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

If distance between $P$ and $Q,PQ=10$ units , then $y=$ ?
Let $P(x_1,y_1),Q(x_2,y_2)$
So $x_1=2,y_1=-3,x_2=10,y_2=y$
$$PQ=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$$
$$=\sqrt{(10-2{)}^2+\{y-(-3){\}}^2}$$
$$=\sqrt{8^2+(y+3{)}^2}$$
$$=\sqrt{64+y^2+6y+9}$$
$$\begin{array}{cc}& PQ=\sqrt{y^2+6y+73}\\ & PQ=\sqrt{y^2+6y+73}=10\end{array}$$
By Squaring on both sides,
$$\begin{array}{cc}& y^2+6y+73=100\\ & y^2+6y+73-100=0\\ & y^2+6y-27=0\\ & y^2+9y-3y-27=0\\ & y(y+9)-3(y+9)=0\\ & (y+9)(y-3)=0\\ & \text{If}y+9=0\text{then OR}y-3=0\text{then}\\ & y=-9,y=0\\ & \therefore y=-9\text{OR}+3\end{array}$$