Given:
The vertices of the triangle are (-10,-3), (1,4) and (-1,7).
To find:
The perimeter of the triangle.
Solution:
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Let the vertices of the triangle are A(-10,-3), B(1,4) and C(-1,7).
Using distance formula, we get
[tex]AB=\sqrt{(1-(-10))^2+(4-(-3))^2}[/tex]
[tex]AB=\sqrt{(1+10)^2+(4+3)^2}[/tex]
[tex]AB=\sqrt{(11)^2+(7)^2}[/tex]
[tex]AB=\sqrt{121+49}[/tex]
[tex]AB=\sqrt{170}[/tex]
Similarly,
[tex]BC=\sqrt{\left(-1-1\right)^2+\left(7-4\right)^2}=\sqrt{13}[/tex]
[tex]AC=\sqrt{\left(-1-\left(-10\right)\right)^2+\left(7-\left(-3\right)\right)^2} =\sqrt{181}[/tex]
Now, the perimeter of the triangle is
[tex]Perimeter=AB+BC+AC[/tex]
[tex]Perimeter=\sqrt{170}+\sqrt{13}+\sqrt{181}[/tex]
[tex]Perimeter\approx 13.038+3.606+13.454[/tex]
[tex]Perimeter=30.098[/tex]
Therefore, the perimeter of the triangle is 30.098 units.
