If lx^2+3xy-2y^2-5x+5y+k=0 represents a pair | vedclass

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

Question

(C) The general equation of a pair of lines is $ax^2+2hxy+by^2+2gx+2fy+c=0$. For the lines to be perpendicular,the sum of the coefficients of $x^2$ an

Vedclass · Accepted Answer

$k=-3, l=2$

Vedclass · Answer

(C) The general equation of a pair of lines is $ax^2+2hxy+by^2+2gx+2fy+c=0$. For the lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a+b=0$. Given $l+ (-2) = 0$,we get $l=2$. Now the equation becomes $2x^2+3xy-2y^2-5x+5y+k=0$. For this to represent a pair of lines,the condition $abc+2fgh-af^2-bg^2-ch^2=0$ must be satisfied. Here $a=2, b=-2, c=k, h=\frac{3}{2}, g=-\frac{5}{2}, f=\frac{5}{2}$. Substituting these values: $(2)(-2)(k) + 2(\frac{5}{2})(-\frac{5}{2})(\frac{3}{2}) - 2(\frac{5}{2})^2 - (-2)(-\frac{5}{2})^2 - k(\frac{3}{2})^2 = 0$. $-4k - \frac{75}{4} - \frac{50}{4} + \frac{50}{4} - \frac{9k}{4} = 0$. $-4k - \frac{9k}{4} - \frac{75}{4} = 0$. $-\frac{25k}{4} = \frac{75}{4}$. $k = -3$.

If $lx^2+3xy-2y^2-5x+5y+k=0$ represents a pair of perpendicular lines,then

Similar Questions

If $p_1$ and $p_2$ denote the lengths of perpendiculars from $(2,3)$ onto the lines given by $15 x^2+31 x y+14 y^2=0$,and if $p_1 > p_2$,then $p_1^2 + \frac{1}{74} - p_2^2 + \frac{1}{13}$ is equal to

The joint equation of two lines passing through the origin,each making an angle of $30^{\circ}$ with the positive $Y$-axis,is

The values of $h$ for which the equation $3x^2 + 2hxy - 3y^2 - 40x + 30y - 75 = 0$ represents a pair of straight lines are:

If one of the lines $ax^2 + 2hxy + by^2 = 0$ bisects the angle between the positive coordinate axes,then

The joint equation of the pair of lines passing through the origin with slopes $(1+\sqrt{2})$ and $\frac{1}{(1+\sqrt{2})}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module