If the straight line passing through P(3,4) m | vedclass

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(D) The equation of a line passing through $P(3,4)$ with an angle $	heta = 30^{\circ}$ is given by $\frac{x-3}{\cos 30^{\circ}} = \frac{y-4}{\sin 30^

Vedclass · Accepted Answer

$\frac{132}{12\sqrt{3} + 5}$

Vedclass · Answer

(D) The equation of a line passing through $P(3,4)$ with an angle $	heta = 30^{\circ}$ is given by $\frac{x-3}{\cos 30^{\circ}} = \frac{y-4}{\sin 30^{\circ}} = r$. Any point $Q$ on this line is $(3 + r\cos 30^{\circ}, 4 + r\sin 30^{\circ}) = (3 + \frac{r\sqrt{3}}{2}, 4 + \frac{r}{2})$. Since $Q$ lies on the line $12x + 5y + 10 = 0$,we substitute the coordinates: $12(3 + \frac{r\sqrt{3}}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$. $36 + 6r\sqrt{3} + 20 + \frac{5r}{2} + 10 = 0$. $66 + r(6\sqrt{3} + 2.5) = 0$. Taking the magnitude for length $PQ = |r|$,we get $r(6\sqrt{3} + 2.5) = 66$. $r = \frac{66}{6\sqrt{3} + 2.5} = \frac{132}{12\sqrt{3} + 5}$.

If the straight line passing through $P(3,4)$ makes an angle $\frac{\pi}{6}$ with the positive $x$-axis in the anticlockwise direction and meets the line $12x + 5y + 10 = 0$ at $Q$,then the length of the segment $PQ$ is

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