Question

Find the perimeter of the triangle whose vertices are
(-2, 1), (4, 6) and (6, -3).​

Answer 1

Answer:

The perimeter of a triangle is the sum of the lengths of its three sides.

To find the length of each side, we use distance formula.

Let A = (-2, 1)

Let B = (4, 6)

Let C = (6, -3)

Distance AB = $\sqrt{(4+2)^{2} +(6-1)^{2} } = \sqrt{36+25} = \sqrt{61}$ units

Distance BC = $\sqrt{(6-4)^{2}+(-3+6)^{2}} = \sqrt{4+9} = \sqrt{13}$ units

Distance CA = $\sqrt{(6+2)^2+(-3-1)^2} = \sqrt{64+16} = \sqrt{80}$ units

The required answer is the sum of these three values:

$\sqrt{61} + \sqrt{13} + \sqrt{80}$ units

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Answer 2

Answer:

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

Step-by-step explanation:

Perimeter means sum of all Sides.

So, distance between two points (a,b) and (c,d) is defined as

$\sqrt{ {(c - a)}^{2} + \: {(d - b)}^{2} }$

Assumptions:

Let A = (-2,1) , B = (4,6) , C = (6,-3) and AB , BC, CA be the Sides of the Triangle

Finding distance between them:

AB = $\sqrt{ {(4 + 2)}^{2} + {(6 - 1)}^{2} }$

= $\sqrt{36 + 25}$

= $\sqrt{61}$

BC = $\sqrt{ {(6 - 4)}^{2} + {( - 3 - 6)}^{2} }$

= $\sqrt{4 + 81}$

= $\sqrt{85}$

CA = $\sqrt{ {(6 + 2)}^{2} + {( - 3 - 1)}^{2} }$

= $\sqrt{64 + 16}$

= $\sqrt{80}$

Result:

Perimeter = Sum of all sides

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