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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

Reduce the following equation into slope-intercept form and find its slope and the $y$-intercept: $y=0$.

Find the equation of the line passing through the point $(2, -3)$ such that the sum of its intercepts on the axes is $-2$.

Find the value of $k$,if the line joining the points $(2, k)$ and $(3, 7)$ is parallel to the line joining points $(-2, 1)$ and $(3, 0)$.

The equation of a line with slope $-\frac{1}{\sqrt{2}}$ and making an intercept of $2 \sqrt{2}$ units on the negative direction of the $y$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module