Question

The coordinates of the point which is equidistant from the vertices of the triangle formed by the points O(0, 0), A(a, 0) and B(0, b), 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
$$\begin{gathered} \Rightarrow {\left(OC\right)}^2={\left(AC\right)}^2={\left(BC\right)}^2 \\ \mathrm{Now},{\left(OC\right)}^2={\left(AC\right)}^2 \\ \Rightarrow {\left(\sqrt{{\left(x-0\right)}^2+{\left(y-0\right)}^2}\right)}^2={\left(\sqrt{{\left(x-a\right)}^2+{\left(y-0\right)}^2}\right)}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(x-a\right)}^2+{\left(y-0\right)}^2 \\ \Rightarrow {\left(x\right)}^2+{\left(y\right)}^2={\left(x-a\right)}^2+{\left(y\right)}^2 \\ \Rightarrow x^2=x^2+a^2-2ax \\ \Rightarrow 2ax=a^2 \\ \Rightarrow 2x=a \\ \Rightarrow x=\frac{a}{2} \\ \mathrm{Also},{\left(OC\right)}^2={\left(BC\right)}^2 \\ \Rightarrow {\left(\sqrt{{\left(x-0\right)}^2+{\left(y-0\right)}^2}\right)}^2={\left(\sqrt{{\left(x-0\right)}^2+{\left(y-b\right)}^2}\right)}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y-0\right)}^2={\left(x-0\right)}^2+{\left(y-b\right)}^2 \\ \Rightarrow {\left(x\right)}^2+{\left(y\right)}^2={\left(x\right)}^2+{\left(y-b\right)}^2 \\ \Rightarrow y^2=y^2+b^2-2by \\ \Rightarrow 2by=b^2 \\ \Rightarrow 2y=b \\ \Rightarrow y=\frac{b}{2} \\ \mathrm{Therefore},x=\frac{a}{2}\mathrm{and}y=\frac{b}{2}. \end{gathered}$$

Hence, the coordinates of the point which is equidistant from the vertices of the triangle formed by the points O(0, 0), A(a, 0) and B(0, b), are $\underline{\left(\frac{a}{2},\frac{b}{2}\right)}.$