Question

If area of the triangle formed by the lines $y^2-9xy+18x^2=0$ and y=9 is A sq. unit find the value of 4A.

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Answer 1

Pair straight lines

$y^2-9xy+18x^2=0$

$18(\frac{x}{y}{)}^2-9(\frac{x}{y})+1=0$

$18(\frac{x}{y}{)}^2-6(\frac{x}{y})+3(\frac{x}{y})+1=0$

$6\left(\frac{x}{y}\right)\left[3\right(\frac{x}{y})-1]-1\left[3\right(\frac{x}{y})-1]=0$

$\left[6\right(\frac{x}{y})-1]\left[3\right(\frac{x}{y})-1]=0$

$\frac{6x}{y}-1=0or\frac{3x}{y}-1=0$

y = 6x and y = 3x

Drawing all three lines on the Cartesian plane

Intersection Points of line y=3x and y=9 is A

Co-ordinates of point A substituting y=9

9=3x

x=3

A(3,9)

Intersection of line y=9 and y=6x

9=6x

$X=\frac{3}{2}$

B$(\frac{3}{2},9)$

Intersection of y = 3x and y = 6x is O(0,0)

Area of triangle OAB

= $\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 3& 9& 1\\ \frac{3}{2}& 9& 1\end{array}\right|$

= $\frac{1}{2}\left[1\right(27-\frac{27}{2}\left)\right]$

= $\frac{1}{2}\times \frac{27}{2}$

A = $\frac{27}{2}$sq.unit

Value of 4A = $4\times \frac{27}{4}=27sq.unit$