Q10 of 66 Page 7

Find the value of k, if the point P(2, 4) is equidistant from the points A(5, k) and B(k, 7).(CBSE 2012)


Given;


PA = PB


P (2, 4)


A (5, k)


B (k, 7)


By squaring both sides;


PA2 = PB2


(5 – 2)2 + (k – 4)2 = (k – 2)2 + (7 – 4)2


9 + (k – 4)2 = (k – 2)2 + 9


9 + (k – 4)2 – (k – 2)2 = 9


(k – 4)2 – (k – 2)2 = 9 – 9


(k – 4)2 – (k – 2)2 = 0


[k2 – (4)2 + 8k]- [k2 – (2)2 + 4k] = 0 ……{by (a-b)2 = a2 – b2 + 2ab}


k2 – 16 + 8k – k2 + 4 – 4k = 0


- 12 + 4k = 0


4k = 12


k = 12/4 = 3


Therefore the value of k is 3.

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