Question

Plot the given points on graph paper. Draw $\Delta \mathit{\text{ABC}}$and $\overline{\mathit{\text{DE}}}$. Find two locations of point $F$such that $\Delta \mathit{\text{ABC}}\cong \Delta \mathit{\text{DEF}}$.

$\mathit{\text{A}}\left(-1,0\right)$$\mathit{\text{B}}\left(-5,4\right)$$\mathit{\text{C}}\left(-6,1\right)$$\mathit{\text{D}}\left(1,0\right)$$\mathit{\text{E}}\left(5,4\right)$.

Short answer

Two points of $F$such that $\mathit{\Delta }\mathit{\text{ABC}}\mathbf{\cong }\mathit{\Delta }\mathit{\text{DEF}}$arerole="math" localid="1638351271212" $\mathrm{(}2,5\mathrm{)}$ and$\left(\mathbf{6}\mathbf{,}\mathbf{1}\right)$.

Step-by-step solution

Step 1. Plot the diagram.

Plot $A\left(-1,0\right)$, $B\left(-5,4\right)$, $C\left(-6,1\right)$, $D\left(1,0\right)$, $E\left(5,4\right)$and $\overline{DE}$on the graph paper as shown below:

Step 2. Show the 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.

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

If $\Delta ABC\cong \Delta DEF$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$\Delta ABC\cong \Delta DEF$.

Step 3. State the conclusion.

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