If d(-1/5,5/2),e(7,3),f(7/2,7/2)are the mid points of …

Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) be the vertices of {tex}\Delta{/tex}ABC.
D is the mid-point of AB.
{tex} \Rightarrow \frac{{{x_1} + {x_2}}}{2} = - \frac{1}{2}{/tex} and {tex}\frac{{{y_1} + {y_2}}}{2} = \frac{5}{2}{/tex}
{tex}\Rightarrow{/tex} x₁ + x₂ = -1 ...(i) and y₁ + y₂ = 5 ...(ii)
E is the mid-point of BC.
{tex} \Rightarrow \frac{{{x_2} + {x_3}}}{2} = 7{/tex} and {tex}\frac{{{y_2} + {y_3}}}{2} = 3{/tex}
{tex}\Rightarrow{/tex} x₂ + x₃ = 14 ...(i) and y₂ + y₃ = 6 ...(iv)
F is the mid-point of AC.
{tex} \Rightarrow \frac{{{x_1} + {x_3}}}{2} = \frac{7}{2}{/tex} and {tex}\frac{{{y_1} + {y_3}}}{2} = \frac{7}{2}{/tex}
{tex}\Rightarrow{/tex} x₁ + x₃ = 7 ...(v) and y₁ + y₃ = 7 ...(vi)
Adding equations (i), (iii) and (v), we get
2(x₁ + x₂ + x₃) = -1 + 14 + 7
{tex}\Rightarrow{/tex} 2(x₁ + x₂ + x₃) = 20
{tex}\Rightarrow{/tex} x₁ + x₂ + x₃ = 10
{tex}\Rightarrow{/tex} -1 + x₃ = 10 ...[From (i)]
{tex}\Rightarrow{/tex} x₃ = 11
{tex}\Rightarrow{/tex} x₁ = 7 - 11 = -4
{tex}\Rightarrow{/tex} x₂ = -1 + 4 = 3
Adding equations (ii), (iv) and (vi), we get
2(y₁ + y₂ + y₃) = 5 + 6 + 7
{tex}\Rightarrow{/tex} 2(y₁ + y₂ + y₃) = 18
{tex}\Rightarrow{/tex} y₁ + y₂ + y₃ = 9
{tex}\Rightarrow{/tex} 5 + y₃ = 9 ...[From (ii)]
{tex}\Rightarrow{/tex} y₃ = 4
{tex}\Rightarrow{/tex} y₁ = 7 - 4 = 3
{tex}\Rightarrow{/tex} y₂ = 6 - 4 = 2
Thus, we have A(-4, 3), B(3, 2) and C(11, 4)
{tex}\therefore{/tex} Area of {tex}\Delta{/tex}ABC {tex} = \frac{1}{2}\left| { - 4\left( {2 - 4} \right) + 3\left( {4 - 3} \right) + 11\left( {3 - 2} \right)} \right|{/tex}
{tex} = \frac{1}{2}\left| {8 + 3 + 11} \right|{/tex}
{tex} = \frac{1}{2} \times 22{/tex}
= 11 sq. units.