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.
Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 14th, 2015
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