Given:
The vertices of the triangle JKL are J(-3, 2), K(2,6), L(8, -1).
To find:
The measures of the sides of triangle JKL and classify it by its sides.
Solution:
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula, we get
[tex]JK=\sqrt{(2-(-3))^2+(6-2)^2}[/tex]
[tex]JK=\sqrt{(2+3)^2+(4)^2}[/tex]
[tex]JK=\sqrt{(5)^2+(4)^2}[/tex]
[tex]JK=\sqrt{25+16}[/tex]
[tex]JK=\sqrt{41}[/tex]
Similarly,
[tex]KL=\sqrt{\left(8-2\right)^2+\left(-1-6\right)^2}=\sqrt{85}[/tex]
[tex]JL=\sqrt{\left(8-\left(-3\right)\right)^2+\left(-1-2\right)^2}=\sqrt{130}[/tex]
Now,
[tex]JK^2+KL^2=(\sqrt{41})^2+(\sqrt{85})^2[/tex]
[tex]JK^2+KL^2=41+85[/tex]
[tex]JK^2+KL^2=126[/tex]
[tex]JK^2+KL^2\neq JL^2[/tex]
Since, measure of all sides are different and sum of squares of two smaller sides is not equal to the square of largest side, therefore the triangle is scalene triangle.
