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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).

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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)

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