Q. 84.1( 310 Votes )

# Find the values of y for which the distance between the points P(2, – 3) and Q (10, y) is 10 units.

Answer :

**Given: Distance between (2, - 3) and (10, y) is 10**

**To find: y**

By distance formula, Distance between two points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) is

AB = √((x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2})

Therefore,

√[(2 - 10)^{2} + (-3- y)^{2}] = 10

√[(-8)^{2} + (3+ y)^{2}] = 10

⇒ 64 + (y + 3)^{2} = 100

⇒ (y + 3)^{2} = 100 - 64

⇒ (y + 3)^{2} = 36

⇒ y + 3 = ±6

⇒ (y + 3) will give two values 6 and -6 as answer because both the values when squared will give 36 as answer.)

y + 3 = 6 or y + 3 = -6

Therefore, y = 3 or -9

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