The combined equation of the bisectors of the | vedclass

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

Question

(A) The given equation is $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = \sqrt{3}$,

Vedclass · Accepted Answer

${x^2} - {y^2} = 0$

Vedclass · Answer

(A) The given equation is $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = \sqrt{3}$,$h = -2$,and $b = \sqrt{3}$. The equation of the bisectors of the angles between the lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$. Substituting the values,we get $\frac{x^2 - y^2}{\sqrt{3} - \sqrt{3}} = \frac{xy}{-2}$. Since the denominator on the left is $0$,the equation becomes $x^2 - y^2 = 0$.

The combined equation of the bisectors of the angle between the lines represented by $({x^2} + {y^2})\sqrt{3} = 4xy$ is

Similar Questions

If the pair of straight lines $xy-x-y+1=0$ and the line $ax+2y-3a=0$ are concurrent,then $a$ is equal to

If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+\lambda y-8=0$ are concurrent,then the value of $\lambda$ is

If $3x^2-11xy+10y^2-7x+13y+k=0$ denotes a pair of straight lines,then the point of intersection of the lines is

If the pair of straight lines $x^2 - 2mxy - y^2 = 0$ and $x^2 - 2nxy - y^2 = 0$ are such that each pair bisects the angle between the other pair,then $mn =$

The combined equation of the bisectors of the angles between the coordinate axes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module