What is the equation of the straight line which is perpendicular to y = x and passes throng (3, 2)?
x + y = 5
Understanding the Problem: Finding a Perpendicular Line
The question asks us to find the equation of a straight line that meets two specific conditions: it must be perpendicular to the line given by the equation \(y = x\), and it must pass through the particular point \((3, 2)\).
To solve this, we need to use our knowledge of the properties of straight lines, specifically their slopes and the point-slope form of the equation of a line.
Step-by-Step Solution to Finding the Line Equation
Step 1: Determine the Slope of the Given Line
The given line has the equation \(y = x\). This equation is in the slope-intercept form, which is \(y = mx + c\), where \(m\) represents the slope and \(c\) represents the y-intercept.
Comparing \(y = x\) with \(y = mx + c\), we can see that the slope (\(m_1\)) of the given line \(y = x\) is \(1\). The y-intercept (\(c\)) is \(0\).
Step 2: Determine the Slope of the Perpendicular Line
Two lines are perpendicular if the product of their slopes is \(-1\). Let the slope of the required line be \(m_2\).
The relationship for perpendicular lines is \(m_1 \times m_2 = -1\).
We know \(m_1 = 1\).
So, \(1 \times m_2 = -1\).
This gives us \(m_2 = -1\).
The slope of the straight line we are looking for is \(-1\).
Step 3: Use the Point-Slope Form of the Equation
We have the slope of the required line (\(m = -1\)) and a point that it passes through \((x_1, y_1) = (3, 2)\). The point-slope form of the equation of a straight line is given by:
\(y - y_1 = m(x - x_1)\)
Substitute the values \(m = -1\), \(x_1 = 3\), and \(y_1 = 2\) into the point-slope form.
\(y - 2 = -1(x - 3)\)
Step 4: Simplify the Equation to Match the Options
Now, we need to simplify the equation and rearrange it into a standard form like \(Ax + By = C\) to compare it with the given options.
Distribute the \(-1\) on the right side:
\(y - 2 = -x + 3\)
Move the \(x\) term to the left side by adding \(x\) to both sides:
\(x + y - 2 = 3\)
Move the constant term (\(-2\)) to the right side by adding \(2\) to both sides:
\(x + y = 3 + 2\)
\(x + y = 5\)
The equation of the straight line perpendicular to \(y = x\) and passing through the point \((3, 2)\) is \(x + y = 5\).
Verification with Options
Let's check the derived equation against the given options:
Option 1: \(x - y = 5\) (Does not match)
Option 2: \(x + y = 5\) (Matches our derived equation)
Option 3: \(x + y = 1\) (Does not match)
Option 4: \(x - y = 1\) (Does not match)
The equation \(x + y = 5\) is indeed one of the provided options.
| Concept | Explanation |
|---|---|
| Slope of \(y = x\) | \(m_1 = 1\) |
| Slope of perpendicular line | \(m_2 = -1\) (since \(m_1 \times m_2 = -1\)) |
| Given point | \((3, 2)\) |
| Point-Slope Form | \(y - y_1 = m(x - x_1)\) |
| Substituting values | \(y - 2 = -1(x - 3)\) |
| Simplified Equation | \(x + y = 5\) |
Revision Table: Key Concepts for Straight Lines
| Concept | Formula/Rule | Notes |
|---|---|---|
| Slope-Intercept Form | \(y = mx + c\) | \(m\) is slope, \(c\) is y-intercept |
| Point-Slope Form | \(y - y_1 = m(x - x_1)\) | \(m\) is slope, \((x_1, y_1)\) is a point on the line |
| Slope from Two Points | \(m = \frac{y_2 - y_1}{x_2 - x_1}\) | Slope of line passing through \((x_1, y_1)\) and \((x_2, y_2)\) |
| Parallel Lines | \(m_1 = m_2\) | Slopes are equal |
| Perpendicular Lines | \(m_1 \times m_2 = -1\) | Product of slopes is \(-1\) (for non-vertical/horizontal lines) |
| General Form | \(Ax + By + C = 0\) | Standard form of a linear equation |
Additional Information: Lines Perpendicular to y = x
The line \(y = x\) represents all points where the x-coordinate is equal to the y-coordinate. It passes through the origin \((0,0)\) and has a slope of \(1\).
Any line perpendicular to \(y = x\) must have a slope \(m_2\) such that \(1 \times m_2 = -1\), which means \(m_2 = -1\).
Lines with a slope of \(-1\) are of the form \(y = -x + c\), where \(c\) is the y-intercept. Substituting the point \((3, 2)\) into this general form: \(2 = -(3) + c\) \(2 = -3 + c\) \(c = 2 + 3\) \(c = 5\) So, the equation is \(y = -x + 5\). Rearranging this gives \(x + y = 5\), confirming our result using the slope-intercept form directly.
The lines perpendicular to \(y=x\) are a family of lines, all having a slope of -1. Each specific line in this family is determined by a point it passes through or its y-intercept.
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