can anyone help?? got this question,, The points ABC have co ordinates (2, 1) (b,3) and (5,5) where b>3 and angle ABC=90degrees ?? I know the answer just have no idea how to work it out.
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(slope of AB)(slope of BC) = -1
In a right triangle $\displaystyle a^2 + b^2 = c^2$. So $\displaystyle (b-2)^2 +(3-1)^2 + (b-5)^2 + (3-5)^2 = (5-2)^2 + (5-1)^2$
does that make b= 6 cause i justtried and got some strange answer sorry i'm a bit rubbish at this Originally Posted by Plato In a right triangle $\displaystyle a^2 + b^2 = c^2$. So $\displaystyle (b-2)^2 +(3-1)^2 + (b-5)^2 + (3-5)^2 = (5-2)^2 + (5-1)^2$
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