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  • If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ.

  If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ.

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Solution.   Given: Point A(2, –4) is equidistant from P(3, 8) and Q(–10, y)
AP = AQ
Square both sides
AP2 = AQ2
(3 – 2)2 + (8 + 4)2 = (– 10 – 2)2 + (y + 4)2    ( using distance formula)
(1)2 + (12)2 = (12)2 + (y)2 + (4)2 + 2 × y × 4 {\because (a+ b)2 = a2 + b2 + 2ab}
1 + 144 = 144 + y2 + 16 + 8y
y2 + 8y + 15 = 0
y2 + 3y + 5y + 15 = 0
y.(y + 3) + 5.(y + 3) = 0
(y + 5) (y + 3) = 0
y = –5, y = –3
Case-I when y = –5
PQ= \sqrt{\left ( -10-3 \right )^{2}+\left ( -5-8 \right )^{2}}= \sqrt{169+169}= 13\sqrt{2}\, units
Case-I when y = –3
PQ= \sqrt{\left ( -10-3 \right )^{2}+\left ( -3-8 \right )^{2}}= \sqrt{169+121}= \sqrt{290} \, units
 

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