We have,
A(x1,y1)=(a,0)
B(x2,y2)=(0,b)
C(x3,y3)=(3,2)
Given that,
Area of triangle =0
12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=0
⇒[a(b−2)+0(2−0)+3(0−b)]=0
⇒ab−2a−3b=0
⇒2a+3b=ab
⇒2a+3bab=1
⇒2aab+3bab=1
⇒2b+3a=1
⇒3a+2b=1
Hence proved.
Hence, this is the
answer.