Question

Determine the length of the perimeter of a triangle whose vertices are L(-1:2) ,M(1:6) and N(4:0)

Answer 1

Answer on Question #44564 – Math – Analytic Geometry

Determine the length of the perimeter of a triangle whose vertices are $L(-1:2), M(1:6)$ and $N(4:0)$.

Solution.

Note that the length of a segment $AB$ with the ends $A(x,y), B(z,t)$ can be computed in the following way:

$$|AB| = \sqrt{(x - z)^2 + (y - t)^2};$$

So:

$$|LM| = \sqrt{(-1 - 1)^2 + (2 - 6)^2} = \sqrt{4 + 16} = 2\sqrt{5};$$ $$|MN| = \sqrt{(1 - 4)^2 + (6 - 0)^2} = \sqrt{9 + 36} = 3\sqrt{5};$$ $$|LN| = \sqrt{(-1 - 4)^2 + (2 - 0)^2} = \sqrt{25 + 4} = \sqrt{29};$$ $$P = |LM| + |MN| + |LN| = 2\sqrt{5} + 3\sqrt{5} + \sqrt{29} = 5\sqrt{5} + \sqrt{29}.$$

Answer.

$$5\sqrt{5} + \sqrt{29}.$$

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