##### checking my work on parallel and perpendicular lines

 Mathematics Tutor: None Selected Time limit: 1 Day

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?

Sep 14th, 2015

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.

Sep 14th, 2015

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Sep 14th, 2015
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Sep 14th, 2015
Dec 3rd, 2016
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