If a line L passing through a point A(2, 3) i | vedclass

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

Question

(D) Let the angle made by line $L$ with the positive $X$-axis be $	heta$. The coordinates of point $B$ can be written as $(2 + 4 \cos 	heta, 3 + 4 \

Vedclass · Accepted Answer

$\pi - \operatorname{Tan}^{-1}\left(\frac{3}{4}\right)$

Vedclass · Answer

(D) Let the angle made by line $L$ with the positive $X$-axis be $	heta$. The coordinates of point $B$ can be written as $(2 + 4 \cos 	heta, 3 + 4 \sin 	heta)$.Since $B$ lies on the line $4x - 3y - 19 = 0$,we substitute these coordinates into the equation:$4(2 + 4 \cos 	heta) - 3(3 + 4 \sin 	heta) - 19 = 0$$8 + 16 \cos 	heta - 9 - 12 \sin 	heta - 19 = 0$$16 \cos 	heta - 12 \sin 	heta = 20$Dividing by $4$,we get $4 \cos 	heta - 3 \sin 	heta = 5$.This can be written as $4 \cos 	heta - 3 \sin 	heta = 5(\cos^2 	heta + \sin^2 	heta)$.Alternatively,using $R \cos(	heta + \alpha) = 5$ where $R = \sqrt{4^2 + (-3)^2} = 5$,we have $5 \cos(	heta + \alpha) = 5$,where $\cos \alpha = 4/5$ and $\sin \alpha = 3/5$.Thus,$\cos(	heta + \alpha) = 1$,which implies $	heta + \alpha = 0$ or $2\pi$.Since $	an \alpha = 3/4$,$	heta = -\alpha = -\operatorname{Tan}^{-1}(3/4)$.In the anti-clockwise direction,the angle is $2\pi - \operatorname{Tan}^{-1}(3/4)$ or simply the angle corresponding to the slope $m = 	an 	heta$. Solving $4 \cos 	heta - 3 \sin 	heta = 5$ gives $	an 	heta = -3/4$. The angle is $\pi - \operatorname{Tan}^{-1}(3/4)$.

If a line $L$ passing through a point $A(2, 3)$ intersects another line $4x - 3y - 19 = 0$ at the point $B$ such that $AB = 4$,then the angle made by the line $L$ with the positive $X$-axis in the anti-clockwise direction is

Similar Questions

The equation of the line passing through the origin and the point $(a \cos \theta, a \sin \theta)$ is .....

$A$ line passing through the point $P(1, 1)$ and parallel to the line $x - y = 5$ cuts the line $x + 3y - 2 = 0$ at $Q$. Then twice the length of the segment $PQ$ is

$A$ line through $A(-5, -4)$ meets the lines $x + 3y + 2 = 0$,$2x + y + 4 = 0$,and $x - y - 5 = 0$ at $B$,$C$,and $D$ respectively. If $\left( \frac{15}{AB} \right)^2 + \left( \frac{10}{AC} \right)^2 = \left( \frac{6}{AD} \right)^2$,then the equation of the line is

The lines $2x + 3y = 6$ and $2x + 3y = 8$ cut the $X$-axis at $A$ and $B$,respectively. $A$ line $L$ drawn through the point $(2, 2)$ meets the $X$-axis at $C$ in such a way that the abscissae of $A, B,$ and $C$ are in arithmetic progression. Then,the equation of the line $L$ is

The equation of the straight line making equal intercepts on the axes and passing through the point $(2, 4)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module