Question

If nonzero numbers $a,b,c$ are in H.P., then the straight line $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=1$ always passes through a fixed point. That point is

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

Answer 1

The correct option is A $(-1,-2)$

$a,b,c$ are in HP thus $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in

AP

thus $\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

$\frac{1}{c}=\frac{2}{b}-\frac{1}{a}$

given line $\Rightarrow \frac{x}{a}+\frac{y}{b}+\frac{1}{c}=1$

$\frac{x}{a}+\frac{y}{b}+\frac{2}{b}-\frac{1}{a}=1$

$\frac{x}{a}+\frac{y}{b}=1+\frac{1}{a}-\frac{2}{b}$

$\frac{1}{a}(n-1)+\frac{1}{b}(y+2)=1$

$(x-1)+\frac{a}{b}(y+2)=a$

The following line does not pass through

any fixed point

There is error in question

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

Instated of this these should be zero.

To get a fixed point