Respuesta :
Answer:
Yes, it is a right triangle
Step-by-step explanation:
Hello, I think I can help you with this
you can solve this finding the lengths and then find the longest length, it would be the hypotenuse, and the lengths must fit to
[tex]side^{2}+side^{2}=hypotenuse^{2}[/tex]
Step 1
you can find the distance between 2 points P1 and P2 using:
[tex]\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2} } \\where\\P1(x_{1},y_{1})\\ P2(x_{2},y_{2})\\\\[/tex]
Let
P1=J(-4,-2)
P2=K(1,-1)
P3=L(-2,-4)
distance JK=P1P2=
[tex]\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2} } \\\sqrt{(-4-1)^{2}+(-2-(-1))^{2} } \\\sqrt{(-5)^{2}+(-1))^{2} } \\\sqrt{25+1} \\\sqrt{26} =5.09\ units[/tex]
distance KL=P2P3=
[tex]\sqrt{(x_{2}-x_{3})^{2}+(y_{2}-y_{3})^{2} } \\\sqrt{(1-(-2))^{2}+(-1-(-4))^{2} } \\\sqrt{(3)^{2}+(3)^{2} } \\\sqrt{9+9} \\\sqrt{18} =4.24\ units[/tex]
distance JL=P1P3=
[tex]\sqrt{(x_{1}-x_{3})^{2}+(y_{1}-y_{3})^{2} } \\\sqrt{(-4-(-2))^{2}+(-2-(-4))^{2} } \\\sqrt{(-2)^{2}+(2)^{2} } \\\sqrt{4+4} \\\sqrt{8} =2.82\ units[/tex]
so the lengths are
[tex]JK=\sqrt{26}\\KL=\sqrt{18} \\JL=\sqrt{8}[/tex]
The longest length is JK, hence this is the hypotenuse
Step 2
Check
[tex]side^{2}+side^{2}=hypotenuse^{2}\\(\sqrt{18} )^{2} +(\sqrt{8} )^{2} =(\sqrt{26} )^{2} \\18+8=26\\26=26,true[/tex]
hence this is a right triangle.
Have a nice day
