A straight line L equiv 0 passing through the | vedclass

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

Question

(C) Given $A=(-5,-4)$ and the slope of line $L$ is $	an 	heta$. The parametric form of the line is $x = -5 + r \cos 	heta$ and $y = -4 + r \sin 	h

Vedclass · Accepted Answer

$\frac{-1}{2}$

Vedclass · Answer

(C) Given $A=(-5,-4)$ and the slope of line $L$ is $	an 	heta$. The parametric form of the line is $x = -5 + r \cos 	heta$ and $y = -4 + r \sin 	heta$. Point $B$ lies on $x+3y+2=0$,so $(-5+r_1 \cos 	heta) + 3(-4+r_1 \sin 	heta) + 2 = 0$. $-5 + r_1 \cos 	heta - 12 + 3r_1 \sin 	heta + 2 = 0 \implies r_1(\cos 	heta + 3 \sin 	heta) = 15 \implies AB = r_1 = \frac{15}{\cos 	heta + 3 \sin 	heta}$. Point $C$ lies on $2x+y+4=0$,so $2(-5+r_2 \cos 	heta) + (-4+r_2 \sin 	heta) + 4 = 0$. $-10 + 2r_2 \cos 	heta - 4 + r_2 \sin 	heta + 4 = 0 \implies r_2(2 \cos 	heta + \sin 	heta) = 10 \implies AC = r_2 = \frac{10}{2 \cos 	heta + \sin 	heta}$. Substituting into the given equation $\frac{100}{AC^2} - \frac{225}{AB^2} = 4 \cos 2	heta + \sin 2	heta$: $(2 \cos 	heta + \sin 	heta)^2 - (\cos 	heta + 3 \sin 	heta)^2 = 4 \cos 2	heta + \sin 2	heta$. $(4 \cos^2 	heta + \sin^2 	heta + 4 \sin 	heta \cos 	heta) - (\cos^2 	heta + 9 \sin^2 	heta + 6 \sin 	heta \cos 	heta) = 4(\cos^2 	heta - \sin^2 	heta) + 2 \sin 	heta \cos 	heta$. $3 \cos^2 	heta - 8 \sin^2 	heta - 2 \sin 	heta \cos 	heta = 4 \cos^2 	heta - 4 \sin^2 	heta + 2 \sin 	heta \cos 	heta$. $4 \sin^2 	heta + 4 \sin 	heta \cos 	heta + \cos^2 	heta = 0$. $(2 \sin 	heta + \cos 	heta)^2 = 0 \implies 2 \sin 	heta = -\cos 	heta \implies 	an 	heta = -\frac{1}{2}$.

$A$ straight line $L \equiv 0$ passing through the point $A=(-5,-4)$ and having slope $\tan \theta$ meets the lines $x+3y+2=0$ and $2x+y+4=0$ respectively at the points $B$ and $C$. If $\frac{100}{AC^2}-\frac{225}{AB^2}=4 \cos 2\theta+\sin 2\theta$,then the slope of the line $L \equiv 0$ is

Similar Questions

Let $A(-1, 1)$,$B(3, 4)$,and $C(2, 0)$ be three given points. $A$ line $y = mx$,$m > 0$,intersects lines $AC$ and $BC$ at points $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $\Delta ABC$ and $\Delta PQC$ respectively,such that $A_1 = 3A_2$. Then the value of $m$ is equal to:

$A$ straight line is drawn through the point $A(1,2)$ such that its point of intersection with the straight line $x+y=4$ is at a distance $\frac{\sqrt{6}}{3}$ from the given point $A$. Find the angle which the line makes with the positive direction of $X$-axis.

If $k_1 > k_2$ are the two values of $k$ such that the lines $y - 3kx + 4 = 0$ and $(2k - 1)x - (8k - 1)y - 6 = 0$ are perpendicular,then the equation of the line passing through $(k_1, k_2)$ and having the slope $\left(\frac{k_2}{k_1}\right)$ is

If a straight line $L$ perpendicular to the line $3x - 4y = 6$ forms a triangle of area $6$ square units with the coordinate axes,then the minimum perpendicular distance from the point $(1, 1)$ to the line $L$ is

For a point $P(x, y)$ in the plane,let $d_1(P)$ and $d_2(P)$ be the distances of the point $P$ from the lines $x-y=0$ and $x+y=0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \leq d_1(P)+d_2(P) \leq 4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module