The pairs of straight lines x^2-3xy+2y^2=0 an | vedclass

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

Question

(C) The given pairs of lines are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$.Factorizing the first equation: $(x-2y)(x-y)=0$,which gives lines $L_1: x-2

Vedclass · Accepted Answer

parallelogram

Vedclass · Answer

(C) The given pairs of lines are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$.Factorizing the first equation: $(x-2y)(x-y)=0$,which gives lines $L_1: x-2y=0$ and $L_2: x-y=0$.Factorizing the second equation: $(x-2y+2)(x-y-1)=0$,which gives lines $L_3: x-2y+2=0$ and $L_4: x-y-1=0$.Comparing the equations,we see that $L_1 \parallel L_3$ and $L_2 \parallel L_4$.Since the opposite sides are parallel,the figure is a parallelogram.To check if it is a rectangle,we find the angle between $L_1$ and $L_2$. The slopes are $m_1 = 1/2$ and $m_2 = 1$. Since $m_1 	imes m_2 
eq -1$,the angle is not $90^{\circ}$.Thus,the figure is a parallelogram.

The pairs of straight lines $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$ form a

Similar Questions

The lines $(lx + my)^2 - 3(mx - ly)^2 = 0$ and $lx + my + n = 0$ form

If $\left(\frac{2}{3}, 0\right)$ is the centroid of the triangle formed by the lines $4x^2-y^2=0$ and $lx+my+n=0$,then $l+m+n=$

If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincides with one of those given by the equation $2x^2 + bxy - 3y^2 = 0$,while the other two lines are perpendicular to each other,then the values of $a$ and $b$ are

If the equations of the opposite sides of a parallelogram are $x^2 - 7x + 6 = 0$ and $y^2 - 14y + 40 = 0$,then the equation of one of its diagonals is

The distance from the origin to the orthocentre of the triangle formed by the lines $x+y-1=0$ and $6x^2-13xy+5y^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module