4.5 Q-4

Question Statement

Solve the equation $2x^2 + 3xy - 5y^2 = 0$ to find the lines it represents, and determine the angle between these lines.

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Background and Explanation

This problem deals with a second-degree equation representing two straight lines. To solve:

1. Convert the equation into a standard form involving $\frac{y}{x}$ (or $\frac{x}{y}$) to find the slopes of the lines.
2. Use the slopes to determine the angle between the lines.

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Solution

Step 1: Rewrite the equation

The given equation is:

$$2x^2 + 3xy - 5y^2 = 0$$

Divide through by $x^2$ (assuming $x \neq 0$):

$$5\left(\frac{y}{x}\right)^2 - 3\left(\frac{y}{x}\right) - 2 = 0$$

Let $\frac{y}{x} = m$, giving:

$$5m^2 - 3m - 2 = 0$$

Step 2: Solve for $m$

Solve the quadratic equation for $m$ using the quadratic formula:

$$m = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(5)(-2)}}{2(5)}$$

Simplify step by step:

1. Compute the discriminant:

$$(-3)^2 - 4(5)(-2) = 9 + 40 = 49$$

2. Apply the quadratic formula:

$$m = \frac{3 \pm \sqrt{49}}{10}$$ $$m = \frac{3 \pm 7}{10}$$

Split into two solutions:

$$m = \frac{3 - 7}{10} = -\frac{4}{10} = -\frac{2}{5}, \quad m = \frac{3 + 7}{10} = 1$$

Step 3: Find the equations of the lines

The slopes of the lines are $m_1 = -\frac{2}{5}$ and $m_2 = 1$. Using the slope-intercept form $y = mx$, rewrite the lines:

1. For $m_1 = -\frac{2}{5}$:

$$y = -\frac{2}{5}x \quad \Rightarrow \quad 2x + 5y = 0$$

2. For $m_2 = 1$:

$$y = x \quad \Rightarrow \quad x - y = 0$$

The equations of the lines are:

$$2x + 5y = 0 \quad \text{and} \quad x - y = 0$$

Step 4: Find the angle between the lines

The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by:

$$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 \cdot m_2} \right|$$

Substitute $m_1 = -\frac{2}{5}$ and $m_2 = 1$:

$$\tan \theta = \left| \frac{1 - \left(-\frac{2}{5}\right)}{1 + \left(1 \cdot -\frac{2}{5}\right)} \right|$$

Simplify:

1. Numerator:

$$1 - \left(-\frac{2}{5}\right) = 1 + \frac{2}{5} = \frac{5}{5} + \frac{2}{5} = \frac{7}{5}$$

2. Denominator:

$$1 + \left(-\frac{2}{5}\right) = 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5}$$

3. Division:

$$\tan \theta = \frac{\frac{7}{5}}{\frac{3}{5}} = \frac{7}{3}$$

Finally, find the angle:

$$\theta = \tan^{-1}\left(\frac{7}{3}\right) \approx 66.8^\circ$$

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Key Formulas or Methods Used

1. Quadratic Formula:

$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

2. Angle Between Lines:

$$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 \cdot m_2} \right|$$

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Summary of Steps

1. Rewrite the given equation as a quadratic in $\frac{y}{x}$.
2. Solve for $\frac{y}{x}$ to find the slopes $m_1 = -\frac{2}{5}$ and $m_2 = 1$.
3. Derive the equations of the lines as $2x + 5y = 0$ and $x - y = 0$.
4. Use the formula for the angle between lines to find $\theta \approx 66.8^\circ$.