Question

A(7, -3), B(5, 3), C(3, -1) are the vertices of a triangle
ABC and AD is its median. Prove that median AD divides triangle ABC into two triangles of equal areas​

Answer 1

Answer:

The vertices of the triangle are A(7,-3),B(5,3) and C(3,-1).

Step-by-step explanation:

Coordinate of D=[5+3/2,3-1/3]=(4,1)

For the area of the triangle ADC,let

A(x1,y1)=A(7,-3)

D(x2,y2)=D(4,1)

C(x3,y3)=C(3,-1). Then

Area of triangle ADC

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

=1/2[7(1+1)+4(-1+3)+3(-3-1)]

=1/2[14+8-12]

=5 sq.unit

Now, for the area of triangle ABD,let

A(x1,y1)=A(7,-3)

B(x2,y2)=B(5,3)

D(x3,y3)=D(4,1). Then

Area of triangle ADC

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

=1/2[7(3-1)+5(1+3)+4(-3-3)]

=1/2[14+20-24]

=5 sq.unit

Thus, Area(triangle ADC)=Area(triangle ABD)=5 sq.unit

Hence,AD divides triangle ABC into two triangles of equal areas.

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