The equation of the pair of lines passing thr | vedclass

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

Question

(A) Let $m_1$ and $m_2$ be the slopes of the lines.Then,$m_1+m_2 = 	ext{arithmetic mean} = \frac{4+9}{2} = \frac{13}{2}$ and $m_1 m_2 = 	ext{geometr

Vedclass · Accepted Answer

$12 x^2-13 x y+2 y^2=0$

Vedclass · Answer

(A) Let $m_1$ and $m_2$ be the slopes of the lines.Then,$m_1+m_2 = 	ext{arithmetic mean} = \frac{4+9}{2} = \frac{13}{2}$ and $m_1 m_2 = 	ext{geometric mean} = \sqrt{4 	imes 9} = \sqrt{36} = 6$.Now,the equation of the pair of lines passing through the origin is given by $(y-m_1 x)(y-m_2 x) = 0$.This simplifies to $y^2 - (m_1+m_2)xy + m_1 m_2 x^2 = 0$.Substituting the values of $m_1+m_2$ and $m_1 m_2$,we get $y^2 - \frac{13}{2}xy + 6x^2 = 0$.Multiplying by $2$,we get $2y^2 - 13xy + 12x^2 = 0$,which is $12x^2 - 13xy + 2y^2 = 0$.

The equation of the pair of lines passing through the origin whose sum and product of slopes are respectively the arithmetic mean and geometric mean of $4$ and $9$ is

Similar Questions

If the slope of one of the lines in the pair of lines $8x^2 + axy + y^2 = 0$ is thrice the slope of the other line,then $a =$

The difference of the slopes of the lines represented by the equation ${x^2}(\sec^2 \theta - \sin^2 \theta) - 2xy \tan \theta + y^2 \sin^2 \theta = 0$ is:

The locus of a point which moves such that its distance from the origin is thrice its distance from the line $y = 2x$ is

If the slope of one of the lines represented by $ax^2 + (2a + 1)xy + 2y^2 = 0$ is the reciprocal of the slope of the other,then the sum of the squares of the slopes is

The equation of the pair of lines perpendicular to the lines represented by $x^2-5xy+4y^2=0$ and passing through the point $(2,1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module