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EAMCET Engineering 2004 Solved Paper

Section: Mathematics

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Question : 130 of 160

Marks: +1, -0

Area of the triangle formed by the lines

3 x^2-4 x y+y^2=0,2 x-y=6

16 sq units
25 sq units
36 sq units
49 sq units

Solution:

Given lines are

\;3 x^2-4 x y+y^2\;=0

\Rightarrow 3 x^2-3 x y-x y+y^2\;=0

(3 x-y)(x-y)\;=0

\Rightarrow\;3 x-y=0, x-y\;=0

\;\text{ and }\;\;2 x-y\;=6

The point of intersection of these lines are

(0,0),(-6,-18)

and

(6,6)

.

\therefore \;\text{ Area of triangle }\;\;=\;\frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}

\;=\;\frac{1}{2}\begin{vmatrix} 0 & 0 & 1 \\ -6 & -18 & 1 \\ 6 & 6 & 1 \end{vmatrix}

\;=\;\frac{1}{2}(36+108)=\;\frac{1}{2}(72)

\;=36 \;\text{ sq unit }\;

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