You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2)

Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).

Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

Let the vertices of the triangle be A (4, - 6), B (3, - 2), and C (5, 2).

Let M be the mid-point of side BC of ∆ABC.

Therefore, AM is the median in ∆ABC.

According to the mid point formula,

O(x, y) = [(x₁ + x₂) / 2, (y₁ + y₂) / 2]

Since M is the mid point of the points joining C(5, 2) and B(3, -2),

Coordinates of point M = [(3 + 5) / 2, (- 2 + 2) / 2] = (4, 0)

Area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃_-(y₁ - y₂)]

Considering ΔABM with A (4, - 6), B (3, - 2) and M(4, 0), by substituting the values of vertices, A, B, M in the formula.

Area of ΔABM = 1/2 [(4){(- 2) - (0)} + (3){(0) - (- 6)} + (4){(- 6) - (- 2)}] square units

= 1/2 (- 8 + 18 - 16) square units

= -3 square units

Considering ΔAMC with A (4, - 6), M(4, 0) and C (5, 2), by substituting the values of vertices, A, M, C in the formula.

Area of ΔAMC = 1/2 [(4){(0) - (2)} + (4){2 - (- 6)} + (5){(- 6) - (0)}] square units

= 1/2 (- 8 + 32 - 30) square units

= - 3 square units

However, the area cannot be negative. Therefore, the area of ∆AMC and ∆ABM is 3 square units.

Hence, clearly, median AM has divided ΔABC in two triangles of equal areas.