Question

If $2x^2+5xy+2y^2+4x+5y+2=0$ and $2x^2+5xy+2y^2-4x-5y+2=0$ represents a parallelogram then

• A. If $\theta$ be the angle between two non-parallel sides then $\sin \theta$ is
• B. If $p_1$ and $p_2$ be the distances between the parallel sides then $p_1p_2=$
• C. Area of the parallelogram is

Answer 1

Answer: (A)

3/5

Answer 2

The two given equations represent pairs of intersecting lines. Equation 1: $2x^2+5xy+2y^2+4x+5y+2=0$ factors as $(2x+y+2)(x+2y+1)=0$, giving lines $L_{11}: 2x+y+2=0$ and $L_{12}: x+2y+1=0$. Equation 2: $2x^2+5xy+2y^2-4x-5y+2=0$ factors as $(2x+y-2)(x+2y-1)=0$, giving lines $L_{21}: 2x+y-2=0$ and $L_{22}: x+2y-1=0$. These four lines form a parallelogram with parallel sides $(L_{11}, L_{21})$ and $(L_{12}, L_{22})$.

For question 25: $\theta$ is the angle between two non-parallel sides. The slopes of the non-parallel sides are $m_1 = -2$ (from $2x+y+2=0$) and $m_2 = -1/2$ (from $x+2y+1=0$). The tangent of the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by: $\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$ Substituting the slopes: $\tan \theta = \left|\frac{-2 - (-1/2)}{1+(-2)(-1/2)}\right| = \left|\frac{-2+1/2}{1+1}\right| = \left|\frac{-3/2}{2}\right| = \left|-\frac{3}{4}\right| = \frac{3}{4}$ To find $\sin \theta$, we can consider a right-angled triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse would be $\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$. Therefore, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$.