P is a point on either of the two lines y - s | vedclass

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

Question

(B) The equation of the lines is $y = \sqrt{3}|x| + 2$. This represents two lines: $y = \sqrt{3}x + 2$ for $x \ge 0$ and $y = -\sqrt{3}x + 2$ for $x <

Vedclass · Accepted Answer

$\left(0, \frac{4 + 5\sqrt{3}}{2}\right)$

Vedclass · Answer

(B) The equation of the lines is $y = \sqrt{3}|x| + 2$. This represents two lines: $y = \sqrt{3}x + 2$ for $x \ge 0$ and $y = -\sqrt{3}x + 2$ for $x The point of intersection $A$ is $(0, 2)$. The angle between the lines and the $y$-axis (which is the angle bisector) can be found from the slope. For $y = \sqrt{3}x + 2$,the slope is $\sqrt{3}$,so the angle with the $x$-axis is $60^\circ$. The angle with the $y$-axis is $90^\circ - 60^\circ = 30^\circ$. Let $P$ be a point on the line at a distance $AP = 5$. The foot of the perpendicular $M$ from $P$ onto the $y$-axis lies on the $y$-axis. The distance $AM = AP \cos(30^\circ) = 5 	imes \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$. Since $A$ is at $(0, 2)$ and the lines extend upwards,the $y$-coordinate of $M$ is $2 + \frac{5\sqrt{3}}{2} = \frac{4 + 5\sqrt{3}}{2}$. Thus,the coordinates of $M$ are $\left(0, \frac{4 + 5\sqrt{3}}{2}ight)$.

$P$ is a point on either of the two lines $y - \sqrt{3}|x| = 2$ at a distance of $5 \ units$ from their point of intersection. The coordinates of the foot of the perpendicular from $P$ on the bisector of the angle between them are

Similar Questions

$A$ straight line passing through $P(3, 1)$ meets the coordinate axes at $A$ and $B$. It is given that the distance of this straight line from the origin $O$ is maximum. The area of triangle $OAB$ is equal to

$A$ straight line which makes equal intercepts on positive $X$ and $Y$ axes and which is at a distance $1$ unit from the origin intersects the straight line $y=2x+3+\sqrt{2}$ at $(x_0, y_0)$. Then $2x_0+y_0$ 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.

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:

The point on the line $3x + y + 4 = 0$ which is equidistant from $(-5, 6)$ and $(3, 2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module