The equation of a pair of lines is given by a | vedclass

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

Question

(B) Let the two lines passing through the origin be $L_1$ and $L_2$.Line $L_1$ is perpendicular to $x+2y+7=0$. The slope of $x+2y+7=0$ is $m = -1/2$.

Vedclass · Accepted Answer

$6x^2+5xy-4y^2=0$

Vedclass · Answer

(B) Let the two lines passing through the origin be $L_1$ and $L_2$.Line $L_1$ is perpendicular to $x+2y+7=0$. The slope of $x+2y+7=0$ is $m = -1/2$. Thus,the slope of $L_1$ is $m_1 = 2$. The equation of $L_1$ is $y = 2x$,or $2x-y=0$.Line $L_2$ is parallel to $3x+4y+5=0$. The slope of $3x+4y+5=0$ is $m = -3/4$. Thus,the slope of $L_2$ is $m_2 = -3/4$. The equation of $L_2$ is $y = -3/4x$,or $3x+4y=0$.The joint equation of the pair of lines is $(2x-y)(3x+4y) = 0$.Expanding this,we get $6x^2 + 8xy - 3xy - 4y^2 = 0$,which simplifies to $6x^2 + 5xy - 4y^2 = 0$.

The equation of a pair of lines is given by a second-degree homogeneous equation. If one of them is perpendicular to the line $x+2y+7=0$ and the other is parallel to the line $3x+4y+5=0$,then the equation of that pair of lines is:

Similar Questions

If the lines $x^2-4xy+y^2=0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis,then $\cot^2 \alpha + \cot^2 \beta = $

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:

For $a, b, h > 0$,if the slope of one of the lines represented by $a^2 x^2 + 2hxy + b^2 y^2 = 0$ is twice that of the other,then the value of $\frac{h}{ab}$ is

The lines represented by the equation $x^2-y^2-x+3y-2=0$ are :

The equation $(x - 5)^2 + (x - 5)(y - 6) - 2(y - 6)^2 = 0$ represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module