Area of the triangle formed by the lines 3x^2 | vedclass

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(C) The given pair of lines is $3x^2-4xy+y^2=0$.Factoring the equation: $3x^2-3xy-xy+y^2=0$ $\Rightarrow 3x(x-y)-y(x-y)=0$ $\Rightarrow (3x-y)(x-y)=0$

Vedclass · Accepted Answer

$36 	ext{ sq units}$

Vedclass · Answer

(C) The given pair of lines is $3x^2-4xy+y^2=0$.Factoring the equation: $3x^2-3xy-xy+y^2=0$ $\Rightarrow 3x(x-y)-y(x-y)=0$ $\Rightarrow (3x-y)(x-y)=0$.So,the two lines are $L_1: 3x-y=0$ and $L_2: x-y=0$.The third line is $L_3: 2x-y=6$.The vertices of the triangle are the intersection points of these lines:$1$. Intersection of $L_1$ and $L_2$: $(0,0)$.$2$. Intersection of $L_1$ and $L_3$: $3x-y=0$ and $2x-y=6$. Subtracting gives $x=-6$,so $y=-18$. Point is $(-6, -18)$.$3$. Intersection of $L_2$ and $L_3$: $x-y=0$ and $2x-y=6$. Subtracting gives $x=6$,so $y=6$. Point is $(6, 6)$.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 $= \frac{1}{2} |0(-18-6) + (-6)(6-0) + 6(0-(-18))| = \frac{1}{2} |0 - 36 + 108| = \frac{1}{2} |72| = 36 	ext{ sq units}$.

Area of the triangle formed by the lines $3x^2-4xy+y^2=0$ and $2x-y=6$ is

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