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 the pair of lines: $4xy + 2x + 6y + 3 = 0$. Factoring the expression: $2x(2y + 1) + 3(2y + 1) = 0$,which gives $(2x + 3)(2y + 1) = 0$. Thus,

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the pair of lines: $4xy + 2x + 6y + 3 = 0$. Factoring the expression: $2x(2y + 1) + 3(2y + 1) = 0$,which gives $(2x + 3)(2y + 1) = 0$. Thus,the 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,and a line perpendicular to $y = -\frac{1}{2}$ is a vertical line. The line passing through $(2, 1)$ perpendicular to $x = -\frac{3}{2}$ is $y = 1$. The line passing through $(2, 1)$ perpendicular to $y = -\frac{1}{2}$ is $x = 2$. Combining these equations: $(y - 1)(x - 2) = 0$. Expanding this gives $xy - 2y - x + 2 = 0$,or $xy - x - 2y + 2 = 0$.

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

If $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$ are two tangents to a circle,then the radius of the circle is

$A$ particle of mass $M$ is moving in a horizontal circle of radius $R$ with uniform speed $v$. When it moves from one point to a diametrically opposite point,its

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi/2$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

If $(1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}$,then $\frac{a_2}{a_1}$ is equal to

Which of the following compounds forms silicones on hydrolysis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module