4.3 Q-26

Question Statement

Find the equations of two parallel lines that are perpendicular to the line $2x - y + 3 = 0$ such that the product of the $x$- and $y$-intercepts of each line is 3.

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Background and Explanation

This problem involves concepts of slopes, intercepts, and perpendicularity in coordinate geometry. Key points to understand:

• A line perpendicular to another has a slope that is the negative reciprocal of the original slope.
• The intercepts of a line can be found by substituting $y = 0$ (for $x$-intercept) and $x = 0$ (for $y$-intercept).
• The product of the intercepts is a condition used to determine the constant in the line’s equation.

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Solution

Step 1: General equation of the line perpendicular to $2x - y + 3 = 0$.

The slope of $2x - y + 3 = 0$ can be calculated by rearranging it into slope-intercept form ($y = mx + c$): $y = 2x + 3$

The slope of this line is 2. The slope of a perpendicular line is the negative reciprocal, which is: $m_{\text{perpendicular}} = -\frac{1}{2}$

The general equation of any line with this slope is: x + 2y + c = 0 \tag{1}

Here, $c$ is a constant to be determined.

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Step 2: Calculate the intercepts of the line.

1. Finding the $x$-intercept: Substitute $y = 0$ into $x + 2y + c = 0$: $x + c = 0 \implies x = -c$

   So, the $x$-intercept is $-c$.

2. Finding the $y$-intercept: Substitute $x = 0$ into $x + 2y + c = 0$: $2y + c = 0 \implies y = -\frac{c}{2}$

   So, the $y$-intercept is $-\frac{c}{2}$.

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Step 3: Use the condition on the product of intercepts.

We are given that the product of the $x$- and $y$-intercepts is 3. Substituting the intercepts: $(-c) \cdot \left(-\frac{c}{2}\right) = 3$

Simplify: $\frac{c^2}{2} = 3$

Multiply through by 2: $c^2 = 6$

Take the square root: $c = \pm \sqrt{6}$

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Step 4: Write the equations of the lines.

Substitute $c = \sqrt{6}$ and $c = -\sqrt{6}$ into the general equation $x + 2y + c = 0$:

1. For $c = \sqrt{6}$: $x + 2y + \sqrt{6} = 0$

2. For $c = -\sqrt{6}$: $x + 2y - \sqrt{6} = 0$

Thus, the equations of the required lines are: $x + 2y + \sqrt{6} = 0 \quad \text{and} \quad x + 2y - \sqrt{6} = 0$

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Key Formulas or Methods Used

1. Slope of a perpendicular line: $m_{\text{perpendicular}} = -\frac{1}{m}$.
2. Intercepts:
   • $x$-intercept: Set $y = 0$.
   • $y$-intercept: Set $x = 0$.
3. Product of intercepts: Relates $x$- and $y$-intercepts through a given condition.

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Summary of Steps

1. Determine the slope of the given line and find the slope of the perpendicular line.

2. Write the general equation of the perpendicular line.

3. Find the $x$- and $y$-intercepts of the line in terms of $c$.

4. Use the condition on the product of intercepts to solve for $c$.

5. Substitute the values of $c$ into the equation to get the two required lines:

   • $x + 2y + \sqrt{6} = 0$
   • $x + 2y - \sqrt{6} = 0$