# Find the perimeter of the triangle whose vetices A(2,3) B(-1, 4) and C (-4,-6)

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Given the triangle ABC such that:

A(2,3) B(-1,4) c(-4,-6) are the vertices's.

To find the perimeter, we need to calculate the length of all 3 sides:

We will use the following distance formula:

D = sqrt[( xA-xB)^2 + ( yA-yB)^2]

==> AB = sqrt( -1-2)^2+ ( 4-3)^2

= sqrt(9+1) = sqrt10

**==> AB= sqrt10**

AC = sqrt[(-4-2)^2 + (-6-3)^2

= sqrt(36+ 81)

= sqrt(117)

**AC = sqrt117)**

BC= sqrt[(-4+1)^2+ ( -6-4)^2

= sqrt( 9 + 100)

= sqrt(109)

**BC = sqrt(109)**

**==> The perimeter is:**

** P = AB + AC + BC**

** = sqrt10 + sqrt117 + sqrt(109)**

** = 24. 42 ( approx.)**

**Then the perimeter is 24.42 **

We have to find perimeter of the triangle ABC with the vertexes at A(2,3) B(-1, 4) and C (-4,-6).

The distance between two points (x1, y1) and (x2 , y2) is given by: sqrt [ ( x2 - x1)^2 + (y2 - y1)^2]

For A(2, 3 and B( -1, 4)

AB =sqrt [( 2 +1)^2 +( 3-4)^2]

= sqrt ( 3^2 + 1^2) = sqrt 10

For B(-1, 4) and C (-4,-6)

For BC =sqrt[ ( -1 +4)^2 + ( 4+ 6)^2]

= sqrt ( 3^2 + 10^2) = sqrt 109

For C (-4,-6) and A(2,3)

CA =sqrt [( -4-2)^2 + (-6-3)^2]

sqrt ( 6^2 + 9^2) = sqrt 117.

**Therefore the perimeter is sqrt 10 + sqrt 109 + sqrt 117.**

**Also, ****sqrt 10 + sqrt 109 + sqrt 117 is approximately equal to 24.419. **

To determine the perimeter of the triangle with vertices A(2,3), B(-1,4) and C (-4,-6).

We use the distance formula d = sqrt{(x2-x1)^2+(y2-y1)^2}, where d is the distance between the points (x1,y1) and (x2,y2).

Therfore the distance AB, BC and CA are given by :

AB = sqrt{(-1-2)^2 +(4-3)^2} = sqrt (9+1) = sqrt10

BC = sqrt{(-4-(-1))^2 +(-6-4)^2 } = sqrt{9+100} = sqrt109

CA = sqrt{(-4-2)^2+(-6-3)^2} = sqrt(36+81) = sqrt117.

Therefore the perimeter of the triangle Ab+BC+CA = sqrt10+sqrt109 +sqrt 117 = 24.42.