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

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Solution

It is given that the distance between (2, - 3) and (10, y) is 10.
And distance between the points is given by
$\sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$
Therefore, $\sqrt{(2 - 10)^{2} + (- 3 - y)^{2}} = 10$
$\sqrt{(- 8)^{2} + (3 + y)^{2}} = 10$
$64 + (y + 3)^{2} = 100$
$(y + 3)^{2} = 36$
$y + 3 = \pm 6$
$y + 3 = + 6 o r y + 3 = - 6$
Therefore, $y = 3 o r - 9$

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