The area (in square units) of the triangle fo | vedclass

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

Question

(D) The given pair of lines is $6 x^2+13 x y+6 y^2=0$. Factoring the quadratic expression: $6 x^2+9 x y+4 x y+6 y^2=0 \implies 3 x(2 x+3 y)+2 y(2 x+3

Vedclass · Accepted Answer

$\frac{45}{8}$

Vedclass · Answer

(D) The given pair of lines is $6 x^2+13 x y+6 y^2=0$. Factoring the quadratic expression: $6 x^2+9 x y+4 x y+6 y^2=0 \implies 3 x(2 x+3 y)+2 y(2 x+3 y)=0 \implies (3 x+2 y)(2 x+3 y)=0$. So,the two lines are $L_1: 3 x+2 y=0$ and $L_2: 2 x+3 y=0$. The third line is $L_3: x+2 y+3=0$. To find the vertices,we solve the systems of equations: $1$) $L_1$ and $L_2$: $(0, 0)$. $2$) $L_1$ and $L_3$: $3 x+2 y=0$ and $x+2 y=-3$. Subtracting gives $2 x=3 \implies x=\frac{3}{2}$. Then $2 y=-3 x=-\frac{9}{2} \implies y=-\frac{9}{4}$. Vertex: $(\frac{3}{2}, -\frac{9}{4})$. $3$) $L_2$ and $L_3$: $2 x+3 y=0$ and $x+2 y=-3$. From $x=-3-2 y$,substitute into $2(-3-2 y)+3 y=0 \implies -6-4 y+3 y=0 \implies y=-6$. Then $x=-3-2(-6)=9$. Vertex: $(9, -6)$. Using the area formula for vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$: $	ext{Area} = \frac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$. $	ext{Area} = \frac{1}{2} |0(-\frac{9}{4}-(-6)) + \frac{3}{2}(-6-0) + 9(0-(-\frac{9}{4}))| = \frac{1}{2} |0 - 9 + \frac{81}{4}| = \frac{1}{2} |\frac{-36+81}{4}| = \frac{1}{2} 	imes \frac{45}{4} = \frac{45}{8}$ square units.

The area (in square units) of the triangle formed by the lines $6 x^2+13 x y+6 y^2=0$ and $x+2 y+3=0$ is

Similar Questions

If $\frac{x^2}{a} + \frac{y^2}{b} + \frac{2xy}{h} = 0$ represents a pair of straight lines and the slope of one line is twice the other,then $ab : h^2$ is

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 $(h, c) =$

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

If the area of the triangle formed by the lines $y=x+c$ and $2x^2+5xy+3y^2=0$ is $\frac{1}{20}$ sq. units,then $c=$

If the lines $ax^2 + 2hxy + by^2 = 0$ represent the adjacent sides of a parallelogram,then the equation of the second diagonal,if one diagonal is $lx + my = 1$,will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module