Chapter 4.7, Problem 6CYU

To determine

The type of transformation in the given graph.

To verify: The transformation is a congruence transformation.

Answer to Problem 6CYU

Translation

Explanation of Solution

Given:

The given graph is:

  

Calculation:

The translation is known as translation if all points of the original figure are moved by the same distance in the same direction. In the given figure, each vertex and its image are in the same position, therefore, triangle JHK is a reflection of triangle MPS.

Consider triangle MPS:

  $\begin{array}{c}MP=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-9-\left(-4\right)\right)}^2+{\left(2-7\right)}^2}\\ =\sqrt{25+25}\\ =\sqrt{50}\\ PS=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-1-\left(-9\right)\right)}^2+{\left(1-2\right)}^2}\\ =\sqrt{64+1}\\ =\sqrt{65}\\ SM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-4-\left(-1\right)\right)}^2+{\left(7-1\right)}^2}\\ =\sqrt{9+36}\\ =\sqrt{45}\end{array}$

Consider triangle JHK:

  $\begin{array}{c}JH=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-1-4\right)}^2+{\left(-5-0\right)}^2}\\ =\sqrt{25+25}\\ =\sqrt{50}\\ HK=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(7-\left(-1\right)\right)}^2+{\left(-6-\left(-5\right)\right)}^2}\\ =\sqrt{64+1}\\ =\sqrt{65}\\ KJ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(4-7\right)}^2+{\left(0-\left(-6\right)\right)}^2}\\ =\sqrt{9+36}\\ =\sqrt{45}\end{array}$

Hence, the triangles are congruent because of SSS postulate.