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Question

Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
  1. -8 , 2
  2. 8, 2
  3. -9, 5
  4. -9, 3

A
8, 2
B
-9, 3
C
-9, 5
D
-8 , 2
Solution
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PQ=10
(102)2+(y+3)2=10
64+y2+9+6y=10
Squaring both sides,
y2+6y27=0
y2+9y3y37=0
y(y+9)3(y+9)=0
(y+9)(y3)=0
y=9,3

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