If the points ( a,0),(0,b)and (3,2) are col linear, prove that $\frac{2}{b}+\frac{3}{a}=1$

We have,

$A\left(x_1,y_1\right)=\left(a,0\right)$

$B\left(x_2,y_2\right)=\left(0,b\right)$

$C\left(x_3,y_3\right)=\left(3,2\right)$

Given that,

Area of triangle $=0$

$\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]=0$

$\Rightarrow \left[a\left(b-2\right)+0\left(2-0\right)+3\left(0-b\right)\right]=0$

$\Rightarrow ab-2a-3b=0$

$\Rightarrow 2a+3b=ab$

$\Rightarrow \frac{2a+3b}{ab}=1$

$\Rightarrow \frac{2a}{ab}+\frac{3b}{ab}=1$

$\Rightarrow \frac{2}{b}+\frac{3}{a}=1$

$\Rightarrow \frac{3}{a}+\frac{2}{b}=1$

Hence proved.

Hence, this is the answer.