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) Given pair of lines: $4xy + 2x + 6y + 3 = 0$. Factorizing the expression: $2x(2y + 1) + 3(2y + 1) = 0$ $(2x + 3)(2y + 1) = 0$. Thus,the individual

Vedclass · Accepted Answer

$xy - x - 2y + 2 = 0$

Vedclass · Answer

(A) Given pair of lines: $4xy + 2x + 6y + 3 = 0$. Factorizing the expression: $2x(2y + 1) + 3(2y + 1) = 0$ $(2x + 3)(2y + 1) = 0$. Thus,the individual lines are $x = -\frac{3}{2}$ (a vertical line) and $y = -\frac{1}{2}$ (a horizontal line). $A$ line perpendicular to $x = -\frac{3}{2}$ is a horizontal line of the form $y = k$. Since it passes through $(2,1)$,$y = 1$ or $y - 1 = 0$. $A$ line perpendicular to $y = -\frac{1}{2}$ is a vertical line of the form $x = h$. Since it passes through $(2,1)$,$x = 2$ or $x - 2 = 0$. The combined equation of these two lines is $(y - 1)(x - 2) = 0$. Expanding this,we get $xy - 2y - x + 2 = 0$,which is $xy - x - 2y + 2 = 0$.

The equation of the pair of lines passing through the point $(2,1)$ and perpendicular to the pair of lines $4xy + 2x + 6y + 3 = 0$ is:

Similar Questions

The length of the $x$-intercept made by the pair of lines $2x^2 + xy - 6y^2 - 2x + 17y - 12 = 0$ is

The joint equation of the pair of lines passing through $(3, -2)$ and parallel to the lines represented by $x^{2} - 4xy + 3y^{2} = 0$ is:

The equation $x^2 - 7xy + 12y^2 = 0$ represents

If $x^2-y^2+2hxy+2gx+2fy+c=0$ is the locus of a point which moves such that it is always equidistant from the lines $x+2y+7=0$ and $2x-y+8=0$,then the value of $g+c+h-f$ equals

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module