ABC ia triangle coordinates of whose …

According to the question, A(0, - 1),D(1, 0) and E(0,1).

Let coordinates of B and C are (x₂, y₂) and (x₃, y₃) respectively.
D is mid-point of AB,
{tex}\therefore{/tex} 1 = {tex}\frac { 0 + x _ { 2 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} x₂ = 2
and 0 = {tex}\frac { - 1 + y _ { 2 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} y₂ = 1
{tex}\therefore{/tex} Coordinates of B are (2, 1)
E is mid-point of AC
{tex}\therefore{/tex} 0 = {tex}\frac { 0 + x _ { 3 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} x₃ = 0
and 1 = {tex}\frac { - 1 + y _ { 3 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} y₃ = 3
{tex}\therefore{/tex}  Coordinates of C are (0, 3)
Area of {tex}\triangle{/tex}ABC
{tex}= \frac { 1 } { 2 }{/tex} {tex}[0(1 - 3) + 2(3 + 1) + 0(-1 + -1)]{/tex}
{tex}= \frac { 1 } { 2 }{/tex} {tex}\times{/tex} 8 = 4 sq. units
F is mid-point of BC
{tex}\therefore{/tex} Coordiantes F are {tex}\left( \frac { 2 + 0 } { 2 } , \frac { 1 + 3 } { 2 } \right){/tex}, i.w, (1, 2).
Area {tex}\triangle{/tex}DEF
{tex}= \frac { 1 } { 2 }{/tex} {tex}[1(2 - 1) + (1 - 0) + 0(0 - 2)]{/tex}
{tex}= \frac { 1 } { 2 }{/tex} [1 + 1] = 1 sq. units