2 x^2-3 x y-2 y^2=0 represents two lines L_1 | vedclass

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

Question

(A) Given the equation $2 x^2-3 x y-2 y^2=0$,we factorize it as $(2 x+y)(x-2 y)=0$. Let $L_1: 2 x+y=0$ and $L_2: x-2 y=0$. Given the second equation $

Vedclass · Accepted Answer

$\frac{3}{10}$

Vedclass · Answer

(A) Given the equation $2 x^2-3 x y-2 y^2=0$,we factorize it as $(2 x+y)(x-2 y)=0$. Let $L_1: 2 x+y=0$ and $L_2: x-2 y=0$. Given the second equation $2 x^2-3 x y-2 y^2-x+7 y-3=0$,we factorize it as $(2 x+y-1)(x-2 y+3)=0$. Let $L_3: x-2 y+3=0$ and $L_4: 2 x+y-1=0$. Solving $L_1$ and $L_3$: $2 x+y=0$ and $x-2 y=-3$. Multiplying the first by $2$ gives $4 x+2 y=0$. Adding to the second gives $5 x=-3$,so $x=-\frac{3}{5}$ and $y=\frac{6}{5}$. Thus,$A = \left(-\frac{3}{5}, \frac{6}{5}ight)$. Solving $L_2$ and $L_4$: $x-2 y=0$ and $2 x+y=1$. Multiplying the second by $2$ gives $4 x+2 y=2$. Adding to the first gives $5 x=2$,so $x=\frac{2}{5}$ and $y=\frac{1}{5}$. Thus,$B = \left(\frac{2}{5}, \frac{1}{5}ight)$. Solving $L_3$ and $L_4$: $x-2 y=-3$ and $2 x+y=1$. Multiplying the second by $2$ gives $4 x+2 y=2$. Adding to the first gives $5 x=-1$,so $x=-\frac{1}{5}$ and $y=\frac{7}{5}$. Thus,$C = \left(-\frac{1}{5}, \frac{7}{5}ight)$. Now,the area of the triangle formed by $A, B, C$ is $\frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|$. Area $= \frac{1}{2} |-\frac{3}{5}(\frac{1}{5}-\frac{7}{5}) + \frac{2}{5}(\frac{7}{5}-\frac{6}{5}) - \frac{1}{5}(\frac{6}{5}-\frac{1}{5})| = \frac{1}{2} |-\frac{3}{5}(-\frac{6}{5}) + \frac{2}{5}(\frac{1}{5}) - \frac{1}{5}(\frac{5}{5})| = \frac{1}{2} |\frac{18}{25} + \frac{2}{25} - \frac{5}{25}| = \frac{1}{2} |\frac{15}{25}| = \frac{1}{2} 	imes \frac{3}{5} = \frac{3}{10}$.

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 pairs of straight lines represented by $3x^2+2hxy-3y^2=0$ and $3x^2+2hxy-3y^2+2x-4y+c=0$ form a square,then $(h, c) =$

Suppose a triangle of area $27$ sq. units is formed by $18x^2 - 9xy + y^2 = 0$ and $y = c$. Then the centroid of the triangle is

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

If the combined equation of the diagonals of the square formed by the pairs of lines $xy+4x-5y-20=0$ and $xy-5x+4y-20=0$ is $x^2-y^2-kx+ly=0$,then $k+l=$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module