find the value of x such that PQ = QR where p, q and r are the points (x,-1) ,(1,3) and (-3,8)
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Tuesday, October 25, 2016 at 8:52am
If PQ = QR , (where P(6,-1), Q(1,3), R(x,8))
√( (3+1)^2 + (1-6)^2) = √( (8-3)^2 + (x-1)^2 )
√41 = √(x^2 - 2x + 26)
square both sides
41 = x^2 - 2x + 26
x^2 - 2x - 15 = 0
(x-5)(x+3) = 0
x = 5 or x = -3
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