Applications and Special Points

The Centroid of a Triangle

The centroid $G$ is the point where the three medians of a triangle meet. For vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$:

$$G = \left( \frac{x_1 + x_2 + x_3}{3},\; \frac{y_1 + y_2 + y_3}{3} \right)$$

Think of it this way: the centroid is just the average of the three vertices.

Key Point: Centroid = average of the three vertices. The centroid divides each median in the ratio $2 : 1$ from the vertex.

[Board Important] Don't confuse the centroid (average of vertices) with the midpoint (average of two points). A frequent error is dividing by 2 instead of 3.

Identifying Types of Triangles

Compute the three side lengths with the distance formula, then classify:

• Equilateral: all three sides equal.
• Isosceles: exactly two sides equal.
• Scalene: all sides different.
• Right-angled: the longest side squared equals the sum of squares of the other two.

A triangle can be both isosceles and right-angled (an isosceles right triangle).

Key Point: Side lengths alone classify a triangle; add the Pythagoras check to detect a right angle.

[Board Important] Always show the squared side lengths in your working — examiners want to see the comparison, not just the conclusion.

Identifying Types of Quadrilaterals

For a quadrilateral $ABCD$ (vertices in order), compute the four sides $AB$, $BC$, $CD$, $DA$ and the two diagonals $AC$, $BD$:

• Parallelogram: opposite sides equal ($AB = CD$, $BC = DA$).
• Rectangle: parallelogram and equal diagonals.
• Rhombus: all four sides equal (diagonals unequal).
• Square: all four sides equal and equal diagonals.

Key Point: Sides decide parallelogram vs rhombus; diagonals then decide rectangle vs square.

[JEE Tip] Equal diagonals signal a rectangle/square; perpendicular diagonals signal a rhombus/square. Combine both facts for a full classification.

Coordinate Geometry Word Problems

Many applied problems reduce to choosing the right formula:

• 'Equidistant from two points' ⇒ set squared distances equal.
• 'Divides in a ratio' / 'trisection' ⇒ section formula.
• 'Find the missing vertex of a parallelogram' ⇒ equate diagonal midpoints.
• 'Area of a plot' ⇒ area-of-triangle (or split into triangles).
• 'Are these points in a line?' ⇒ zero-area collinearity test.

Key Point: Translate the words into one of the four core tools — distance, section/midpoint, area, or collinearity.

[Board Important] Read carefully whether the question wants a point, a ratio, a length, or an area — that decides which formula to use.

Solved Examples

Example 1: Centroid

Find the centroid of the triangle $A(1, 2)$, $B(3, 4)$, $C(5, 0)$.

Solution:

1. $G = \left( \dfrac{1+3+5}{3}, \dfrac{2+4+0}{3} \right) = (3, 2)$.

Final Answer: $(3, 2)$.

Takeaway: Average all three vertices — divide by 3, not 2.

Example 2: Third vertex from centroid

Two vertices of a triangle are $(2, 1)$ and $(4, 3)$, and its centroid is $(3, 2)$. Find the third vertex.

Solution:

1. $\dfrac{2 + 4 + x}{3} = 3 \Rightarrow 6 + x = 9 \Rightarrow x = 3$.
2. $\dfrac{1 + 3 + y}{3} = 2 \Rightarrow 4 + y = 6 \Rightarrow y = 2$.

Final Answer: $(3, 2)$.

Takeaway: Use the centroid equations to back out a missing vertex.

Example 3: Classify the triangle

Classify the triangle with vertices $A(3, 0)$, $B(6, 4)$, $C(-1, 3)$.

Solution:

1. $AB^2 = 3^2 + 4^2 = 25$; $BC^2 = 7^2 + 1^2 = 50$; $CA^2 = 4^2 + 3^2 = 25$.
2. $AB^2 = CA^2 = 25$ ⇒ isosceles.
3. $AB^2 + CA^2 = 25 + 25 = 50 = BC^2$ ⇒ right-angled at $A$.

Final Answer: Isosceles right triangle.

Takeaway: Equal sides + Pythagoras relation ⇒ isosceles right triangle.

Example 4: Show it's a square

Show that $A(1, 2)$, $B(5, 2)$, $C(5, 6)$, $D(1, 6)$ form a square.

Solution:

1. $AB = 4$, $BC = 4$, $CD = 4$, $DA = 4$ — all sides equal.
2. Diagonals: $AC = \sqrt{16+16} = 4\sqrt2$; $BD = \sqrt{16+16} = 4\sqrt2$ — equal.
3. Equal sides and equal diagonals ⇒ square.

Final Answer: $ABCD$ is a square.

Takeaway: All sides equal + equal diagonals ⇒ square.

Example 5: Parallelogram, not rectangle

Show that $A(0,0)$, $B(4,0)$, $C(6,3)$, $D(2,3)$ form a parallelogram but not a rectangle.

Solution:

1. $AB = 4$, $CD = 4$; $BC = \sqrt{4+9} = \sqrt{13}$, $DA = \sqrt{4+9} = \sqrt{13}$ ⇒ opposite sides equal, so parallelogram.
2. Diagonals: $AC = \sqrt{36+9} = \sqrt{45}$; $BD = \sqrt{4+9} = \sqrt{13}$ ⇒ unequal.
3. Diagonals unequal ⇒ not a rectangle.

Final Answer: Parallelogram, not a rectangle.

Takeaway: Equal opposite sides ⇒ parallelogram; equal diagonals are needed for a rectangle.

Example 6: Centroid lies on a median

The vertices of $\triangle ABC$ are $A(0, 6)$, $B(8, 12)$, $C(8, 0)$. Find the centroid and the midpoint of $BC$.

Solution:

1. Centroid $G = \left( \dfrac{0+8+8}{3}, \dfrac{6+12+0}{3} \right) = \left( \dfrac{16}{3}, 6 \right)$.
2. Midpoint of $BC = (8, 6)$.
3. $G$ lies on the median $AD$ where $D = (8, 6)$; it divides $AD$ as $2:1$.

Final Answer: $G = \left( \dfrac{16}{3}, 6 \right)$, $D = (8, 6)$.

Takeaway: The centroid sits two-thirds of the way along each median from the vertex.

Example 7: Rhombus check

Show that $A(3, 0)$, $B(4, 5)$, $C(-1, 4)$, $D(-2, -1)$ form a rhombus.

Solution:

1. $AB = \sqrt{1+25} = \sqrt{26}$; $BC = \sqrt{25+1} = \sqrt{26}$; $CD = \sqrt{1+25} = \sqrt{26}$; $DA = \sqrt{25+1} = \sqrt{26}$.
2. All four sides equal ⇒ rhombus.
3. (Diagonals: $AC = \sqrt{16+16} = \sqrt{32}$, $BD = \sqrt{36+36} = \sqrt{72}$ — unequal, so not a square.)

Final Answer: Rhombus (all sides equal, diagonals unequal).

Takeaway: All four sides equal but unequal diagonals ⇒ rhombus.

Example 8: Equidistant word problem

Find the point on the x-axis equidistant from $A(-2, 5)$ and $B(2, -3)$.

Solution:

1. Let $P(x, 0)$, then $PA^2 = PB^2$.
2. $(x+2)^2 + 25 = (x-2)^2 + 9$.
3. $x^2 + 4x + 4 + 25 = x^2 - 4x + 4 + 9$.
4. $8x = -16 \Rightarrow x = -2$.

Final Answer: $(-2, 0)$.

Takeaway: 'Equidistant' problems reduce to setting squared distances equal.

Example 9: Area of a triangular plot

A triangular park has corners at $(0, 0)$, $(6, 0)$, $(4, 5)$ (in units of 100 m). Find its area in square metres.

Solution:

1. Area (in units²) $= \tfrac12|0(0-5) + 6(5-0) + 4(0-0)| = \tfrac12|30| = 15$.
2. Each unit $= 100$ m, so 1 unit² $= 10000$ m².
3. Area $= 15 \times 10000 = 150000$ m².

Final Answer: $150000$ m² (15 hectares).

Takeaway: Compute the area in coordinate units, then scale by the real-world unit squared.

Example 10: Collinear vertices fail to form a triangle

Can $A(1, 4)$, $B(3, 8)$, $C(5, 12)$ be the vertices of a triangle?

Solution:

1. Area $= \tfrac12|1(8-12) + 3(12-4) + 5(4-8)| = \tfrac12|{-4 + 24 - 20}| = 0$.
2. Zero area ⇒ collinear.

Final Answer: No — they are collinear, so no triangle exists.

Takeaway: Always check that three given points are non-collinear before treating them as a triangle.