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

It is known that the distance between the points x1,y1,x2,y2
is x2-x12+y2-y12

According to question,

PQ=10=x-92+4-102x-92+36=100x-92=64x2+81-18x-64=0x2-18x+17=0x2-17x-x+17=0xx-17-x-17=0x-1x-17=0x=1,17

Hence, the required values of x will be 1 and 17.

  • 0

(PQ)2 = 102 =  100

(9-x)2 + (10-4)2 = 100

81 - x2 - 18x  + 36 = 100

x2 - 18x + 17 = 0

by splitting the middle term

(x-1)(x-18)

   
  • 1
sorry it was left incomplete then we get x=1 or x= 18   
  • 2
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