Question

Find all possible values of y for which the distance between the points $A\left(2,-3\right)\mathrm{and}B\left(10,y\right)$ is 10 units.

Answer 1

The given points are $A\left(2,-3\right)\mathrm{and}B\left(10\mathit{,}y\right)$.
$$\begin{gathered} \therefore AB=\sqrt{{\left(2-10\right)}^2+{\left(-3-y\right)}^2} \\ =\sqrt{{\left(-8\right)}^2+{\left(-3-y\right)}^2} \\ =\sqrt{64+9+y^2+6y} \end{gathered}$$
$$\begin{gathered} \because AB=10 \\ \therefore \sqrt{64+9+y^2+6y}=10 \\ \Rightarrow 73+y^2+6y=100\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow y^2+6y-27=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow y^2+9y-3y-27=0 \\ \Rightarrow y\left(y+9\right)-3\left(y+9\right)=0 \\ \Rightarrow \left(y+9\right)\left(y-3\right)=0 \\ \Rightarrow y+9=0\mathrm{or}y-3=0 \end{gathered}$$
$\Rightarrow y=-9\mathrm{or}y=3$
Hence, the possible values of y are $-9\mathrm{and}3$.