Distance from a Point to a Line

Distance from a Point to a Line#

Line Equations Revision#

Theory#

Before finding distances from points to lines, recall:

  • General form of a line: \(ax + by + c = 0\)

  • Point-slope form: \(y - y_1 = m(x - x_1)\)

  • The coefficients \(a\) and \(b\) determine the line’s direction

Application#

Converting between forms helps us apply the distance formula effectively.

Distance from a Point to a Line#

Theory#

The perpendicular distance from a point \((x_0, y_0)\) to a line \(ax + by + c = 0\) is given by:

\[d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\]

Key concepts:

  • This gives the shortest distance from the point to the line

  • The distance is always measured perpendicular to the line

  • The absolute value ensures distance is positive

  • The denominator \(\sqrt{a^2 + b^2}\) normalizes the line equation

Special cases:

  • Distance from origin \((0, 0)\): \(d = \frac{|c|}{\sqrt{a^2 + b^2}}\)

  • For vertical line \(x = k\): distance from \((x_0, y_0)\) is \(|x_0 - k|\)

  • For horizontal line \(y = k\): distance from \((x_0, y_0)\) is \(|y_0 - k|\)

Interactive Visualization: Point-to-Line Distance Explorer#

Application#

Examples#

Example 1: Basic distance calculation

Find the distance from the point \((4, 3)\) to the line \(3x - 4y + 5 = 0\).

Solution: Using the distance formula: $\(d = \frac{|3(4) - 4(3) + 5|}{\sqrt{3^2 + (-4)^2}}\)\( \)\(d = \frac{|12 - 12 + 5|}{\sqrt{9 + 16}}\)\( \)\(d = \frac{|5|}{\sqrt{25}} = \frac{5}{5} = 1\)$

The distance is 1 unit.

Example 2: Finding parallel lines at a given distance

Find equations of lines parallel to \(2x + y - 3 = 0\) at a distance of 2 units from it.

Solution: Parallel lines have the form \(2x + y + k = 0\). The distance between parallel lines \(2x + y - 3 = 0\) and \(2x + y + k = 0\) is: $\(d = \frac{|k - (-3)|}{\sqrt{2^2 + 1^2}} = \frac{|k + 3|}{\sqrt{5}}\)$

Setting \(d = 2\): $\(\frac{|k + 3|}{\sqrt{5}} = 2\)\( \)\(|k + 3| = 2\sqrt{5}\)\( \)\(k + 3 = \pm 2\sqrt{5}\)\( \)\(k = -3 \pm 2\sqrt{5}\)$

The two parallel lines are:

  • \(2x + y + (-3 + 2\sqrt{5}) = 0\)

  • \(2x + y + (-3 - 2\sqrt{5}) = 0\)

Example 3: Point equidistant from two lines

Find the locus of points equidistant from the lines \(3x + 4y - 12 = 0\) and \(3x + 4y + 8 = 0\).

Solution: Let \((x, y)\) be equidistant from both lines: $\(\frac{|3x + 4y - 12|}{\sqrt{9 + 16}} = \frac{|3x + 4y + 8|}{\sqrt{9 + 16}}\)$

Since denominators are equal: $\(|3x + 4y - 12| = |3x + 4y + 8|\)$

This gives us two cases:

  1. \(3x + 4y - 12 = 3x + 4y + 8\) (impossible)

  2. \(3x + 4y - 12 = -(3x + 4y + 8)\)

From case 2: $\(3x + 4y - 12 = -3x - 4y - 8\)\( \)\(6x + 8y = 4\)\( \)\(3x + 4y = 2\)$

The locus is the line \(3x + 4y - 2 = 0\), which is parallel to and midway between the given lines.

Multiple Choice Questions#

Sector Specific Questions: Distance Applications#

Key Takeaways#

Important

Essential Distance Concepts:

  1. Distance Formula: For point \((x_0, y_0)\) to line \(ax + by + c = 0\): $\(d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}\)$

  2. Key Properties:

    • Always gives the perpendicular (shortest) distance

    • Result is always positive (due to absolute value)

    • Denominator normalizes the line equation

  3. Special Cases:

    • Distance from origin: \(d = \frac{|c|}{\sqrt{a^2 + b^2}}\)

    • Vertical/horizontal lines have simpler formulas

  4. Applications:

    • Finding parallel lines at given distances

    • Optimization problems

    • Physics (fields, waves)

    • Engineering (clearances, tolerances)

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