# find the distance between the points of the triangle A(2,1) B(-1,4) and C(-4,3).

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I believe that you need to find the length of the sides of the triangle ABC.

Given the vertices's A(2,1), B(-1,4), and C(-4,3).

Let us use the distance formula to find the length of the sides AB, BC, and AC.

==> AB = sqrt( xA-xB)^2 + (yA-yB)^2

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

= sqrt(18) = 3sqrt2

**==> AB = 3sqrt2.**

==> AC = sqrt(xA-xC)^2 + (yA-yC)^2

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

= sqrt(6^2 + -2^2)

= sqrt(36+ 4) = sqrt40.= 2sqrt10

**==> AC = 2sqrt10.**

==> BC = sqrt(xB-xC)^2 + (yB-yC)^2

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

= sqrt(3^2 + 1)

= sqrt10.

**==> BC = sqrt10**

To find the distances between the 3 points: A(2,1), B(-1,4) and C(-4,3).

The distance d between the points A and B is given by:

**AB** = sqrt{(xB-xA)^2+(yB-yA)^2} = sqrt{(-1-2)^2+(4-1)^2} = sqrt{9+9} = sqrt18 = **3sqrt 2**.

**BC**= sqrt{(xC-xB)^2+(yC-yB)^2} = sqrt{(-4-(-1))^2+(3-4)^2} = sqrt{9+1} = **sqrt10**.

**CA** = sqrt{(xA-xC)^2+(yA-yC)^2} = sqrt{(2-(-4))^2+(1-3)^2} = sqrt{36+4} = sqrt40 = **2sqrt10**.

Therefore the distances **AB = 3sqrt2, BC = sqrt10 and CA = 2sqrt10.**