If L is a line passing through the point (-1, | vedclass

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

Question

(C) First,we factorize the given pairs of lines to find the common line.\$6x^2 - xy - 12y^2 = 6x^2 - 9xy + 8xy - 12y^2 = 3x(2x - 3y) + 4y(2x - 3y) = (

Vedclass · Accepted Answer

$x^2 - y^2 = 0$

Vedclass · Answer

(C) First,we factorize the given pairs of lines to find the common line.\$6x^2 - xy - 12y^2 = 6x^2 - 9xy + 8xy - 12y^2 = 3x(2x - 3y) + 4y(2x - 3y) = (3x + 4y)(2x - 3y) = 0$.\$15x^2 + 14xy - 8y^2 = 15x^2 + 20xy - 6xy - 8y^2 = 5x(3x + 4y) - 2y(3x + 4y) = (5x - 2y)(3x + 4y) = 0$.\The common line is $3x + 4y = 0$.\Line $L$ passes through $(-1, 1)$ and is parallel to $3x + 4y = 0$,so its equation is $3x + 4y + k = 0$. Substituting $(-1, 1)$,we get $3(-1) + 4(1) + k = 0$ $\Rightarrow -3 + 4 + k = 0$ $\Rightarrow k = -1$. Thus,$L: 3x + 4y - 1 = 0$,or $3x + 4y = 1$.\To find the pair of lines joining the origin to the intersection of $2x^2 - xy - y^2 + x - y = 0$ and $3x + 4y = 1$,we homogenize the curve equation using $1 = 3x + 4y$:\$2x^2 - xy - y^2 + (x - y)(3x + 4y) = 0$.\Expanding this: $2x^2 - xy - y^2 + 3x^2 + 4xy - 3xy - 4y^2 = 0$.\Combining like terms: $(2 + 3)x^2 + (-1 + 4 - 3)xy + (-1 - 4)y^2 = 0$.\$5x^2 - 5y^2 = 0 \Rightarrow x^2 - y^2 = 0$.

Similar Questions

$A$ pair of perpendicular straight lines passes through the origin and also through the point of intersection of the curve $x^2+y^2=4$ with $x+y=a$. The set containing the value of $a$ is

The lines joining the origin to the points of intersection of the curves $ax^2 + 2hxy + by^2 + 2gx = 0$ and $a'x^2 + 2h'xy + b'y^2 + 2g'x = 0$ will be mutually perpendicular,if

The pair of lines joining the origin to the points of intersection of the line $x-y=2$ with the curve $5x^2+12xy-8y^2+8x-4y+12=0$ are equally inclined to the pair of lines

The combined equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+y^2+xy+x+3y+1=0$ and the line $x+y+2=0$ is

The acute angle formed between the lines joining the origin to the points of intersection of the curves $x^2 + y^2 - 2x - 1 = 0$ and $x + y = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module