Question

JKL has vertices at J(-4,-2) K(1,-1) and L (-2,-4). Determine if JKL is a right triangle.

Answer 1

Yes, it is a right triangle.

Answer 2

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

$side^{2}+side^{2}=hypotenuse^{2}$

Step 1

you can find the distance between 2 points P1 and P2 using:

$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2} } \\where\\P1(x_{1},y_{1})\\ P2(x_{2},y_{2})$

Let

P1=J(-4,-2)

P2=K(1,-1)

P3=L(-2,-4)

distance JK=P1P2=

$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2} } \\\sqrt{(-4-1)^{2}+(-2-(-1))^{2} } \\\sqrt{(-5)^{2}+(-1))^{2} } \\\sqrt{25+1} \\\sqrt{26} =5.09\ units$

distance KL=P2P3=

$\sqrt{(x_{2}-x_{3})^{2}+(y_{2}-y_{3})^{2} } \\\sqrt{(1-(-2))^{2}+(-1-(-4))^{2} } \\\sqrt{(3)^{2}+(3)^{2} } \\\sqrt{9+9} \\\sqrt{18} =4.24\ units$

distance JL=P1P3=

$\sqrt{(x_{1}-x_{3})^{2}+(y_{1}-y_{3})^{2} } \\\sqrt{(-4-(-2))^{2}+(-2-(-4))^{2} } \\\sqrt{(-2)^{2}+(2)^{2} } \\\sqrt{4+4} \\\sqrt{8} =2.82\ units$

so the lengths are

$JK=\sqrt{26}\\KL=\sqrt{18} \\JL=\sqrt{8}$

The longest length is JK, hence this is the hypotenuse

Step 2

Check

$side^{2}+side^{2}=hypotenuse^{2}\\(\sqrt{18} )^{2} +(\sqrt{8} )^{2} =(\sqrt{26} )^{2} \\18+8=26\\26=26,true$

hence this is a right triangle.

Have a nice day