If the lines y + px = 1 and y - qx = 2 are perpendicular, then which one of the following is correct?
pq - 1 = 0
When two lines in a coordinate plane are perpendicular, their slopes have a special relationship. The product of the slopes of two perpendicular lines is always -1, provided neither line is vertical or horizontal. Vertical lines have undefined slopes, and horizontal lines have a slope of 0. A vertical line is perpendicular to a horizontal line.
Let's consider the given equations of the lines:
To find the slopes of these lines, we need to rewrite them in the slope-intercept form, which is \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.
For the first line, \(y + px = 1\), we isolate \(y\) by subtracting \(px\) from both sides:
\(\qquad y = -px + 1\)
Comparing this to \(y = mx + c\), the slope of the first line, let's call it \(m_1\), is:
\(\qquad m_1 = -p\)
For the second line, \(y - qx = 2\), we isolate \(y\) by adding \(qx\) to both sides:
\(\qquad y = qx + 2\)
Comparing this to \(y = mx + c\), the slope of the second line, let's call it \(m_2\), is:
\(\qquad m_2 = q\)
The problem states that the two lines are perpendicular. The condition for two non-vertical perpendicular lines is that the product of their slopes is -1. That is, \(m_1 \times m_2 = -1\).
Substitute the slopes we found (\(m_1 = -p\) and \(m_2 = q\)) into this condition:
\(\qquad (-p) \times (q) = -1\)
\(\qquad -pq = -1\)
To find the relationship between \(p\) and \(q\), we can multiply both sides of the equation by -1:
\(\qquad pq = 1\)
Alternatively, we can move the constant term to the left side by subtracting 1 from both sides:
\(\qquad pq - 1 = 0\)
Let's compare our derived relationship, \(pq - 1 = 0\), with the provided options:
Our derived condition \(pq - 1 = 0\) matches Option 3.
| Step | Description | Calculation |
|---|---|---|
| 1 | Identify the equations of the lines. | Line 1: \(y + px = 1\) Line 2: \(y - qx = 2\) |
| 2 | Convert equations to slope-intercept form (\(y=mx+c\)). | Line 1: \(y = -px + 1\) Line 2: \(y = qx + 2\) |
| 3 | Identify the slopes (\(m_1\) and \(m_2\)). | \(m_1 = -p\) \(m_2 = q\) |
| 4 | Apply the perpendicularity condition (\(m_1 \times m_2 = -1\)). | \((-p) \times (q) = -1\) |
| 5 | Simplify the equation. | \(-pq = -1\) \(pq = 1\) \(pq - 1 = 0\) |
| 6 | Compare the result with the options. | Matches \(pq - 1 = 0\) |
| Concept | Description | Condition |
|---|---|---|
| Perpendicular Lines | Two lines that intersect at a right angle (90 degrees). | Product of slopes is -1 (\(m_1 \times m_2 = -1\)), unless one line is vertical and the other is horizontal. |
| Slope of a Line | A measure of the steepness of a line; represented as the change in y divided by the change in x (\(\frac{\Delta y}{\Delta x}\)). In \(y=mx+c\), the slope is \(m\). | |
| Slope-Intercept Form | A way to write the equation of a line: \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. | \(y = mx + c\) |
| Vertical Line | A line parallel to the y-axis. | Equation is \(x=a\); slope is undefined. Perpendicular to horizontal lines. |
| Horizontal Line | A line parallel to the x-axis. | Equation is \(y=b\); slope is 0. Perpendicular to vertical lines. |
Besides perpendicular lines, lines can also be parallel or intersecting non-perpendicularly.
Understanding the slopes of lines is fundamental in coordinate geometry for determining the relationship between lines.
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