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

The centroid of the triangle formed by the pair of straight lines $12x^2 - 20xy + 7y^2 = 0$ and the line $2x - 3y + 4 = 0$ is $(\alpha, \beta)$. Then,$\alpha + 2\beta =$

The equations $x-y=4$ and $x^2+4xy+y^2=0$ represent the sides of a/an

Suppose the slopes $m_1$ and $m_2$ of the lines represented by $ax^2+2hxy+by^2=0$ satisfy $3(m_1-m_2)-7=0$ and $m_1m_2-2=0$. Then,which of the following is true?

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

Suppose that the sides passing through the vertex $(\alpha, \beta)$ of a triangle are bisected at right angles by the lines $y^2-8xy-9x^2=0$. Then,the centroid of the triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module