Question

Find the perimeter of the triangles whose vertices have the following coordinates $(-2,1),(4,6),(6,-3)$.

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let the given vertices be $A(-2,1)$, $B(4,6)$ and $C(6,-3)$

We first find the distance between $A(-2,1)$ and $B(4,6)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(4-(-2\left)\right)}^2+{(6-1)}^2}$
$=\sqrt{{(4+2)}^2+5^2}=\sqrt{6^2+5^2}=\sqrt{36+25}=\sqrt{61}$

Similarly, the distance between $B(4,6)$ and $C(6,-3)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(6-4)}^2+{(-3-6)}^2}$
$=\sqrt{2^2+{(-9)}^2}=\sqrt{4+81}=\sqrt{85}$

Now, the distance between $C(6,-3)$ and $A(-2,1)$ is:

$CA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(6-(-2\left)\right)}^2+{(-3-1)}^2}$
$=\sqrt{{(6+2)}^2+{(-4)}^2}=\sqrt{8^2+{(-4)}^2}=\sqrt{64+16}$

$=\sqrt{80}$

Since the perimeter $P$ of a triangle $ABC$ is $AB+BC+CA$, therefore,

$P=\sqrt{61}+\sqrt{85}+\sqrt{80}$

Hence, the perimeter of the triangle is $(\sqrt{61}+\sqrt{85}+\sqrt{80})$ units.