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Absolutely clueless on how to do this... Geometry Help 
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#19
Dec1711, 12:31 AM

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For the perpendicular bisector to the segment with vertices (a,0) and (b,c): [itex]\displaystyle y=\frac{b+a}{c}\,x+k_1[/itex]k_{1} is the yintercept for this line, so that's where this line intersects the line, x=0. Do similar for the other bisector. 


#20
Dec1711, 11:30 AM

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Okay so lets use the first perpendicular bisector: and the corresponding midpoint
((ba)/2, c/2) We know that y = (b+a)/c*x + K1 So to substitute the x value we get c/2 = (b^{2}a^{2})/c + K1 multplying the c we recieve: c^{2}/2 = k b^{2} + a^{2} Which leaves us finally with K = 1/2(c^{2}+b^{2}a^{2} = k And since that is the Y coordinate of the point of intersection we get the intersection of three perpendicular bisectors as: (0, 1/2c^{2} + b^{2}a^{2}) which unfortunately doesn't match up with my book's answer (I wanted to do this from scratch without working backwards) which is: (0, (a^{2} + b^{2}+^{2})/2c) The two expressions are not equal either, yet this one that I have attained seems quite logically deduced 


#21
Dec1711, 07:00 PM

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You failed to multiply k by c. Anyway there should have been a 2 in both denominator. So, after multiplying by 2c you should have: [itex]\displaystyle c^2=2ckb^2+a^2[/itex]Solving for k gives: [itex]\displaystyle k=\frac{b^2+c^2a^2}{2c}[/itex] 


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