If the straight line through the point P(3, 4 | vedclass

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Question

(A) The equation of a line passing through $P(3, 4)$ with an inclination of $	heta = \frac{\pi}{6}$ is given by the parametric form: $\frac{x - 3}{\c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a line passing through $P(3, 4)$ with an inclination of $	heta = \frac{\pi}{6}$ is given by the parametric form: $\frac{x - 3}{\cos(\pi/6)} = \frac{y - 4}{\sin(\pi/6)} = r$. 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: $12(3 + \frac{r\sqrt{3}}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$. $36 + 6r\sqrt{3} + 20 + 2.5r + 10 = 0$. $66 + r(6\sqrt{3} + 2.5) = 0$. Since $PQ$ is a length,we take the magnitude: $r = |\frac{-66}{6\sqrt{3} + 2.5}| = \frac{66}{6\sqrt{3} + 2.5} = \frac{132}{12\sqrt{3} + 5}$.

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

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