If the lines x^2+kxy+y^2=0 and x+y=1 form the | vedclass

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

Question

(B) The equation $x^2+kxy+y^2=0$ represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m_1$ and $m_2$. The

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The equation $x^2+kxy+y^2=0$ represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m_1$ and $m_2$. The equation of the third line is $x+y=1$,which can be written as $y=-x+1$,so its slope is $m_3=-1$.Since the lines form an equilateral triangle,the angle between any two lines is $60^{\circ}$.The angle between the line $y=mx$ and the line $x+y=1$ (slope $-1$) is given by:$	an 60^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} ight|$$\sqrt{3} = \left| \frac{m+1}{1-m} ight|$Squaring both sides:$3 = \frac{(m+1)^2}{(1-m)^2}$$3(1-2m+m^2) = m^2+2m+1$$3-6m+3m^2 = m^2+2m+1$$2m^2-8m+2 = 0$$m^2-4m+1 = 0$Since $m = y/x$,we substitute this into the quadratic equation:$(y/x)^2 - 4(y/x) + 1 = 0$$y^2 - 4xy + x^2 = 0$Comparing this with $x^2+kxy+y^2=0$,we get $k=-4$.Therefore,$k^2 = (-4)^2 = 16$.

If the lines $x^2+kxy+y^2=0$ and $x+y=1$ form the sides of an equilateral triangle,then the value of $k^2$ is

Similar Questions

The angle between the lines represented by the equation $x^2 + 2xy \sec \theta + y^2 = 0$ is

The lines represented by $ax^2 + 2hxy + by^2 = 0$ are perpendicular to each other,if ........

The equation $x^2-3xy+\lambda y^2+3x-5y+2=0$,where $\lambda$ is a real number,represents a pair of lines. If $\theta$ is the acute angle between the lines,then $\frac{\operatorname{cosec}^2 \theta}{\sqrt{10}} = $

If $ax^2 + 6xy + by^2 - 10x + 10y - 6 = 0$ represents a pair of perpendicular lines,then $|a| =$

Which of the following pairs of straight lines intersect at right angles?

Vedclass Test Series

Exam Paper Generator

Online Exam Module