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The line on which the lines $a x+b y=1$ and $b x+a y=1$ (with $a \neq 0 \neq b$ ) intersect for any real values of $a$ and $b$ is
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The correct answer is:
$x=y$
Lines : $a x+b y=1$...(i)
$b x+a y=1$...(ii)
[Multiply Eq. (i) by $b$ ] - [Multiply Eq. (ii) by $a$ ] $a b x+b^2 y-a b x-a^2 y=b-a$
$\Rightarrow \quad y=\frac{b-a}{b^2-a^2}=\frac{1}{a+b}$
Put $y=\frac{1}{a+b}$ in Eq. (i),
$a x=1-\frac{b}{a+b}=\frac{a}{a+b}$
$\therefore \quad x=\frac{1}{a+b}$
$\therefore$ Point of intersection of both lines are of type
$\left(\frac{1}{a+b}, \frac{1}{a+b}\right)$.
$\therefore$ Hence, such points lie on $x=y$ only.
$b x+a y=1$...(ii)
[Multiply Eq. (i) by $b$ ] - [Multiply Eq. (ii) by $a$ ] $a b x+b^2 y-a b x-a^2 y=b-a$
$\Rightarrow \quad y=\frac{b-a}{b^2-a^2}=\frac{1}{a+b}$
Put $y=\frac{1}{a+b}$ in Eq. (i),
$a x=1-\frac{b}{a+b}=\frac{a}{a+b}$
$\therefore \quad x=\frac{1}{a+b}$
$\therefore$ Point of intersection of both lines are of type
$\left(\frac{1}{a+b}, \frac{1}{a+b}\right)$.
$\therefore$ Hence, such points lie on $x=y$ only.
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