Question

Find the midpoint of the longest side of a triangle withvertices at W(0,0), X(5,0), and Y(0,-4).

Answer 1

Let's begin by identifying key information given to us:

`\begin{gathered} W\mleft(0,0\mright),X\mleft(5,0\mright),Y(0,-4) \\ \\ \end{gathered}`

We will find the longest side as shown below:

`\begin{gathered} WX=(5-0,0-0)=(5,0) \\ XY=(5-0,0--4)=(5,4) \\ WY=(0-0,0--4)=(0,4) \\ We\text{ will calculate using the formula:} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ d_(wx)=\sqrt[]{(5-0)^2+(0-0)^2}=\sqrt[]{5^2+0^2}=\sqrt[]{25+0}=\sqrt[]{25} \\ d_(wx)=5 \\ \\ d_(xy)=\sqrt[]{(5-0)^2+(0--4)^2}=\sqrt[]{5^2+4^2}=\sqrt[]{25+16}=\sqrt[]{41} \\ d_(xy)=\sqrt[]{41} \\ \\ d_(wy)=\sqrt[]{(0-0)^2+(0--4)^2}=\sqrt[]{0^2+4^2}=\sqrt[]{0+16}=\sqrt[]{16} \\ d_(wy)=4 \end{gathered}`

Therefore, the longest side is XY. The midpoint of XY is given by:

`\begin{gathered} Midpoint(XY)=((5+0)/(2),(0+4)/(2)) \\ Midpoint(XY)=((5)/(2),(4)/(2)) \\ Midpoint(XY)=((5)/(2),2) \\ Midpoint(XY)=(2.5,2) \end{gathered}`

The midpoint is (2.5, 2)