A is a point on either of two lines y + sqrt{ | vedclass

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

Question

(D) The given equation is $y + \sqrt{3} |x| = 2$. This represents two lines: $L_1: y + \sqrt{3}x = 2$ (for $x \ge 0$) and $L_2: y - \sqrt{3}x = 2$ (fo

Vedclass · Accepted Answer

$(0, 4)$

Vedclass · Answer

(D) The given equation is $y + \sqrt{3} |x| = 2$. This represents two lines: $L_1: y + \sqrt{3}x = 2$ (for $x \ge 0$) and $L_2: y - \sqrt{3}x = 2$ (for $x Both lines intersect at the point $P(0, 2)$.The angle bisectors of these lines are the $y$-axis $(x=0)$ and the line $y=2$.The point $A$ lies on the lines at a distance of $d = \frac{4}{\sqrt{3}}$ from $P(0, 2)$.For $L_1$,the slope is $m = -\sqrt{3}$,so the angle with the $x$-axis is $120^\circ$. The coordinates of $A$ are $(0 + d \cos 120^\circ, 2 + d \sin 120^\circ) = \left( \frac{4}{\sqrt{3}} \cdot \left( -\frac{1}{2} \right), 2 + \frac{4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \right) = \left( -\frac{2}{\sqrt{3}}, 4 \right)$.However,checking the geometry,the foot of the perpendicular from $A$ to the angle bisector $x=0$ is $(0, 4)$.

$A$ is a point on either of two lines $y + \sqrt{3} |x| = 2$ at a distance of $\frac{4}{\sqrt{3}}$ units from their point of intersection. The coordinates of the foot of the perpendicular from $A$ on the bisector of the angle between them are

Similar Questions

The area of the triangle formed by the coordinate axes and the line $px + qy = r$ (where $p, q$ and $r$ are positive real numbers) is $\frac{r^2}{2pq}$. If this area is $\frac{1}{54}$ sq. units, then which of the following is true?

$A$ straight line through the origin $O(0, 0)$ meets the lines $4x + 3y - 10 = 0$ and $8x + 6y + 5 = 0$ at points $A$ and $B$ respectively. Then $O$ divides the segment $AB$ in the ratio:

If ${x_1}, {x_2}, {x_3}$ and ${y_1}, {y_2}, {y_3}$ are both in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations to the sides $AB$ and $AC$ are respectively $px + qy = 1$ and $qx + py = 1$. Then the equation to the median through $A$ is

$A$ pair of straight lines drawn through the origin form an isosceles right-angled triangle with the line $2x + 3y = 6$. Find the equations of the lines and the area of the triangle thus formed.

Vedclass Test Series

Exam Paper Generator

Online Exam Module