The lines joining the origin to the points of | vedclass

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

Question

(B) The given line is $3x - 2y = 1$,which can be written as $3x - 2y = 1$. To make the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ homogeneous,we multiply

Vedclass · Accepted Answer

Perpendicular to each other

Vedclass · Answer

(B) The given line is $3x - 2y = 1$,which can be written as $3x - 2y = 1$. To make the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ homogeneous,we multiply the terms of degree $1$ by $(3x - 2y)$ and the constant term by $(3x - 2y)^2$. $3x^2 + 5xy - 3y^2 + (2x + 3y)(3x - 2y) = 0$. Expanding this,we get $3x^2 + 5xy - 3y^2 + 6x^2 - 4xy + 9xy - 6y^2 = 0$. Combining like terms,we get $9x^2 + 10xy - 9y^2 = 0$. For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the lines are perpendicular if $a + b = 0$. Here,$a = 9$ and $b = -9$. Since $a + b = 9 + (-9) = 0$,the lines are perpendicular to each other.

The lines joining the origin to the points of intersection of the line $3x - 2y = 1$ and the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ are:

Similar Questions

The distance between the lines represented by $16x^2 - 24xy + 9y^2 + 48x - 36y + 35 = 0$ is ...... units.

The equation of the pair of straight lines joining the origin to the points of intersection of the curve $x^2 + y^2 = 4$ and the line $y - x = 2$ is

The lines joining the origin to the points of intersection of the circle $x^2 + y^2 = 3$ and the line $x + y = 2$ are:

If the distance between the parallel lines given by the equation $x^2+4xy+4y^2+3x+6y-4=0$ is $\lambda$,then $\lambda^2=$

Let $L$ be the line joining the origin to the point of intersection of the lines represented by $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. If $L$ is perpendicular to the line $kx + y + 3 = 0$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module