Area of a Triangle and Collinearity

Area of a Triangle from Vertices

Given a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, its area is

$$\text{Area} = \frac{1}{2}\left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

The vertical bars mean absolute value — area is always taken as positive.

Key Point: Each x-coordinate is multiplied by the difference of the other two y-coordinates, in cyclic order $1\to2\to3$. Always take the modulus at the end.

[Board Important] This area formula is removed from the rationalised NCERT but is retained by many State Boards, so we include it fully. Write the formula first, then substitute.

Collinearity Test (Zero Area)

Three points are collinear (lie on one straight line) exactly when the 'triangle' they form has zero area:

$$x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$$

This is far faster than the distance method from Section 2.

Key Point: Collinear $\iff$ area $= 0$. Drop the $\tfrac12$ and the modulus — just set the bracket to 0.

[Board Important] A common question: 'Find $k$ so that the three points are collinear.' Set the area expression to 0 and solve for $k$.

Area of a Quadrilateral

To find the area of a quadrilateral $ABCD$, split it along a diagonal into two triangles (say $ABC$ and $ACD$), find each area, and add:

$$\text{Area}(ABCD) = \text{Area}(ABC) + \text{Area}(ACD)$$

Make sure you take the vertices in order (going around the quadrilateral), or the split won't be valid.

Key Point: Split a quadrilateral into two triangles using a diagonal, then sum the two areas.

[JEE Tip] For polygons, there's a 'shoelace' shortcut that generalises this, but for boards the two-triangle split is enough.

Reading the Area Result

A few quick checks help avoid errors:

• If the area comes out 0, the three points are collinear (no genuine triangle).
• The formula gives the same answer regardless of the order you list the vertices, because of the modulus.
• Always include the $\tfrac12$ — forgetting it doubles the area.

Key Point: Zero area = collinear; non-zero area = a real triangle. Keep the $\tfrac12$ and the modulus.

[Board Important] Units of area are square units (e.g., sq. units or cm²) — state them in the final answer.

Solved Examples

Example 1: Area of a triangle

Find the area of the triangle with vertices $A(1, 1)$, $B(4, 2)$, $C(3, 5)$.

Solution:

1. Area $= \tfrac12 |1(2-5) + 4(5-1) + 3(1-2)|$.
2. $= \tfrac12 |1(-3) + 4(4) + 3(-1)| = \tfrac12|{-3 + 16 - 3}| = \tfrac12(10) = 5$.

Final Answer: 5 square units.

Takeaway: Substitute carefully and take the modulus before halving.

Example 2: Collinearity check

Are $A(1, 2)$, $B(2, 4)$, $C(3, 6)$ collinear?

Solution:

1. Expression $= 1(4-6) + 2(6-2) + 3(2-4)$.
2. $= 1(-2) + 2(4) + 3(-2) = -2 + 8 - 6 = 0$.

Final Answer: Yes, collinear (zero area).

Takeaway: A zero result means the points lie on one line.

Example 3: Find k for collinearity

Find $k$ so that $A(2, 3)$, $B(4, k)$, $C(6, -3)$ are collinear.

Solution:

1. Set area expression to 0: $2(k+3) + 4(-3-3) + 6(3-k) = 0$.
2. $2k + 6 - 24 + 18 - 6k = 0 \Rightarrow -4k + 0 = 0 \Rightarrow k = 0$.

Final Answer: $k = 0$.

Takeaway: Set the area bracket to zero and solve for the unknown.

Example 4: Triangle with a vertex at the origin

Find the area of the triangle with vertices $O(0, 0)$, $A(4, 0)$, $B(0, 6)$.

Solution:

1. Area $= \tfrac12|0(0-6) + 4(6-0) + 0(0-0)| = \tfrac12|24| = 12$.

Final Answer: 12 square units.

Takeaway: For a right triangle on the axes, area $= \tfrac12 \times$ base $\times$ height $= \tfrac12(4)(6) = 12$ — a good cross-check.

Example 5: Area of a quadrilateral

Find the area of quadrilateral $ABCD$ with $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$, $D(2, 3)$.

Solution:

1. Split into $\triangle ABC$ and $\triangle ACD$.
2. $\triangle ABC = \tfrac12|{-4(-5+2) -3(-2+2) +3(-2+5)}| = \tfrac12|12 - 0 + 9| = \tfrac{21}{2}$.
3. $\triangle ACD = \tfrac12|{-4(-2-3) +3(3+2) +2(-2+2)}| = \tfrac12|20 + 15 + 0| = \tfrac{35}{2}$.
4. Total $= \tfrac{21}{2} + \tfrac{35}{2} = 28$.

Final Answer: 28 square units.

Takeaway: Add the two triangle areas obtained from a diagonal split.

Example 6: Verify collinearity by area

Show that $A(-5, 1)$, $B(1, -1)$, $C(4, -2)$ are collinear.

Solution:

1. Expression $= -5(-1+2) + 1(-2-1) + 4(1+1)$.
2. $= -5(1) + 1(-3) + 4(2) = -5 - 3 + 8 = 0$.

Final Answer: Collinear.

Takeaway: Zero area confirms collinearity quickly.

Example 7: Area with a negative coordinate

Find the area of the triangle $A(2, -1)$, $B(3, 4)$, $C(-2, 3)$.

Solution:

1. Area $= \tfrac12|2(4-3) + 3(3+1) + (-2)(-1-4)|$.
2. $= \tfrac12|2 + 12 + 10| = \tfrac12(24) = 12$.

Final Answer: 12 square units.

Takeaway: Mind the signs carefully with negative coordinates.

Example 8: Find k given area

If the area of the triangle with vertices $(1, -1)$, $(-4, 2k)$, $(-k, -5)$ is 24 square units, set up the equation for $k$ (one value).

Solution:

1. Area $= \tfrac12|1(2k+5) -4(-5+1) -k(-1-2k)|= 24$.
2. $|2k + 5 + 16 + k + 2k^2| = 48$.
3. $|2k^2 + 3k + 21| = 48$. Taking $2k^2 + 3k + 21 = 48$: $2k^2 + 3k - 27 = 0 \Rightarrow (2k+9)(k-3)=0$.
4. $k = 3$ or $k = -\tfrac92$.

Final Answer: $k = 3$ (or $-\tfrac92$).

Takeaway: Set the area formula equal to the given value and solve the resulting quadratic.

Example 9: Median divides into equal areas

The vertices of $\triangle ABC$ are $A(4, -6)$, $B(3, -2)$, $C(5, 2)$. Verify that the median from $A$ divides it into two triangles of equal area.

Solution:

1. Midpoint of $BC$: $D = (4, 0)$.
2. $\triangle ABD = \tfrac12|4(-2-0) + 3(0+6) + 4(-6+2)| = \tfrac12|{-8 + 18 - 16}| = \tfrac12(6) = 3$.
3. $\triangle ACD = \tfrac12|4(2-0) + 5(0+6) + 4(-6-2)| = \tfrac12|8 + 30 - 32| = \tfrac12(6) = 3$.

Final Answer: Both areas $= 3$ — equal.

Takeaway: A median always bisects the area of a triangle.

Example 10: Collinear, so no triangle

What is the area of the 'triangle' with vertices $(2, 3)$, $(4, 6)$, $(6, 9)$?

Solution:

1. Area $= \tfrac12|2(6-9) + 4(9-3) + 6(3-6)| = \tfrac12|{-6 + 24 - 18}| = 0$.

Final Answer: 0 — the points are collinear, so no triangle is formed.

Takeaway: Zero area signals that the three points lie on a straight line.