Q13 of 49 Page 1

Find that value(s) of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.

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

To find distance between P and Q,


X1=x , y1=4
x2=9 , y2=10


So,√ (x-9)2 + (4-10)2 = 10


⇒ (x-9)2 + (-6)2 = (10)2


⇒ (x-9)2 + 36 = 100


⇒ (x-9)2 =64


⇒ x-9 =8 or x-9 =-8 are the two possible solutions.


x – 9 = 8 and x - 9 = -8
x = 17 and x = 1


are the two possible values of x.


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