Find all possible values of y for which the distance between the points A(2, – 3) and B(10, y) is 10 units.

Given, the distance AB = 10 units


By distance formula, as shown below:



AB = √{(10 – 2)2 + (y – (– 3))2}


10 = √{(8)2 + (y + 3)2}


10 = √{64 + y2 + 6y + 9}


10 = √{73 + y2 + 6y}


Squaring both sides we get


100 = 73 + y2 + 6y


On solving the equation, 100 = 73 + y2 + 6y


27 + y2 + 6y = 0


y2 + 6y + 27 = 0


(y – 3)(y + 9) = 0


y = 3 or y = – 9


The values of y can be 3 or – 9


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