checking my work on parallel and perpendicular lines

label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

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

Thank you for the opportunity to help you with your question!

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

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
Click to visit
The Notebank
...
Sep 14th, 2015
...
Sep 14th, 2015
Oct 18th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer