Question

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

Answer 1

We have,

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

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

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

Given that,

Area of triangle $=0$

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

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

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

$\Rightarrow 2a+3b=ab$

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

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

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

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

Hence proved.

Hence, this is the answer.