equation of a perpendicular line passing through a point

Drawing upon the provided sources and our conversation history, we can determine the equation of a line perpendicular to a given line and passing through a specific point, often starting with the General Form of the given line.

The general equation of a line is expressed as Ax + By + C = 0. This form is particularly useful as it can represent any straight line, including vertical lines, which have an undefined slope. For this to represent a line, A and B cannot be zero simultaneously.

The relationship between two perpendicular lines (that are not vertical or horizontal) is characterised by their slopes. If two non-vertical lines have slopes m₁ and m₂, they are perpendicular if and only if the product of their slopes is -1, i.e., m₁ * m₂ = -1.

For a non-vertical line in the general form Ax + By + C = 0 (where B ≠ 0), its slope (m) is given by m = -A/B.

To find the equation of a line perpendicular to a given line Ax + By + C = 0 and passing through a point (x₀, y₀), we can follow these steps, as illustrated by an example in the sources:

1. Find the slope of the given line: Let the given line be A₁x + B₁y + C₁ = 0. If B₁ ≠ 0, its slope is m₁ = -A₁/B₁.
   • Example: Given line is x - 2y + 3 = 0. Here, A₁=1, B₁=-2, C₁=3. The slope m₁ = -(1)/(-2) = 1/2.
2. Find the slope of the perpendicular line: The slope m₂ of the line perpendicular to the given line satisfies m₁ * m₂ = -1.
   • If m₁ ≠ 0, then m₂ = -1/m₁.
   • Example: Since m₁ = 1/2, m₂ = -1 / (1/2) = -2.
3. Use the Point-Slope Form: We now have the slope (m₂) of the perpendicular line and a point (x₀, y₀) it passes through. The equation of a line with slope m and passing through (x₀, y₀) is given by the Point-Slope Form: (y - y₀) = m(x - x₀).
   • Example: The perpendicular line has slope m₂ = -2 and passes through (-1, 2). Substituting these values: (y - 2) = -2(x - (-1)).
   • This simplifies to (y - 2) = -2(x + 1).
4. Convert to General Form: Rearrange the equation from the point-slope form into the general form Ax + By + C = 0.
   • Example: (y - 2) = -2(x + 1) becomes y - 2 = -2x - 2.
   • Adding 2x and 2 to both sides gives 2x + y = 0 [Conversation history based on example result]. This can be written in general form as 2x + y + 0 = 0. The source also presents the equivalent form -2x + y = 0, which is obtained by rearranging y = -2x.

Direct Condition for Perpendicular Lines in General Form:

The sources also provide a direct condition for two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be perpendicular, provided b₁ and b₂ are not zero. The lines are perpendicular if: a₁ × a₂ = −b₁ × b₂

Let’s verify this condition with the example:

• Line 1: x - 2y + 3 = 0 (a₁=1, b₁=-2).
• Perpendicular Line (derived): 2x + y + 0 = 0 (a₂=2, b₂=1) [Conversation history based on example result].
• Checking the condition a₁a₂ = -b₁b₂: (1) * (2) = -(-2) * (1) => 2 = -(-2) => 2 = 2. This matches the condition for perpendicular lines.

Special Cases (Vertical and Horizontal Lines):

The method using slopes and the point-slope form works for non-vertical lines. However, the General Form Ax + By + C = 0 naturally handles vertical and horizontal lines as well.

• A vertical line is represented by Ax + C = 0 (where B=0, A≠0), which simplifies to x = -C/A. Vertical lines have an undefined slope.
• A horizontal line is represented by By + C = 0 (where A=0, B≠0), which simplifies to y = -C/B. Horizontal lines have a slope of 0.
• If the given line is vertical (e.g., x = k), a line perpendicular to it must be horizontal (y = constant). You would then find the constant using the point the perpendicular line passes through.
• If the given line is horizontal (e.g., y = k), a line perpendicular to it must be vertical (x = constant). You would find the constant using the given point.

The general form allows for a unified algebraic representation, and while finding slopes and using the point-slope form is a common method, the direct condition on coefficients can also be applied when applicable (i.e., when the y-coefficients are non-zero).