Fig. 7.23(A)
Now OA = 10 units

Squaring
100 = 4α² + 8α + 68
4α² + 8α - 32 = 0
α² + 2α - 8 = 0
α² + 4α + 2α - 8 = 0
α(α + 4) + 2(α - 8) = 0
(α - 4) (α - 2) = 0
α = -4, α = 2.

Let the vertices of the triangle are A(0, 0), B(6,0) and C(4, 3).

Here, we have
x₁ = 0,    y₁ = 0
x₂ = 6,    y₂ = 0
and    x₃ = 4, y₃ = 3
Now, Area of ∆ABC

The vertices of the triangle are given as A(5, 2), B(4, 7) and C(7, -4)
Here, we have
x₁ = 5,    y₁ = 2
x₂ = 4,    y₂ = 7
and    x₃ = 7,    y₃ = -4
Now, area of triangle ABC

We cannot take area as -ve.
So, area of the triangle = 2 sq. units.

The vertices of the triangle are given as A(3, 8), B(-4, 2) and C(5, 1)
Here, we have
x₁ = 3,    y₁ = 8
x₂ = -4,    y₂ = 2
and    x₃ = 5,    y₃ = 1
Now, area of triangle ABC