The two pairs of straight lines 12x^2+7xy-12y | vedclass

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

Question

(A) Factorizing $12x^2+7xy-12y^2=0$:$12x^2+16xy-9xy-12y^2=0$$4x(3x+4y)-3y(3x+4y)=0$$(4x-3y)(3x+4y)=0$So,the lines are $4x-3y=0$ and $3x+4y=0$. Since t

Vedclass · Accepted Answer

area of square $\frac{1}{25}$ sq units

Vedclass · Answer

(A) Factorizing $12x^2+7xy-12y^2=0$:$12x^2+16xy-9xy-12y^2=0$$4x(3x+4y)-3y(3x+4y)=0$$(4x-3y)(3x+4y)=0$So,the lines are $4x-3y=0$ and $3x+4y=0$. Since their slopes are $4/3$ and $-3/4$,they are perpendicular.Factorizing $12x^2+7xy-12y^2-x+7y-1=0$:Let the equation be $(4x-3y+c_1)(3x+4y+c_2)=0$.Expanding this: $12x^2+7xy-12y^2 + (4c_2+3c_1)x + (4c_1-3c_2)y + c_1c_2 = 0$.Comparing with $12x^2+7xy-12y^2-x+7y-1=0$:$4c_2+3c_1 = -1$$4c_1-3c_2 = 7$Solving these,we get $c_1=1$ and $c_2=-1$.So the lines are $4x-3y+1=0$ and $3x+4y-1=0$.The distance between parallel lines $4x-3y=0$ and $4x-3y+1=0$ is $d_1 = \frac{|1-0|}{\sqrt{4^2+(-3)^2}} = \frac{1}{5}$.The distance between parallel lines $3x+4y=0$ and $3x+4y-1=0$ is $d_2 = \frac{|-1-0|}{\sqrt{3^2+4^2}} = \frac{1}{5}$.Since the lines are perpendicular and the distance between parallel pairs is equal,the figure is a square with side length $1/5$.Area $= (1/5)^2 = 1/25$ sq units.

The two pairs of straight lines $12x^2+7xy-12y^2=0$ and $12x^2+7xy-12y^2-x+7y-1=0$ constitute a

Similar Questions

The line $x+y=1$ meets the lines represented by the equation $y^3-6xy^2+11x^2y-6x^3=0$ at the points $P, Q, R$. If $O$ is the origin,then $(OP)^2+(OQ)^2+(OR)^2=$

If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2=0, k>0$ is $45^{\circ}$,then the area (in square units) of the triangle formed by the pair of bisectors of angles between the given lines and the line $x+2 y+1=0$ is

The area of the region enclosed between the pair of lines $xy = 0$ and the lines $xy + 5x - 4y - 20 = 0$ is .....

The centroid of the triangle formed by the lines $x+3y=10$ and $6x^2+xy-y^2=0$ is

The triangle formed by the lines $2x^2+xy-6y^2=0$ and $x+y-1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module