Let B and C be two points on line y+x=0 such that B and C are symmetric with respect to origin. Supp

English

Gujarati

Hindi

9.Straight Line

hard

Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin. Suppose $A$ is a point on $y -2 x =2$ such that $\triangle ABC$ is an equilateral triangle. Then, the area of the $\triangle ABC$ is

A

$3 \sqrt{3}$

B

$2 \sqrt{3}$

C

$\frac{8}{\sqrt{3}}$

D

$\frac{10}{\sqrt{3}}$

(JEE MAIN-2023)

Solution

At A $x=y$

$Y-2 x=2$

$(-2,-2)$

Height from line $x + y =0$

$h=\frac{4}{\sqrt{2}}$

Area of $\Delta=\frac{\sqrt{3}}{4} \frac{ h ^2}{\sin ^2 60}=\frac{8}{\sqrt{3}}$

Standard 11

Mathematics

Share

Similar Questions

$ABC$ is an isosceles triangle . If the co-ordinates of the base are $(1, 3)$ and $(- 2, 7) $, then co-ordinates of vertex $A$ can be :

normal

The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

hard

(JEE MAIN-2024)

In a rectangle $A B C D$, the coordinates of $A$ and $B$ are $(1,2)$ and $(3,6)$ respectively and some diameter of the circumscribing circle of $A B C D$ has equation $2 x-y+4=0$. Then, the area of the rectangle is

normal

(KVPY-2011)

The locus of a point so that sum of its distance from two given perpendicular lines is equal to $2$ unit in first quadrant, is

easy

A variable straight line passes through the points of intersection of the lines, $x + 2y = 1$ and $2x – y = 1$ and meets the co-ordinate axes in $A\,\, \&\,\, B$ . The locus of the middle point of $AB$ is :

normal