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 the slope of one of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$ is $\lambda$ times that of the other,then:

The three sides of a triangle are given by $(x^2+7xy+2y^2)(y-1)=0$. Then the centroid of that triangle is

The joint equation of the pair of lines passing through the origin and forming an equilateral triangle with the line $y=3$ is

If $ad \neq 0$ and two of the lines represented by $ax^3+3bx^2y+3cxy^2+dy^3=0$ are perpendicular,then

If one of the lines of the pair of straight lines $ax^2 + 2hxy + by^2 = 0$ bisects the angle between the coordinate axes,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module