The lines represented by the equation a{x^2}( | vedclass

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

Question

(C) The given equation is $a(b - c)x^2 - xy(a(b - c) + c(a - b)) + c(a - b)y^2 = 0$.This is a quadratic equation in $x$ and $y$ of the form $Ax^2 + Bx

Vedclass · Accepted Answer

$a(b - c)x - c(a - b)y = 0$,$x - y = 0$

Vedclass · Answer

(C) The given equation is $a(b - c)x^2 - xy(a(b - c) + c(a - b)) + c(a - b)y^2 = 0$.This is a quadratic equation in $x$ and $y$ of the form $Ax^2 + Bxy + Cy^2 = 0$.We can factorize this as $(a(b - c)x - c(a - b)y)(x - y) = 0$.Expanding this: $a(b - c)x^2 - a(b - c)xy - c(a - b)xy + c(a - b)y^2 = 0$.$a(b - c)x^2 - xy(a(b - c) + c(a - b)) + c(a - b)y^2 = 0$.Thus,the lines are $a(b - c)x - c(a - b)y = 0$ and $x - y = 0$.

The lines represented by the equation $a{x^2}(b - c) - xy(a(b - c) + c(a - b)) + c{y^2}(a - b) = 0$ are

Similar Questions

If one of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is coincident with one of the lines represented by $a'x^2 + 2h'xy + b'y^2 = 0$,then:

If $L{x^2} - 10xy + 12{y^2} + 5x - 16y - 3 = 0$ represents a pair of straight lines,then the value of $L$ is:

If the slopes of the lines represented by $K x^2 + 5 x y + y^2 = 0$ differ by $1$,then $K =$

If the equation $ax^{2}+2hxy+by^{2}+2gx+2fy=0$ has one line as the bisector of the angle between the coordinate axes,then

Find the equation of the pair of lines passing through the origin and perpendicular to the pair of lines given by $5x^2 - 7xy - 3y^2 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module