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Q.
The equation of one of the lines parallel to $4x - 3y = 5$ and at a unit distance from the point $(- 1, - 4)$ is
Straight Lines
Solution:
Equation of any lines parallel to
$4x-3 y = 5$ is $4x-3y + \lambda = 0\quad\ldots\left(1\right)$
Since distance between the lines is one unit.
$\therefore $ Length of the perpendicular from $\left(-1, -4\right)$ to $4x - 3y + \lambda = 0$ is $1$.
$\Rightarrow \left|\frac{4\left(-1\right)-3\left(-4\right)+\lambda}{\sqrt{4^{2}+3^{2}}}\right|=1$
$\Rightarrow \left|8+\lambda\right|=5$
$\Rightarrow 8 + \lambda = 5$ or $8 + \lambda = -5$
$\Rightarrow \lambda=-3$, or $\lambda=-13$
$\Rightarrow 4 x -3 y -3 = 0$ or $4x-3y-13 = 0$