Question

find the perimeter of triangle whose vertices have the coordinates (3, 10) (5, 2) and (4, 12)​

Answer 1

(3, 10) (5, 2) (4, 12)

(3,10) (5,2) (4,12)

let us assume that

3=x1 10=y1

5=x2 5=y2

4=x2 12=y3

∆=1/2|x1(y2-y3) +x2(y3-y1) +x3(y1-y2) |

∆=1/2|3(2-12) +5(12-10) +4(10-2) |

∆=1/2|3(-10) +5(2) +4(8) |

∆=1/2|-30+10+32|

∆=1/2|-30+42|

∆=1/2|+12|

∆=1/2×12

∆=6