# Calculate the perimeter of the triangle whose vertices are (1,2) (-1,4) and * (-2,3). To find the perimeter we need the distances between the points (1,2) and  (-1,4), and the points (-1, 4) and (-2,3) and the points (-2,3) and (1,2).

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

The distance between (1,2) and ...

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To find the perimeter we need the distances between the points (1,2) and  (-1,4), and the points (-1, 4) and (-2,3) and the points (-2,3) and (1,2).

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

The distance between (1,2) and  (-1,4)

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

=> sqrt ( 2^2 + 2^2)

=> sqrt 8

The distance between (-1, 4) and (-2,3)

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

=> sqrt (1^2 + 1^2)

=> sqrt 2

The distance between (-2,3) and (1,2)

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

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

=> sqrt 10

Therefore the perimeter is sqrt 10 + sqrt 8 + sqrt 2.

Approved by eNotes Editorial Team Let ABC be a triangle.

Given the vertices's of a triangle are:

A(1,2) , B(-1,4), and C(-2,3).

We need to find the perimeter of the triangle.

First we need to find the length of the sides.

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

= sqrt( 2^2 + 2^2 )

= sqrt( 4+4)

= sqrt8

==> AB = 2sqrt2.

==> AC = sqrt( 1+ 2)^2 + ( 2-3)^2

= sqrt( 3^2 + 1^2)

= sqrt(9+ 1)

= sqrt10.

==> AC= sqrt10.

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

= sqrt( 1^2 + 1^2 )

= sqrt2

==> BC = sqrt2.

==> The perimeter of the triangle is:

P = AB + AC + BC

= 2sqrt2 + sqrt10 + sqrt2

= 7.4 units. ( approx.)

Approved by eNotes Editorial Team