The equation of the bisectors of the angle be | vedclass

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

Question

(A) The given equation is $(y - mx)^2 = (x + my)^2$.Expanding both sides,we get $y^2 + m^2x^2 - 2mxy = x^2 + m^2y^2 + 2mxy$.Rearranging the terms,we g

Vedclass · Accepted Answer

$mx^2 + (m^2 - 1)xy - my^2 = 0$

Vedclass · Answer

(A) The given equation is $(y - mx)^2 = (x + my)^2$.Expanding both sides,we get $y^2 + m^2x^2 - 2mxy = x^2 + m^2y^2 + 2mxy$.Rearranging the terms,we get $(m^2 - 1)x^2 - 4mxy + (1 - m^2)y^2 = 0$.This is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = m^2 - 1$,$H = -2m$,and $B = 1 - m^2$.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 $\frac{x^2 - y^2}{(m^2 - 1) - (1 - m^2)} = \frac{xy}{-2m}$.$\frac{x^2 - y^2}{2(m^2 - 1)} = \frac{xy}{-2m}$.$-m(x^2 - y^2) = (m^2 - 1)xy$.$-mx^2 + my^2 = (m^2 - 1)xy$.$mx^2 + (m^2 - 1)xy - my^2 = 0$.

The equation of the bisectors of the angle between the lines represented by the equation $(y - mx)^2 = (x + my)^2$ is

Similar Questions

Find the point of intersection of the lines represented by $2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$.

If $ax^2+2hxy-2ay^2+3x+15y-9=0$ represents a pair of lines intersecting at $(1,1)$,then $ah=$

If the line $y = mx$ bisects the angle between the lines $ax^2 + 2hxy + by^2 = 0$,then $m$ is a root of the quadratic equation:

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

The point of intersection of the pair of lines $x^2+xy+2y^2-3x+2y+4=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module