Question

the figure below shows \\( \triangle a b c \\) and one side of \\( \triangle d e f \\). find the coordinates of point \\( f \\) so that \\( \triangle a b c \cong \triangle d e f \\).

\\( f(\square, \square) \\)

Explanation:

Step1: Analyze the properties of congruent triangles

Since \(\triangle ABC\cong\triangle DEF\), the corresponding sides are equal.

Step2: Determine the length of sides in \(\triangle ABC\)

In \(\triangle ABC\), \(AB\) has a vertical length from \(y = 2\) to \(y=-8\), so \(AB=\vert2 - (-8)\vert=10\). \(BC\) has a horizontal length from \(x = 4\) to \(x = 5\), so \(BC = 1\). \(AC\) can be found using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), but for congruence, we focus on the transformation.

Step3: Use the transformation from \(\triangle ABC\) to \(\triangle DEF\)

Point \(D\) is at \((-5,2)\). Since \(AB = DE\) (vertical side). If we consider the vertical movement from \(A(5,2)\) to \(B(5,-8)\) (a change of \(y\) - coordinate by \(- 10\)), for \(\triangle DEF\) with \(D(-5,2)\), moving down \(10\) units for the vertical side (corresponding to \(AB\)) gives \(E(-5,-8)\). Now, for the horizontal side (corresponding to \(BC\)). In \(\triangle ABC\), \(BC\) is from \(x = 4\) to \(x = 5\) (right - 1). For \(\triangle DEF\), starting from \(E(-5,-8)\), moving right - 1 (since the triangles are congruent and we assume the same orientation based on the given side \(DE\)) gives \(F(-4,-8)\).

Answer:

\(F(-4,-8)\)