Question

Find the ordinate of a point whose abscissa is 10 and which is at a distance of 10 units from the points P(2, −3).

Answer 1

Let Q be the other point whose coordinates are Q(10, y).
Given, PQ = 10 units
Therefore, (PQ)² = 100
$$\begin{gathered} \Rightarrow {\left(10-2\right)}^2+{\left\{y-\left(-3\right)\right\}}^2=100 \\ \Rightarrow {\left(8\right)}^2+{\left(y+3\right)}^2=100 \\ \Rightarrow 64+{\left(y+3\right)}^2=100 \\ \Rightarrow {\left(y+3\right)}^2=100-64 \\ \Rightarrow {\left(y+3\right)}^2=36 \\ \Rightarrow \left(y+3\right)=\pm \sqrt{36} \\ \Rightarrow y+3=\pm 6 \\ \Rightarrow y+3=6\mathrm{or}y+3=-6 \\ \Rightarrow y=3\mathrm{or}-9 \end{gathered}$$
Therefore, y = 3 or −9.