Analytic Geometry: Distance, Triangles, Medians

Analytic Geometry: Distance, Triangles, Medians

Analytic geometry translates geometric shapes into coordinate language, making it possible to compute distances, midpoints, and triangle types using exact formulas. On this page we will learn the distance formula and the midpoint formula, identify isosceles and right triangles from coordinates, and explore the median, altitude, and midsegment of a triangle.

Background and Basic Definitions

Distance between two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) (derived from the Pythagorean theorem):

\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Midpoint of segment \(AB\) — average of the coordinates:

\[ M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) \]

Special lines in a triangle:

  • Median — a segment from a vertex to the midpoint of the opposite side. Every triangle has \(3\) medians, and they meet at one point (the centroid).
  • Altitude — a segment from a vertex perpendicular to the opposite side. The three altitudes meet at a point called the orthocentre. In a right triangle the orthocentre is located at the vertex of the right angle.
  • Midsegment — a segment joining the midpoints of two sides. It is parallel to the third side and its length is half of that side.

Classifying a triangle from coordinates: compute the lengths of all three sides. If two sides are equal — isosceles; if all three are equal — equilateral; if the sum of the squares of the two shorter sides equals the square of the longest — right triangle.

Solution Steps

  1. Step 1 — Write down the coordinates of every relevant point and label which is \((x_1,y_1)\) and which is \((x_2,y_2)\).
  2. Step 2 — For distance: substitute into \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\); pay attention to minus signs when coordinates are negative.
  3. Step 3 — For the midpoint: average the \(x\)-coordinates separately and the \(y\)-coordinates separately.
  4. Step 4 — To classify the triangle: compute all three side lengths, compare them, and apply the converse of the Pythagorean theorem to check for a right angle.
  5. Step 5 — For a median or altitude: first find the relevant point (midpoint of a side for the median; foot of the perpendicular for the altitude), then compute the required distance or expression.

Worked Examples

Example 1: Distance Between Two Points

Problem: Find the distance between points \(A(-3, 2)\) and \(B(1, 5)\).

Solution:

  1. Use the distance formula: \(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
  2. Coordinate differences: \(x_2 - x_1 = 1 - (-3) = 4\), \(y_2 - y_1 = 5 - 2 = 3\).
  3. Substitute: \(AB = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25}\).
  4. Take the square root: \(AB = 5\).

Answer: The distance is \(5\) units.

Example 2: Identifying a Right Triangle from Coordinates

Problem: A triangle has vertices \(A(1, 1)\), \(B(5, 1)\), \(C(1, 4)\). Determine whether it is a right triangle.

Solution:

  1. Compute all three sides. \(AB = \sqrt{(5-1)^2 + (1-1)^2} = \sqrt{16} = 4\).
  2. \(AC = \sqrt{(1-1)^2 + (4-1)^2} = \sqrt{9} = 3\).
  3. \(BC = \sqrt{(1-5)^2 + (4-1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\).
  4. Apply the converse of the Pythagorean theorem: \(AB^2 + AC^2 = 16 + 9 = 25 = BC^2\).
  5. The equality holds, so the triangle is a right triangle (right angle at \(A\)).

Answer: Yes, the triangle is a right triangle, with the right angle at vertex \(A\).

Example 3: Midpoint of a Segment and Length of a Median

Problem: In a triangle with vertices \(A(0, 0)\), \(B(8, 0)\), \(C(4, 6)\), find the midpoint of side \(AB\) and the length of the median from \(C\) to \(AB\).

Solution:

  1. Midpoint of \(AB\): \(M = \left(\dfrac{0+8}{2}, \dfrac{0+0}{2}\right) = (4, 0)\).
  2. The median from \(C\) is segment \(CM\), i.e. the distance from \(C(4,6)\) to \(M(4,0)\).
  3. \(CM = \sqrt{(4-4)^2 + (0-6)^2} = \sqrt{0 + 36} = 6\).
  4. The median is vertical because both points share the same \(x = 4\).

Answer: The midpoint of \(AB\) is \((4, 0)\), and the length of the median from \(C\) is \(6\) units.

Example 4: Isosceles Triangle — Finding the Altitude to the Base

Problem: An isosceles triangle has legs of length \(5\) cm and a base of length \(6\) cm. Find the altitude to the base.

Solution:

  1. In an isosceles triangle the altitude to the base bisects it, creating a right triangle whose one leg is half the base \(= 3\).
  2. The equal side (leg of the isosceles triangle) is the hypotenuse \(= 5\), and the altitude \(h\) is the other leg.
  3. By the Pythagorean theorem: \(h^2 = 5^2 - 3^2 = 25 - 9 = 16\).
  4. Take the square root: \(h = \sqrt{16} = 4\) cm.

Answer: The altitude to the base is \(4\) cm.

Example 5: Area of a Triangle from a Side and Its Altitude

Problem: Find the area of a triangle in which one side is \(12\) cm and the altitude to that side is \(7\) cm.

Solution:

  1. Triangle area formula: \(S = \dfrac{1}{2}\cdot \text{base} \cdot \text{height}\).
  2. Substitute base \(= 12\) and height \(= 7\): \(S = \dfrac{1}{2}\cdot 12\cdot 7\).
  3. Calculate: \(\dfrac{1}{2}\cdot 84 = 42\).

Answer: The area is \(42\) cm\(^2\).

Common Mistakes

✗ Common mistake: In the distance formula, multiplying \((x_2 - x_1)\cdot(y_2 - y_1)\) instead of summing their squares.

✓ The correct way: The formula is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) — square each difference separately and then add. This is essentially the Pythagorean theorem in the coordinate plane.

✗ Common mistake: Making a sign error when subtracting negative coordinates, e.g. writing \(1 - (-3) = -2\) instead of \(4\).

✓ The correct way: Subtracting a negative number is the same as adding: \(1 - (-3) = 1 + 3 = 4\). Squaring will eliminate the sign anyway, but it is important to be precise.

✗ Common mistake: Confusing a median with an altitude — computing the distance to the midpoint of a side when an altitude is required.

✓ The correct way: A median goes to the midpoint of the opposite side; an altitude is perpendicular to a side and goes to the foot of the perpendicular. They coincide only in isosceles triangles (to the base) and equilateral triangles.

Practice Tips

  • Tip — The distance formula is Pythagoras: \(\Delta x\) and \(\Delta y\) are the legs, and the distance is the hypotenuse. Picture a right triangle between the two points.
  • Tip — For the midpoint, simply take the average: the \(x\)-coordinate of the midpoint is the average of the \(x\)-values, and similarly for \(y\).
  • Tip — The midsegment is always parallel to the third side and exactly half its length, so it is never longer than that side.
  • Tip — In a right triangle the orthocentre is exactly at the vertex of the right angle — a useful fact to remember in exams.
  • Tip — To classify a triangle, always compute all three side lengths before drawing conclusions, and use the converse of the Pythagorean theorem to check for a right angle.

Summary and Key Formulas

Distance: \(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Midpoint: \(\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)\).

  • Median — from a vertex to the midpoint of the opposite side (3 medians meeting at the centroid).
  • Altitude — perpendicular to a side; altitudes meet at the orthocentre (at the right-angle vertex in a right triangle).
  • Midsegment — parallel to the third side and equal to half its length.
  • Triangle type: compare side lengths; converse of Pythagoras \(\Rightarrow\) right triangle.
  • Area: \(S = \dfrac{1}{2}\cdot \text{base}\cdot \text{height}\).