Chapter 13.9, Problem 6ST2

To determine

To state the coordinates of point J without introducing any new letters.

  

Answer to Problem 6ST2

The coordinates of point J is $(2e,0)$

Explanation of Solution

Given information :

The vertices of isosceles triangle are A (e, f), O (0,0) and J.

Concept used:

In isosceles triangle two sides are equal. The sides can be find using distance formula;

  $\text{Distance = }\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let’s say, J (a, 0)

  

In isosceles triangle ΔAOJ, two sides OA and JA are equal.

  $\begin{array}{l}OA\text{ = }\sqrt{{(e-0)}^2+{(f-0)}^2}\\ OA\text{ = }\sqrt{e^2+f^2}\\ JA\text{ = }\sqrt{{(e-a)}^2+{(f-0)}^2}\\ JA\text{ = }\sqrt{{(e-a)}^2+f^2}\\ OA=JA\\ \text{squaring both sides,}\\ O{A}^2=J{A}^2\\ $ e^2+f^2={(e-a)}^2+f^2$$ e^2={(e-a)}^2$e^2=e^2-2ae+a^2\\ a^2=2ae\\ a=2e\\ J=(2e,0)\end{array}$

Hence,

The coordinates of point J is $(2e,0)$