The area of the triangle formed by the lines | vedclass

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(C) The given equation of the pair of lines is $y^2 - 9xy + 18x^2 = 0$.This can be factored as $(y - 3x)(y - 6x) = 0$.Thus,the two lines are $y = 3x$

Vedclass · Accepted Answer

$6.75$

Vedclass · Answer

(C) The given equation of the pair of lines is $y^2 - 9xy + 18x^2 = 0$.This can be factored as $(y - 3x)(y - 6x) = 0$.Thus,the two lines are $y = 3x$ and $y = 6x$.These lines pass through the origin $(0, 0)$.Substitute $y = 9$ into the equations of the lines:For $y = 3x$,$9 = 3x \Rightarrow x = 3$. So,the vertex is $(3, 9)$.For $y = 6x$,$9 = 6x \Rightarrow x = 1.5$. So,the vertex is $(1.5, 9)$.The vertices of the triangle are $(0, 0)$,$(3, 9)$,and $(1.5, 9)$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |0(9 - 9) + 3(9 - 0) + 1.5(0 - 9)|$Area $= \frac{1}{2} |0 + 27 - 13.5| = \frac{1}{2} |13.5| = 6.75 \, sq. \, units$.

The area of the triangle formed by the lines $y^2 - 9xy + 18x^2 = 0$ and $y = 9$ is ............ $sq. \, units$.

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