Question

The coordinates of the point equidistant from the vertices O(0, 0), A(6, 0) and B(0, 8) of ΔOAB are ___________.

Answer 1

Circumcentre is the point which is equidistant from the vertices of the triangle.

Let the point be C(x, y).

Thus, OC = AC = BC
$$\begin{gathered} \Rightarrow {\left(OC\right)}^2={\left(AC\right)}^2={\left(BC\right)}^2 \\ \mathrm{Now},{\left(OC\right)}^2={\left(AC\right)}^2 \\ \Rightarrow {\left(\sqrt{{\left(x-0\right)}^2+{\left(y-0\right)}^2}\right)}^2={\left(\sqrt{{\left(x-6\right)}^2+{\left(y-0\right)}^2}\right)}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(x-6\right)}^2+{\left(y-0\right)}^2 \\ \Rightarrow {\left(x\right)}^2+{\left(y\right)}^2={\left(x-6\right)}^2+{\left(y\right)}^2 \\ \Rightarrow x^2=x^2+36-12x \\ \Rightarrow 12x=36 \\ \Rightarrow x=3 \\ \mathrm{Also},{\left(OC\right)}^2={\left(BC\right)}^2 \\ \Rightarrow {\left(\sqrt{{\left(x-0\right)}^2+{\left(y-0\right)}^2}\right)}^2={\left(\sqrt{{\left(x-0\right)}^2+{\left(y-8\right)}^2}\right)}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(x-0\right)}^2+{\left(y-8\right)}^2 \\ \Rightarrow {\left(x\right)}^2+{\left(y\right)}^2={\left(x\right)}^2+{\left(y-8\right)}^2 \\ \Rightarrow y^2=y^2+64-16y \\ \Rightarrow 16y=64 \\ \Rightarrow y=4 \\ \mathrm{Therefore},x=3\mathrm{and}y=4. \end{gathered}$$

Hence, coordinates of the point equidistant from the vertices O(0, 0), A(6, 0) and B(0, 8) of ΔOAB are (3, 4).