A line L perpendicular to the line 5x - 12y + | vedclass

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

Question

(C) The given line is $5x - 12y + 6 = 0$. The slope of this line is $m = \frac{5}{12}$.Since line $L$ is perpendicular to this line,its slope $m_L$ mu

Vedclass · Accepted Answer

$\frac{169}{60}$

Vedclass · Answer

(C) The given line is $5x - 12y + 6 = 0$. The slope of this line is $m = \frac{5}{12}$.Since line $L$ is perpendicular to this line,its slope $m_L$ must satisfy $m_L 	imes \frac{5}{12} = -1$,so $m_L = -\frac{12}{5}$.The equation of line $L$ in normal form is $x \cos 	heta + y \sin 	heta = p$,where $p = 2$ is the distance from the origin.The slope of this line is $-\cot 	heta = -\frac{12}{5}$,which implies $\cot 	heta = \frac{12}{5}$.Thus,$	an 	heta = \frac{5}{12}$.Since the line makes a positive intercept on the $Y$-axis,we check the intercept form: $y = -\frac{12}{5}x + \frac{2}{\sin 	heta}$.Using $\cot 	heta = \frac{12}{5}$,we have $\sin 	heta = \frac{5}{13}$ and $\cos 	heta = \frac{12}{13}$.The intercept is $\frac{2}{\sin 	heta} = \frac{2}{5/13} = \frac{26}{5} > 0$,which is positive.Finally,$	an 	heta + \cot 	heta = \frac{5}{12} + \frac{12}{5} = \frac{25 + 144}{60} = \frac{169}{60}$.

Similar Questions

Let a line intersect the co-ordinate axes in points $A$ and $B$ such that the area of the triangle $OAB$ is $12$ sq. units. If the line passes through the point $(2,3)$,then the equation of the line is

Find the value of $k$ for which the line $(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$ is parallel to the $x$-axis.

Find the equation of a line that makes an intercept of $-3$ on the $y$-axis and makes an angle of ${\tan ^{ - 1}}\frac{3}{5}$ with the $x$-axis.

The equation of the line which cuts off the intercepts $2a \sec \theta$ and $2a \csc \theta$ on the axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module