What is the perimeter of triangle with vertices of (12) (3-4) and (-45)?

Perimeter is ##23.558##
To find the perimeter of a triangle with vertices of ##(12)## ##(34)## and ##(45)## we have to first find distance between each pair of points which will give length of sides. For this we use distance formula between two points ##(x_1y_1)## and ##(x_2y_2)## is ##sqrt((x_2-x_1)^2+(y_2-y_1)^2)##. Hence if lengths of sides are ##L_1L_2L_3## these are as follows:
##L_1=sqrt((3-1)^2+((-4)-(2))^2)=sqrt(2^2+(-6)^2)=sqrt(4+36)=sqrt40=2sqrt10=6.325##
##L_2=sqrt((-4-(3))^2+(5-(-4))^2)=sqrt((-7)^2+9^2)=sqrt(49+81)=sqrt130=11.402##
##L_3=sqrt((-4-1)^2+(5-2)^2)=sqrt((-5)^2+3^2)=sqrt(25+9)=sqrt34=5.831##
Hence Perimeter is ##6.325+11.402+5.831=23.558##