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प्रश्न
Find value of x for which the distance between the points P(x,4) and Q(9,10) is 10 units.
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उत्तर
The given points are `P (x,4) and Q (9,10)
`∴ PQ = sqrt((x-9)^2 +(4-10)^2)`
`=sqrt((x-9)^2 +(-6)^2)`
`=sqrt(x^2 -18x+81+36)`
`=sqrt(x^2-18x+117)`
∵ PQ = 10
`∴ sqrt(x^2 -18x +117)`
`⇒x^2 -18x +117=100` (Squaring both sides)
` ⇒ x^2 -18x +17 `
`x^2 -17x -x +17=0`
⇒ ( x - 17 ) ( x - 1) = 0
⇒ ( x - 17 ) (x - 1 ) = 0
⇒ x- 17 = 0 or x - 1 = 0
⇒ x = 17 or x = 1
Hence, the values of x are 1 and 17.
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