Question

In EFG, E(-2,7), F(7,-8) and G(-6,-3). Is EFG a right triangle. provide proof of your yes/no answer.

Answer 1

Proof of a Right-Angled Triangle.

The correct answer is triangle EFG is not a right-angled triangle.

Step1: Find the distances: |EF| , |FG| and |GE| by using the formula below:

`\text{Distance =}\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}`

`\begin{gathered} E(-2,7)\Rightarrow x_1=-2;y_1=7 \\ F(7,-8)\Rightarrow x_2=7;y_2=-8 \end{gathered}`
`\begin{gathered} |EF|^2=(x_2-x_1)^2+(y_2-y_1)^2 \\ |EF|^2=(7--2)^2+(-8-7)^2=9^2+(-15)^2=306 \end{gathered}`
`\begin{gathered} F(7,-8)\Rightarrow x_1=7;y_1=-8 \\ G(-6,-3)\Rightarrow x_2=-6;y_2=-3 \end{gathered}`
`\begin{gathered} |FG|^2=(-6-7)^2+(-3--8)^2 \\ |FG|^2=(-13)^2+5^2=169+25=194 \end{gathered}`
`\begin{gathered} G(-6,-3)\Rightarrow x_1=-6;y_1=-3 \\ E(-2,7)\Rightarrow x_2=-2;y_2=7 \end{gathered}`
`\begin{gathered} |GE|^2=(x_2-x_1)^2+(y_2-y_1)^2 \\ |GE|^2=(-2--6)^2+(7--3)^2=4^2+10^2 \\ |GE|^2=16+100=116 \end{gathered}`

Step2: By Pythagoras Theorem, if

`\begin{gathered} |FG|^2+|GE|^2=|EF|^2\ldots eqn(1) \\ \text{If the above expression labeled eqn(1) is true, then we conclude that triangle EFG is a } \\ \text{right}-\text{angled triangle, otherwise, it is not a right-angled triangle.} \end{gathered}`
`\begin{gathered} |FG|^2+|GE|^2=|EF|^2 \\ 194+116=|EF|^2 \\ 310=|EF|^2,but|EF|^2=306 \\ \text{Clearly, }310\\e306,\text{ we conclude that triangle EFG is not a right-angled triangle.} \end{gathered}`

Hence, the correct answer is triangle EFG is not a right-angled triangle.