Area of the triangle bounded by the lines giv | vedclass

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

Question

(D) The given pair of lines is $12 x^2-20 x y+7 y^2=0$.Factoring the quadratic expression: $12 x^2-14 x y-6 x y+7 y^2=0$.This simplifies to $2 x(6 x-7

Vedclass · Accepted Answer

$\frac{4}{39}$

Vedclass · Answer

(D) The given pair of lines is $12 x^2-20 x y+7 y^2=0$.Factoring the quadratic expression: $12 x^2-14 x y-6 x y+7 y^2=0$.This simplifies to $2 x(6 x-7 y)-y(6 x-7 y)=0$, which gives $(2 x-y)(6 x-7 y)=0$.Thus, the two lines are $y=2 x$ and $y=\frac{6 x}{7}$.The third line is $x+y=1$, or $y=1-x$.To find the vertices, we solve the equations pairwise:$1$. Intersection of $y=2 x$ and $y=\frac{6 x}{7}$ is $(0,0)$.$2$. Intersection of $y=2 x$ and $x+y=1$: $x+2 x=1 \Rightarrow 3 x=1 \Rightarrow x=\frac{1}{3}, y=\frac{2}{3}$. Vertex is $(\frac{1}{3}, \frac{2}{3})$.$3$. Intersection of $y=\frac{6 x}{7}$ and $x+y=1$: $x+\frac{6 x}{7}=1 \Rightarrow \frac{13 x}{7}=1 \Rightarrow x=\frac{7}{13}, y=\frac{6}{13}$. Vertex is $(\frac{7}{13}, \frac{6}{13})$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.Area $= \frac{1}{2} |0(\frac{2}{3}-\frac{6}{13}) + \frac{1}{3}(\frac{6}{13}-0) + \frac{7}{13}(0-\frac{2}{3})|$.Area $= \frac{1}{2} |\frac{6}{39} - \frac{14}{39}| = \frac{1}{2} |-\frac{8}{39}| = \frac{4}{39}$.

Area of the triangle bounded by the lines given by the equations $12 x^2-20 x y+7 y^2=0$ and $x+y-1=0$ is

Similar Questions

If two pairs of straight lines with combined equations $xy+4x-3y-12=0$ and $xy-3x+4y-12=0$ form a square,then the combined equation of its diagonals is

From the point $(3,-4)$,perpendicular lines $L_1$ and $L_2$ are drawn to each of the lines represented by $S \equiv 2x^2+3xy-2y^2-7x+y+3=0$. The area of the quadrilateral formed by the pair of lines $S=0$,$L_1$,and $L_2$ is (in square units):

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

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module