AD Median is drawn to side BC which is the vertex of ∆ABC.

Median AD divides AABC into two triangles ∆ABD and ∆ADC which are equal in area.

Now Median AD bisects BC.

∴ BD = DC.

(i) As per Mid-Points formula,

Coordinates of D are

= \(\sqrt{(4)^2 - (0)^2}\)

= \(\sqrt{16, 0}\)

= (4,0)

∴ Coordinates of D are (4, 0)

(ii) Now, Area of ∆ABD:

(iiii) Now, Area of ADC:

= – 3 sq. units.

∴ Area of ∆ADC = -3 sq. units.

∴ Area of ∆ABC : = Area of ∆ABD + Area of ∆ADC

= (-3) +(-3)

= -6 sq. units.

(iv) Now, Area of ∆ABC : (Direct Method)