Coordinate Geometry · Topic 3 of 6

🔺 Triangle Area & Collinearity

2 exam question types, fully solved

One formula gives the area of a triangle straight from its three corner points — no base or height needed. And the moment that area comes out as zero, the three points must lie on a single straight line.

The shoelace formula

• Area of the triangle with vertices is . Keep the absolute value — area is never negative.

• Collinearity. If that area is , the three points are collinear (they make no triangle). This is the fastest collinearity test.

Tip: if a vertex is the origin the formula shrinks to .

📐Coordinate Plane ExplorerLive calculations as you move points

P1 (x₁, y₁)

x₁

y₁

P2 (x₂, y₂)

x₂

y₂

Distance P1P2√((4−1)² + (5−2)²) = √(9+9)4.243 units

Midpoint M((1+4)/2, (2+5)/2)(2.5, 3.5)

Slope (m)(5−2) / (4−1) = 3/31

Line Equationy = mx + c → y = 1x + (1)y = 1x + 1

Triangle (0, 0), (4, 0), (0, 3) has area .

Q1The area of the triangle with vertices (0, 0), (4, 0) and (0, 6) is:

Q2The points (1, 1), (2, 2) and (3, 3) are:

Q3The area of the triangle with vertices (0, 0), (0, 5) and (5, 0) is:

📚

Real exam questions — Triangle Area & Collinearity

2 question types · 8 solved examples from real SSC papers

Coordinate-geometry area questions reward pattern-spotting: if vertices share a coordinate or sit on the axes, the triangle is right-angled, so the area is just half base $\times$ height. Collinearity is the same idea inside-out — three points line up exactly when the triangle they form has zero area.

How to solve this type

The all-purpose formula for vertices

(x_1,y_1),(x_2,y_2),(x_3,y_3)

is Area

=\dfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

. But the real time-saver: if two vertices share an

x

(or a

y

), or the triangle sits on the axes, it is right-angled, so Area

=\dfrac12\times\text{base}\times\text{height}

. For a line with both axes, the base and height are simply its intercepts (set

y=0

for the

x

-intercept,

x=0

for the

y

-intercept).

Find the area of the triangle with vertices $(1, 2)$ , $(4, 2)$ and $(1, 6)$ .

6

5

7

4

Find the area (in square units) of the triangle formed by the line $x+y=4$ and the two coordinate axes.

8

12

4

16

The lines $8x+3y=24$ , $y=2x+8$ and the $x$ -axis form a triangle. Find its area (in square units).

15

28

14

24

The line $5x+12y=60$ forms a triangle with the two axes. Find its area and circumradius $R$ .

AArea

=36

R=6.5

BArea

=30

R=6.5

CArea

=30

R=5

DArea

=60

R=13