Perpendicular Bisector & Altitude of Triangle: A(4;2) B(11;6) C(-3;-1)

AI Thread Summary
The discussion focuses on finding the equations of the perpendicular bisector of side a and the altitude of side c for triangle ABC with vertices A(4,2), B(11,6), and C(-3,-1). The proposed equation for the perpendicular bisector of side a is -7x + 14y = 7, while the altitude of side c is suggested to be -4x + 7y = 5. Participants clarify the definition of side a, indicating it is typically the side opposite angle A. Additionally, there is a request for guidance on how to find the intersection of the two lines. The conversation emphasizes the need for clear problem-solving steps in geometry.
adod
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The vertices of a triangle are A(4;2) ; B(11;6) ; C(-3;-1)
What is the equation of the perpendicular bisector of side a?
(i think -7x+14y=7)
What is the equation of the altitude of side c?
(i think -4x+7y=5)
What is the intersection of the previous lines?
i have no idea about this,please help,and also write down how you solved it
 
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adod said:
The vertices of a triangle are A(4;2) ; B(11;6) ; C(-3;-1)
What is the equation of the perpendicular bisector of side a?
(i think -7x+14y=7)
What is the equation of the altitude of side c?
(i think -4x+7y=5)
What is the intersection of the previous lines?
i have no idea about this,please help,and also write down how you solved it

Welcome to PH (=
what is side a? AB? AC ? BC?
 
Side a is usually the side across from angle A.

How does one normally find the point of intersection of two lines?
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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