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If $\alpha$ is the angle made by the perpendicular drawn from origin to the line $3 x-4 y+5=0$ with positive $X$-axis in positive direction and $a x+b y=1$ is the equation of a line passing through the point $(1,-1)$ with $\tan \alpha$ as its slope, then $a+a b+b=$
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Verified Answer
The correct answer is:
$19$
Slope of perpendicular to $3 x-4 y+5=0$ is $\frac{-4}{3}$
$\therefore \quad \tan \alpha=\frac{-4}{3}$
Equation of line through $(1,-1) ; y+1=\frac{-4}{3}(x-1)$
$$
\begin{aligned}
& \Rightarrow \quad 3 y+3=-4 x+4 \Rightarrow 4 x+3 y=1 \\
& \therefore \quad a=4, b=3 \\
& \therefore \quad a+a b+b=4+12+3=19 .
\end{aligned}
$$
$\therefore \quad \tan \alpha=\frac{-4}{3}$
Equation of line through $(1,-1) ; y+1=\frac{-4}{3}(x-1)$
$$
\begin{aligned}
& \Rightarrow \quad 3 y+3=-4 x+4 \Rightarrow 4 x+3 y=1 \\
& \therefore \quad a=4, b=3 \\
& \therefore \quad a+a b+b=4+12+3=19 .
\end{aligned}
$$
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