Find the values of y for which the distance between
Question:

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

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

The distance between two points P(2,3) and Q(10,y) is given as 10 units. Substituting these values in the formula for distance between two points we have,

$10=\sqrt{(2-10)^{2}+(-3-y)^{2}}$

Now, squaring the above equation on both sides of the equals sign

$100=(-8)^{2}+(-3-y)^{2}$

$100=64+9+y^{2}+6 y$

$27=y^{2}+6 y$

Thus we arrive at a quadratic equation. Let us solve this now,

$y^{2}+6 y-27=0$

$y^{2}+9 y-3 y-27=0$

$y(y+9)-3(y+9)=0$

$(y-3)(y+9)=0$

The roots of the above quadratic equation are thus 3 and −9.

Thus the value of ‘y’ could either be.