with the point (-2,5); giving us...
y - 5 = -1 (x - (-2))
y - 5 = -1 (x + 2)
y - 5 = -1x - 2
y = -1x + 3; which is what you had (less the "x") Then you could graph this.
For the parallel line, the slope would be the opposite reciprocal of -1; so that would be positive 1.
y - 5 = 1(x - (-2))
y - 5 = x + 2
y = x + 7; This is the equation of the perpendicular line.
I hope this helps. If you have any questions, please let me know.
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Sep 14th, 2015
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