Answer :

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

Let D be the mid-point of side BC of ΔABC Therefore, AD is the median in ΔABC**To prove: Area ΔABC = Area ΔABD + Area ΔADC**

**Midpoint of the line joining the points A(x _{1, }y_{1}) and B(x_{2}, y_{2}) is given by**

Therefore, by applying the formula we can get the coordinate of D

Coordinates of point D = () = (4, 0)

We also know that if **A(x _{1, }y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) are vertices of a triangle then**

Area of

**triangle is given by,**

Area of the Δ (ABC) = = [4 (-2 - 2) + 3 (0+ 8) + 5 (-6+ 2)]

For A (4, - 6), B (3, - 2), and C (5, 2)

For A (4, - 6), B (3, - 2), and C (5, 2)

= (-16 + 24 – 20)

= - 6 Square units

However, area cannot be negative. Therefore, **area of ΔABC is 6 square units**

Now for A(4, -6), D(4, 0), C(5, 2)

Area of the Δ (ADC) = = [4 (0 - 2) + 4 (2 + 6) + 5 (-6 - 0)]

= (-8 + 32 – 30)

= - 3 Square units

However, area cannot be negative. Therefore, **area of ΔADC is 3 square units**

Now for for A(4, -6), B(3, -2) and D(4, 0)

Area of Δ ABD = [4(-2 - 0) + 3(0 + 6) + 4(-6 +2)]

Area = [-8 + 18 - 16]

Area = -3 square units

Now the Area cannot be negative so **Area of ΔABD = 3 units**

Area of Δ ABD + Area of Δ ADC = 3 + 3 square units

Area of Δ ABD + Area of Δ ADC = 6 = Area of Δ ABC

Clearly, median AD has divided ΔABC in two triangles of equal areas

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