# What type of triangle is formed by joining the points D(7, 3), E(8, 1), and F(4,-1)?

Oct 1, 2016

A right-angled triangle is formed.

#### Explanation: 1) calculate d the distance between the points of the triangles, use the distance formula :
d=sqrt((x2−x1)^2+(y2−y1)^2

$D \left(7 , 3\right) , E \left(8 , 1\right) , F \left(4 , - 1\right)$

=> DE=sqrt((8-7)^2+(1-3)^2))=sqrt(1^2+2^2)=sqrt5
$\implies E F = \sqrt{{\left(4 - 8\right)}^{2} + {\left(- 1 - 1\right)}^{2}} = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{20}$
$\implies D F = = \sqrt{{\left(4 - 7\right)}^{2} + {\left(- 1 - 3\right)}^{2}} = \sqrt{{3}^{2} + {4}^{2}} = \sqrt{25} = 5$

Pythagorean theorem : ${c}^{2} = {a}^{2} + {b}^{2}$

The theorem states that: The sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of the hypotenuse.

Now we use the theorem to check and see if the triangle is right-angled or not :

$\implies {\sqrt{25}}^{2} = {\sqrt{5}}^{2} + {\sqrt{20}}^{2}$
$\implies 25 = 5 + 20$

As ${\left(D F\right)}^{2} = {\left(D E\right)}^{2} + {\left(E F\right)}^{2}$, triangle $D E F$ is right-angled.