Question: Isosceles triangle \( EFG \) is shown below. Give the coordinates of \( G \). \[…

Isosceles triangle \( EFG \) is shown below. Give the coordinates of \( G \).

\[ \begin{array}{c} \text{Isosceles triangle } EFG \text{ is shown below. Give the coordinates of } G. \\ \begin{tikzpicture} \draw[->] (-1, 0) — (3, 0) node[right] {$x$}; \draw[->] (0, -1) — (0, 3) node[above] {$y$}; \draw (0, 0) — (2, 0) — (1, 2) — cycle; \draw (0, 0) node[below left] {$F$} node[above right] {$E$} node[below right] {$G$}; \draw (2, 0) node[below right] {$G$}; \draw (1, 2) node[above right] {$E$}; \end{tikzpicture} \end{array} \]

Solution

To find the coordinates of point \( G \) in the isosceles triangle \( EFG \), we need to use the properties of an isosceles triangle. In an isosceles triangle, the two legs are equal. Here, \( EF = EG \). Given: - \( F(0, 0) \) - \( E(j, k) \) Using the distance formula, we set the distances \( EF \) and \( EG \) equal: \[ EF = \sqrt{(j - 0)^2 + (k - 0)^2} = \sqrt{j^2 + k^2} \] \[ EG = \sqrt{(j - x)^2 + (k - 0)^2} = \sqrt{(j - x)^2 + k^2} \] Since \( EF = EG \), we have: \[ \sqrt{j^2 + k^2} = \sqrt{(j - x)^2 + k^2} \] Square both sides to eliminate the square roots: \[ j^2 + k^2 = (j - x)^2 + k^2 \] Simplify: \[ j^2 + k^2 = j^2 - 2jx + x^2 + k^2 \] Cancel \( j^2 \) and \( k^2 \) from both sides: \[ 0 = -2jx + x^2 \] Reorganize: \[ x^2 - 2jx = 0 \] Factor out \( x \): \[ x(x - 2j) = 0 \] The solutions are: \[ x = 0 \quad \text{or} \quad x = 2j \] Since \( F = (0, 0) \) and \( G \) should not be the same as \( F \), the x-coordinate of \( G \) is \( 2j \). Thus, the coordinates of \( G \) are \( (2j, 0) \).