Section formula

Let the coordinates of point A be $({ x }_{ 1 },\quad { y }_{ 1 })$ , B be $({ x }_{ 2 },\quad { y }_{ 2 })$ and C be $({ x }_{ 3 },\quad { y }_{ 3 })$ . Given the midpoint of BC is $(1,0)$ , that of CA is $(3,5)$ and that of AB is $(-2,4)$ . Then we obtain three equations in ${ x }_{ 1 }$ , ${ x }_{ 2 }$ and ${ x }_{ 3 }$ :

$\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =-2$
$\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =1$ $\frac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } =3$

Adding all the above equations together, we get ${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=2$ Which gives us ${ x }_{ 1 }=0$ , ${ x }_{ 2 }=-4$ and ${ x }_{ 3 }=6$

Similarly, we can solve for the y-coordinates:

$\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =4$ $\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =0$ $\frac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } =5$

Adding all the equations, we have ${ y }_{ 1 }=9$ , ${ y }_{ 2 }=-1$ and ${ y }_{ 3 }=1$

Hence we have the coordinates $A(0,9)$ , $B(-4,-1)$ and $C(6,1)$ of the vertices of the $\Delta ABC$