What are the coordinates of a point on the li | vedclass

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(B) The slope of the given line $2x + 3y + 7 = 0$ is $m = -\frac{2}{3}$.Let the angle of inclination be $	heta$,so $	an 	heta = -\frac{2}{3}$.Since

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$\left( 1 - \frac{9}{\sqrt{13}}, -3 + \frac{6}{\sqrt{13}} \right)$

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(B) The slope of the given line $2x + 3y + 7 = 0$ is $m = -\frac{2}{3}$.Let the angle of inclination be $	heta$,so $	an 	heta = -\frac{2}{3}$.Since $	an 	heta The parametric form of the line passing through $(x_1, y_1) = (1, -3)$ is $\frac{x - 1}{\cos 	heta} = \frac{y + 3}{\sin 	heta} = r$.Substituting $r = 3$,we get $\frac{x - 1}{-3/\sqrt{13}} = \frac{y + 3}{2/\sqrt{13}} = 3$.Thus,$x - 1 = 3 	imes \left( -\frac{3}{\sqrt{13}} ight) = -\frac{9}{\sqrt{13}}$ and $y + 3 = 3 	imes \left( \frac{2}{\sqrt{13}} ight) = \frac{6}{\sqrt{13}}$.Therefore,$x = 1 - \frac{9}{\sqrt{13}}$ and $y = -3 + \frac{6}{\sqrt{13}}$.The coordinates are $\left( 1 - \frac{9}{\sqrt{13}}, -3 + \frac{6}{\sqrt{13}} ight)$.

What are the coordinates of a point on the line $2x + 3y + 7 = 0$ which is at a distance of $3$ units from the point $(1, -3)$?

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