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Analytic Geometry: Lines and Circles · Ages 16–17

Distance from a Point to a Line Worksheet

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Designed for 11th-grade analytic geometry students. Rewrite all line equations into Ax + By + C = 0 before calculating distances and triangle areas.

  1. Find the distance from the point (0, 0) to the line 3x + 4y - 10 = 0.

  2. Find the distance from the point (2, 3) to the vertical line x = -1.

  3. Calculate the distance from (1, -2) to the line 5x - 12y + 3 = 0.

  4. Convert y = (3/4)x + 2 to general form Ax + By + C = 0, then find the distance from (-2, 4) to this line.

  5. A line has the equation 4x + 3y - 12 = 0. A triangle has its base along this line between points (0, 4) and (3, 0), and its third vertex is (5, 6). Find the height and the area of the triangle.

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Distance from a Point to a Line Worksheet

Answer key

  1. Find the distance from the point (0, 0) to the line 3x + 4y - 10 = 0.

    d = |3(0) + 4(0) - 10| / √(3² + 4²) = |-10| / 5 = 2.

  2. Find the distance from the point (2, 3) to the vertical line x = -1.

    Rewrite as x + 1 = 0. d = |1(2) + 1| / √(1² + 0²) = 3 / 1 = 3 (or count horizontally from x = -1 to x = 2).

  3. Calculate the distance from (1, -2) to the line 5x - 12y + 3 = 0.

    d = |5(1) - 12(-2) + 3| / √(5² + (-12)²) = |5 + 24 + 3| / 13 = 32 / 13 (approx. 2.46).

  4. Convert y = (3/4)x + 2 to general form Ax + By + C = 0, then find the distance from (-2, 4) to this line.

    General form: 3x - 4y + 8 = 0. d = |3(-2) - 4(4) + 8| / √(3² + (-4)²) = |-6 - 16 + 8| / 5 = |-14| / 5 = 2.8.

  5. A line has the equation 4x + 3y - 12 = 0. A triangle has its base along this line between points (0, 4) and (3, 0), and its third vertex is (5, 6). Find the height and the area of the triangle.

    Base length = √((3-0)² + (0-4)²) = 5. Height is the distance from (5, 6) to 4x + 3y - 12 = 0: h = |4(5) + 3(6) - 12| / √(4² + 3²) = |20 + 18 - 12| / 5 = 26 / 5 = 5.2. Area = 0.5 * 5 * 5.2 = 13 square units.