If the lines x^2-4xy+y^2=0 and x+y=10 contain | vedclass

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

Question

(C) The pair of lines $x^2-4xy+y^2=0$ passes through the origin $(0,0)$. The angle $	heta$ between these lines is given by $	an 	heta = \left|\frac

Vedclass · Accepted Answer

$\frac{50}{\sqrt{3}}$ sq. units

Vedclass · Answer

(C) The pair of lines $x^2-4xy+y^2=0$ passes through the origin $(0,0)$. The angle $	heta$ between these lines is given by $	an 	heta = \left|\frac{2\sqrt{h^2-ab}}{a+b}ight|$.Here $a=1, h=-2, b=1$,so $	an 	heta = \left|\frac{2\sqrt{(-2)^2-1	imes 1}}{1+1}ight| = \sqrt{3}$,which means $	heta = 60^{\circ}$.Since the lines form an equilateral triangle with the line $x+y=10$,the origin is one vertex of the triangle.The perpendicular distance from the origin $(0,0)$ to the line $x+y-10=0$ is the altitude $h$ of the equilateral triangle.$h = \frac{|0+0-10|}{\sqrt{1^2+1^2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.For an equilateral triangle with side length $s$,the altitude $h = \frac{\sqrt{3}}{2}s$,so $s = \frac{2h}{\sqrt{3}} = \frac{2(5\sqrt{2})}{\sqrt{3}} = \frac{10\sqrt{2}}{\sqrt{3}}$.The area of the equilateral triangle is $\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4} 	imes \left(\frac{10\sqrt{2}}{\sqrt{3}}ight)^2 = \frac{\sqrt{3}}{4} 	imes \frac{200}{3} = \frac{50\sqrt{3}}{3} = \frac{50}{\sqrt{3}}$ sq. units.

If the lines $x^2-4xy+y^2=0$ and $x+y=10$ contain the sides of an equilateral triangle,then the area of the equilateral triangle is

Similar Questions

If the equation of the pair of straight lines passing through the point $(1,1)$ and perpendicular to the pair of lines $3x^2+11xy-4y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+12=0$,then $2(a-h+b-g+f-12)=$

The combined equation of the diagonals of the square formed by the two pairs of straight lines given by $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ is

The equations to a pair of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations to its diagonals are

If $ax^2+6xy-2y^2=0$ represents a pair of perpendicular lines and $9x^2+2hxy+4y^2=0$ $(h>0)$ represents a pair of coincident lines,then $h=$

If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2=0, k>0$ is $45^{\circ}$,then the area (in square units) of the triangle formed by the pair of bisectors of angles between the given lines and the line $x+2 y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module