http://www.haeseandharris.com.au/sam...a11gt-5_02.pdf

PLEASE HELP ME!

Can you go to this link and go to chapter 2, page 89 question 7. I really really need help with this question, I know how to parts a and b, but i need help with c!!

THANKS!

xx

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- Mar 19th 2011, 01:38 AMTessarinaCo-ordinate geometry!
http://www.haeseandharris.com.au/sam...a11gt-5_02.pdf

PLEASE HELP ME!

Can you go to this link and go to chapter 2, page 89 question 7. I really really need help with this question, I know how to parts a and b, but i need help with c!!

THANKS!

xx - Mar 19th 2011, 02:14 AMemakarov
What did you get in b? The correct answer is $\displaystyle b$, i.e., the same as the x-coordinate of R. This means that RS is perpendicular to OB.

If you did not get the answer $\displaystyle b$ in b, then what are the equations of the perpendicular bisectors you got in a? - Mar 19th 2011, 02:38 AMTessarina
No, I got x=b in b :)

But it is c that I am having trouble with, how you need to show that RS is perpendicular to OB.

I am not sure how to show that. - Mar 19th 2011, 03:14 AMemakarov
The x-coordinate of both R and S is b. This means that RS is vertical, i.e., perpendicular to OB. If this is not what you think is expected, then could you explain the definition of "perpendicular" that you use and the way you showed that some lines are perpendicular in your course?

- Mar 19th 2011, 06:18 AMbjhopperco-ordinate geometry
Hello Tessarina,

S is the intersection of the perpendicular bisectors of two sides.It is therefore the circumcenter of circum scribed circle of triangle.All points from S to vertices are equal S lies on the perpendicular bisector of base.M the midpoint of base lies on the same line

bjh