Question

if non zero numbers $a,b,c$ are in HP then the straight line $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$ always passes through a fixed point

• A. $\left(1,–\frac{1}{2}\right)$
• B. $\left(1,-2\right)$
• C. $(-1,-2)$
• D. $(-1,2)$

Answer 1

The correct option is B

$\left(1,-2\right)$

Explanation for correct answer

Given $a,b,c$ are in H.P.

So$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP.

So $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\frac{1}{c}=\frac{2}{b}–\frac{1}{a}$

Given $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$

$\Rightarrow \frac{x}{a}+\frac{y}{b}+\frac{2}{b}–\frac{1}{a}=0$

$\Rightarrow \left(\frac{1}{a}\right)\left(x–1\right)+\left(\frac{1}{b}\right)\left(y+2\right)=0$

$\Rightarrow x=1$and $y=-2$

So the line passes through $(1,-2).$

Hence, option B is the answer.