Let two straight lines drawn from the origin | vedclass

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

Question

(C) Given the line $3x + 4y = 12$. Let the coordinates of $P$ be $(r \cos 	heta, r \sin 	heta)$ and $Q$ be $(r \cos(90^{\circ} + 	heta), r \sin(90^

Vedclass · Accepted Answer

$46$

Vedclass · Answer

(C) Given the line $3x + 4y = 12$. Let the coordinates of $P$ be $(r \cos 	heta, r \sin 	heta)$ and $Q$ be $(r \cos(90^{\circ} + 	heta), r \sin(90^{\circ} + 	heta)) = (-r \sin 	heta, r \cos 	heta)$ since $	riangle OPQ$ is isosceles with $OP = OQ = r$ and $\angle POQ = 90^{\circ}$.Since $P$ and $Q$ lie on the line $3x + 4y = 12$:For $P$: $3(r \cos 	heta) + 4(r \sin 	heta) = 12 \Rightarrow r(3 \cos 	heta + 4 \sin 	heta) = 12 \ldots(1)$For $Q$: $3(-r \sin 	heta) + 4(r \cos 	heta) = 12 \Rightarrow r(4 \cos 	heta - 3 \sin 	heta) = 12 \ldots(2)$Squaring and adding $(1)$ and $(2)$:$r^2(3 \cos 	heta + 4 \sin 	heta)^2 + r^2(4 \cos 	heta - 3 \sin 	heta)^2 = 12^2 + 12^2$$r^2(9 \cos^2 	heta + 16 \sin^2 	heta + 24 \sin 	heta \cos 	heta + 16 \cos^2 	heta + 9 \sin^2 	heta - 24 \sin 	heta \cos 	heta) = 288$$r^2(25 \cos^2 	heta + 25 \sin^2 	heta) = 288$ $\Rightarrow 25r^2 = 288$ $\Rightarrow r^2 = \frac{288}{25}$.In $	riangle OPQ$,$PQ^2 = OP^2 + OQ^2 = r^2 + r^2 = 2r^2$.Then $l = OP^2 + PQ^2 + QO^2 = r^2 + 2r^2 + r^2 = 4r^2$.$l = 4 	imes \frac{288}{25} = \frac{1152}{25} = 46.08$.The greatest integer less than or equal to $l$ is $\lfloor 46.08 floor = 46$.

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^{\circ}$. If $l = OP^2 + PQ^2 + QO^2$,then the greatest integer less than or equal to $l$ is:

Similar Questions

If the intercept of a straight line $L$ made between the straight lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$,then the equation of $L$ is

Let $L$ be the line passing through the point $P(1, 2)$ such that its intercepted segment between the coordinate axes is bisected at $P$. If $L_1$ is the line perpendicular to $L$ and passing through the point $(-2, 1)$,then the point of intersection of $L$ and $L_1$ is

The line $2x + 3y = 12$ meets the $x$-axis at $A$ and $y$-axis at $B$. The line through $(5, 5)$ perpendicular to $AB$ meets the $x$-axis,$y$-axis,and the $AB$ at $C, D,$ and $E$ respectively. If $O$ is the origin of coordinates,then the area of $OCEB$ is

Suppose $P$ and $Q$ lie on $3x + 4y - 4 = 0$ and $5x - y - 4 = 0$ respectively. If the mid-point of $PQ$ is $(1, 5)$,then the slope of the line passing through $P$ and $Q$ is

If $l, m, n$ are in arithmetic progression,then the straight line $lx + my + n = 0$ will always pass through the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module