Question

In triangle ABC,ad is the median a(4,-6)(3,-2)and(5,2).prove that area of∆adc =area of∆abd

Answer 1

=In ∆ABC, median (4-6) (3-2) and (5-2)

Prove that area∆adc =area ∆abd

Let,

∆ABC be shown

AD be the median which divide BC into two - :

eqal parts BD & CD

= (x₁+x₂/2 y₁+y₂)

= (3+5/2,-2+2/2)

= (8/2,0/2)

= (4,0)

RHS

area of triangle ABD

=1/2[x₁(y₁-y₃)+x₃(y₁-y₂)

=1/2[4(-2-0)+(3) (0-(6)+4(-6-(-2)]

=1/2[4(-2)+3(6)+4(-4)]

=1/2[-8+18-16]

=1/2[-6]

=-3

area of triangle ABC=3 square

RHS

area of triangle ACD

=1/2[x₁(y₁-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]

=x₁=4,y₁=6

=x₂=5,y₂=2

=x₃=4,y₃=0

Area of triangle ACD

=1/2[4(2-0)+(5)(0-(-6)+log)4(-6-4)]

=1/2[4(2)+5(0+6)+4(-8)]

=1/2[4(2)+5(6)+4(-8)]

=1/2[8+30]

=1/2(6)

=3s squar unit

RHS..

Area of ∆ADC=Area of ∆ABD

hence proved..