4.5 Q-5

Question Statement

Solve the quadratic equation:
$6x^2 - 19xy + 15y^2 = 0$
Find the equations of the lines it represents and the angle between the lines.

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Background and Explanation

This is a homogeneous quadratic equation in $x$ and $y$, representing two straight lines passing through the origin. To determine these lines, we rewrite the equation in terms of $\frac{y}{x} = m$ (slope). The solution involves solving a quadratic equation to find the slopes of the lines. The angle between the lines can then be calculated using the tangent formula.

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Solution

Step 1: Rewrite the Equation in Terms of $\frac{y}{x}$

Substitute $\frac{y}{x} = m$ into the equation $6x^2 - 19xy + 15y^2 = 0$:

$$15m^2 - 19m + 6 = 0.$$

This is now a quadratic equation in $m$.

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Step 2: Solve for $m$ Using the Quadratic Formula

The quadratic formula is:

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

Here, $a = 15$, $b = -19$, and $c = 6$. Substituting these values:

$$m = \frac{-(-19) \pm \sqrt{(-19)^2 - 4(15)(6)}}{2(15)}.$$

Simplify the discriminant:

$$m = \frac{19 \pm \sqrt{361 - 360}}{30}.$$ $$m = \frac{19 \pm \sqrt{1}}{30}.$$ $$m = \frac{19 \pm 1}{30}.$$

Thus:

$$m = \frac{20}{30} = \frac{2}{3}, \quad m = \frac{18}{30} = -\frac{3}{5}.$$

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Step 3: Write the Equations of the Lines

The slopes of the lines are $m = \frac{2}{3}$ and $m = -\frac{3}{5}$. Using the slope-intercept form $y = mx$, the equations of the lines are:

$$y = \frac{2}{3}x \quad \text{or} \quad 2x - 3y = 0,$$ $$y = -\frac{3}{5}x \quad \text{or} \quad 3x + 5y = 0.$$

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Step 4: Find the Angle Between the Lines

The general formula for the tangent of the angle $\theta$ between two lines represented by a homogeneous quadratic equation $ax^2 + 2hxy + by^2 = 0$ is:

$$\tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b}.$$

Here:

• $a = 6$,
• $b = 15$,
• $h = -\frac{19}{2}$.

Substitute these values:

$$\tan\theta = \frac{2\sqrt{\left(-\frac{19}{2}\right)^2 - (6)(15)}}{6 + 15}.$$

Simplify step by step:

1. Compute $h^2 - ab$:

$$h^2 - ab = \left(-\frac{19}{2}\right)^2 - 6 \cdot 15 = \frac{361}{4} - 90 = \frac{1}{4}.$$

2. Substitute into the formula:

$$\tan\theta = \frac{2\sqrt{\frac{1}{4}}}{21}.$$

3. Simplify:

$$\tan\theta = \frac{2 \cdot \frac{1}{2}}{21} = \frac{1}{21}.$$

Thus:

$$\theta = \tan^{-1}\left(\frac{1}{21}\right).$$

Using a calculator:

$$\theta \approx 2.73^\circ.$$

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Key Formulas or Methods Used

1. Quadratic Substitution: $\frac{y}{x} = m$ reduces the equation to a quadratic in $m$.
2. Quadratic Formula: $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$
3. Angle Formula: $\tan\theta = \frac{2\sqrt{h^2 - ab}}{a + b}.$

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Summary of Steps

1. Substitute $\frac{y}{x} = m$ into the given equation to simplify it.
2. Solve the resulting quadratic equation for $m$ to find the slopes of the lines.
3. Write the equations of the lines using the slopes.
4. Use the tangent formula to calculate the angle between the lines.
5. Conclude that the angle is approximately $2.73^\circ$.