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

The joint equation of two lines passing through the origin,each of which makes an angle of $30^{\circ}$ with the line $x+y=0$,is

The nature of the straight lines represented by the equation $4x^2 + 12xy + 9y^2 = 0$ is

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:

If the products of the perpendiculars from the origin to the pairs of lines $xy+x+y+1=0$,$x^2-y^2+2x+1=0$,and $2x^2+3xy-2y^2+2x+1=0$ are $p_1, p_2$,and $p_3$ respectively,then:

Let $PQR$ be a right-angled isosceles triangle,with the right angle at $Q(2, 1)$. If the equation of the line $PR$ is $2x + y = 3$,then the combined equation representing the pair of lines $PQ$ and $QR$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module