Chapter 4.1, Problem 19WE

To determine

To plot: the given points on graph paper, and to draw $\Delta ABC$ and $\overline{DE}$ and to find two locations of point $F$ such that $\Delta ABC\cong \Delta DEF$

Answer to Problem 19WE

The graph conclude that point $F$ can be $\left(2,5\right)$ and $\left(6,1\right)$ such that $\Delta ABC\cong \Delta DEF$ .

Explanation of Solution

Given information:

Given points,

  $A\left(-1,0\right),B\left(-5,4\right),C\left(-6,1\right),D\left(1,0\right),E\left(5,4\right)$

Calculation:

The two triangles are said to be congruent if they are copies of each other and if their vertices are superposed, then can say that the corresponding angles and the sides of the triangles are congruent.

Plot the given points on the graph paper and draw $\Delta ABC$ and $\overline{DE}$ as shown below:

  $A\left(-1,0\right),B\left(-5,4\right),C\left(-6,1\right),D\left(1,0\right),E\left(5,4\right)$

  

Now, to locate point $F$ so that $\Delta ABC\cong DEF$ , and location of point $F$ can be rather on right side or on left side of $\overline{DE}$ .

If $\Delta ABC\cong DEF$ then, corresponding parts of congruent triangles are congruent.

Therefore,

  $\begin{array}{l}AB=DE\\ BC=EF\\ AC=DF\end{array}$

In the above graph, to reach point $C$ from point $A$ , move 5 boxes to the left and then, 1 box vertically above. In the same manner, to reach point $F$ from point $D$ , move 5 boxes vertically above and then, 1 box to the right. If from point $D$ , will move 5 boxes to the right and then, 1 box vertically above, will get another location of point $F$ .

See the graph given below:

  

Thus, from the graph conclude that point $F$ can be $\left(2,5\right)$ and $\left(6,1\right)$ such that $\Delta ABC\cong \Delta DEF$ .