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(A) Given that the line makes an intercept of $3$ units on the negative direction of the $y$-axis,so $c = -3$.The angle of inclination is $	heta = {\

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$3x - 5y - 15 = 0$

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(A) Given that the line makes an intercept of $3$ units on the negative direction of the $y$-axis,so $c = -3$.The angle of inclination is $	heta = {	an ^{ - 1}}\left( {\frac{3}{5}} ight)$,which implies the slope $m = 	an 	heta = \frac{3}{5}$.The slope-intercept form of the equation of a line is $y = mx + c$.Substituting the values of $m$ and $c$:$y = \frac{3}{5}x - 3$Multiply by $5$ to clear the fraction:$5y = 3x - 15$Rearranging the terms,we get:$3x - 5y - 15 = 0$.

Find the equation of a straight line that makes an intercept of $3$ units on the negative direction of the $y$-axis and is inclined at an angle of ${\tan ^{ - 1}}\left( {\frac{3}{5}} \right)$ with the $x$-axis.

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