Perpendicular bisector

cosmic

Junior Member
Joined
Mar 3, 2014
Messages
84
Hi guys,

I'm asked to find the equation of the perpendicular bisector given two points, (-8, -7) and (-2, 5).

I obtained the answer y= -1/2x - 7/2

Would really appreciate if someone could check to see whether that is indeed the correct answer or point me in the right direction if it's not.

Thanks in advance. :smile:
 
Hi guys,

I'm asked to find the equation of the perpendicular bisector given two points, (-8, -7) and (-2, 5).
I presume you mean the perpendicular bisector of the line segment having those points as endpoints. What is the midpoint of that line segment? What is its slope? What is the slope of a line perpendicular to the line segment?

I obtained the answer y= -1/2x - 7/2

Would really appreciate if someone could check to see whether that is indeed the correct answer or point me in the right direction if it's not.

Thanks in advance. :smile:

You can check it your self. Does it have the right slope? Does it pass through the midpoint?
 
What equation did you get for the line being bisected?
What coordinates did you get for the intersection point?

Hi,

Thanks for your reply. Not sure about the equation of the line being bisected wasn't aware that it needed to be calculated. The coordinate I got for the intersection point however was (-5, -1).

I presume you mean the perpendicular bisector of the line segment having those points as endpoints. What is the midpoint of that line segment? What is its slope? What is the slope of a line perpendicular to the line segment?



You can check it your self. Does it have the right slope? Does it pass through the midpoint?

Hi,

Thanks for your reply. Yes I mean the equation of the perpendicular bisector of the line segment joining the two points. I calculated the midpoint to be (-5, -1) and the slope to be 2. The slope of the perpendicular bisector is -1/2 from my calculations. I have not drawn the figure accurately so can't check it accurately as a result. Just wondering if it's correct or not.

Thanks. :smile:
 
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