Question

the mid point of the side of ∆ are (2,-1)(4,5)and(3,3) find co ordinates of vertices​

Answer 1

★ Question

the mid point of the side of ∆ are (2,-1)(4,5)and(3,3) find co ordinates of vertices

Answer ★

$\implies \tt 3=\frac{x1+x2}{2}\\ \implies 6=x1+x2-----equ(1)\\ \implies\tt 2=\frac{x2+x3}{2}\\ \implies\tt=4=x2+x3---equ(2)\\ \implies \tt 4=\frac{x1+x3}{2}\\ \implies\tt 8=x1+x1-----equ(3)\\ \implies\tt adding\: equation\:(1)+(2)+(3)\\ we\:get\\ \implies 6+4+8=2(x1+x2+x3)\\ \implies\tt 18=2(x1+x2+x3)\\ \implies 9=(x1+x2+x3------equ(4)\\ \:from(4)-(1)\\ \implies\tt 5=x1\\ from\:(4)-(3)\\ \implies\tt 1=x2\\ \implies \tt 3=\frac{y1+y2}{2}---equ(5)\\ \implies \tt -1=\frac{y2+y3}{2}\\ \tt -2=y2+y3-----equ(6)\\ \implies \tt 5=\frac{y1+y3}{2}\\ \implies\tt 10=y1+y3------equ(7)\\ from\: equation (5)+(6)+(7)\\ \implies\tt 6-2+10=2(y1+y2+y3)\\ \implies\tt 14= 2(y1+y2+y3)\\ \implies\tt 7=y1+y2+y3-----equ(8)\\ subtract\:eqution(8)-(5)\\ \implies \tt y3=1\\ equ(8)-(6)\\ \implies\tt y1=9\\ equ(8)-(7)\\ \implies\tt y=-3$