Question

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

Answer 1

It is given that the distance between (2,-3) and (10,y) is 10.
Therefore, \(\sqrt{(2-10)^2+(-3-y)^2}=10\)
\(\sqrt{(-8)^2+(3+y)^2}=10\)
\(64+(y+3)^2=100\)
\((y+3)^2=36\)
\(y+3=\pm 6\)
\(y+3=6\) or \(y+3=-6\)
Therefore, \(y=3\) or \(-9\)