Question

If the point $(a,0),(0,b)and(3,2)$ are collinear, prove that

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

Answer 1

We have,

$(x_1,y_1)=(a,0)$

$(x_2,y_2)=(0,b)$

$(x_3,y_3)=(3,2)$

Given that, it is collinear,

So,

$Areaoftriangle=0$

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

$\Rightarrow \frac{1}{2}[a(b-2)+0(2-0)+3(0-b)]=0$

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

$\Rightarrow 2a+3b=ab$

On dividing $ab$ both side and we get,

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

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

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

Hence, this is the answer.