Question

If $\theta$ be the angle between two non-parallel sides then sin $\theta$ is

• A. 3/5
• B. 1/3
• C. $2\sqrt{2}/3$
• D. 4/5

Answer 1

Answer: (A)

3/5

Answer 2

The given equations are: $2x^2 + 5xy + 2y^2 + 4x + 5y + 2 = 0$ $2x^2 + 5xy + 2y^2 - 4x - 5y + 2 = 0$ Let these equations be $E_1=0$ and $E_2=0$. The quadratic part $2x^2 + 5xy + 2y^2$ can be factored as $(2x+y)(x+2y)$. The equations can be rewritten as: $E_1: (2x+y)(x+2y) + (4x+5y) + 2 = 0$ $E_2: (2x+y)(x+2y) - (4x+5y) + 2 = 0$ Let $X = 2x+y$ and $Y = x+2y$. Then $4x+5y = 2(2x+y) + (x+2y) = 2X+Y$. Substituting these into the equations: $E_1: XY + (2X+Y) + 2 = 0 \implies XY + 2X + Y + 2 = 0 \implies X(Y+2) + (Y+2) = 0 \implies (X+1)(Y+2) = 0$. This means $E_1=0$ represents the union of the lines $X+1=0$ and $Y+2=0$. So, $2x+y+1=0$ and $x+2y+2=0$.

$E_2: XY - (2X+Y) + 2 = 0 \implies XY - 2X - Y + 2 = 0$. This equation represents the other pair of parallel sides. To find the lines, we can compare it with the standard form $(2x+y+c_1)(x+2y+d_1) = k$. Expanding this, we get $2x^2+5xy+2y^2 + (2d_1+c_1)x + (d_1+2c_1)y + c_1d_1 - k = 0$. Comparing with $E_2 = 2x^2+5xy+2y^2-4x-5y+2=0$: $2d_1+c_1 = -4$ $d_1+2c_1 = -5$ $c_1d_1 - k = 2$ Solving the first two equations: Multiply the first by 2: $4d_1+2c_1 = -8$. Subtract the second: $(4d_1+2c_1) - (d_1+2c_1) = -8 - (-5) \implies 3d_1 = -3 \implies d_1 = -1$. Substitute $d_1=-1$ into $2d_1+c_1=-4 \implies 2(-1)+c_1 = -4 \implies -2+c_1 = -4 \implies c_1 = -2$. Now, check the third equation: $c_1d_1 - k = (-2)(-1) - k = 2 - k = 2 \implies k=0$. So, $E_2=0$ represents the union of the lines $2x+y-2=0$ and $x+2y-1=0$.

The four lines forming the parallelogram are: $L_1: 2x+y+1=0$ $L_2: x+2y+2=0$ $L_3: 2x+y-2=0$ $L_4: x+2y-1=0$

The pairs of parallel sides are given by $(L_1, L_3)$ and $(L_2, L_4)$. The angle $\theta$ between the non-parallel sides $2x+y+1=0$ and $x+2y+2=0$. The slopes are $m_1 = -2$ and $m_2 = -1/2$. The tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by: $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$ $\tan\theta = \left|\frac{-2 - (-1/2)}{1 + (-2)(-1/2)}\right| = \left|\frac{-2 + 1/2}{1 + 1}\right| = \left|\frac{-3/2}{2}\right| = \frac{3}{4}$. If $\tan\theta = 3/4$, we can form a right triangle with opposite side 3 and adjacent side 4. The hypotenuse is $\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$. Therefore, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$.