Lines are drawn parallel to the line 4x - 3y | vedclass

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Question

(A) The equation of any line parallel to $4x - 3y + 2 = 0$ is of the form $4x - 3y + \lambda = 0$.The perpendicular distance from the origin $(0, 0)$

Vedclass · Accepted Answer

$\left( -\frac{1}{4}, \frac{2}{3} \right)$

Vedclass · Answer

(A) The equation of any line parallel to $4x - 3y + 2 = 0$ is of the form $4x - 3y + \lambda = 0$.The perpendicular distance from the origin $(0, 0)$ to this line is given by $\frac{|\lambda|}{\sqrt{4^2 + (-3)^2}} = \frac{|\lambda|}{5}$.Given that this distance is $\frac{3}{5}$,we have $\frac{|\lambda|}{5} = \frac{3}{5}$,which implies $|\lambda| = 3$,so $\lambda = \pm 3$.Thus,the equations of the lines are $4x - 3y + 3 = 0$ and $4x - 3y - 3 = 0$.Now,we check the given points:For option $A$: $4(-\frac{1}{4}) - 3(\frac{2}{3}) + 3 = -1 - 2 + 3 = 0$. This point satisfies the equation $4x - 3y + 3 = 0$.

Lines are drawn parallel to the line $4x - 3y + 2 = 0$ at a distance of $\frac{3}{5}$ from the origin. Which one of the following points lies on any of these lines?

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