Q24 of 49 Page 1

If P(2, 4) is equidistant from Q(7, 0) and R(x, 9), find the values of x. Also, find the distance PQ.

Given that PQ=PR

Distance between two points (x1, y1) and (x2, y2) is given by √ (x1-x2)2 + (y1-y2)2


Length of PQ=√(7-2)2 + (0-4)2 =√ 25 + 16 =√41


Length of PR=√(x-2)2 + (9-4)2 =√ (x-2)2 + 25


PQ=PR


√ 41=√ (x-2)2 + 25


Squaring both sides,


41=(x-2)2 + 25


16=(x-2)2


(x-2) = �4


x-2=4 or x-2=-4


x=6 or x=-2 are the possible values of x


Length of PQ=Length of PR=√ 41


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