let's do PA first
PA = √( (a-b)^2 + (0-c)^2 )
= √( a^2 - 2ab + b^2 + c^2 ) , that was easy
now ..
TR = √( (-b + a)^2 + (c-0)^2 )
= √(b^2 - 2ab + a^2 + c^2 )
= PA
I don't see why you didn't get that, all I did was apply the distance formula
I think your problem was in expanding (a-b)^2
= (a-b)(a-b)= a^2 - ab - ab + b^2
= a^2 - 2ab + b^2
When I tried answering a problem then looked at the answer in the back of my book, I was confused how to get to the correct answer. The problem is there is a trapezoid on a coordinate plane that is centered at the origin. The coordinates of the points have variables: T(-a,0), R(-b,c),A(b,c), P(a,0). The non parallel sides are TR and PA. I have to use the distance formula to find the length of segment TR and segment PA. In the selected answers sections of the book it says the answer is Yes;TR=PA= square root of a^2-2ab+b^2+c^2. How do I get this answer? The answer I got as the square root of a^2+b^2+c^2.
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