Coordinate Geometry • Topic 3 of 3

Area of Triangle, Midpoint, and Reflection of Points

What is the area of a triangle using coordinates? For a triangle with vertices A($x_{1}$, $y_{1}$), B($x_{2}$, $y_{2}$), and C($x_{3}$, $y_{3}$), the area is:

Area = (1/2) | $x_{1}$($y_{2} - y_{3}$) + $x_{2}$($y_{3} - y_{1}$) + $x_{3}$($y_{1} - y_{2}$) |

The vertical bars | | mean absolute value (area is always positive). If the result is 0, the three points are collinear (lie on a straight line).

What is the reflection of a point in coordinate axes? Reflection means flipping a point over a line (like a mirror). The key rules:

Reflection in	Original (x, y)	Reflected point
x-axis	(x, y)	(x, −y)
y-axis	(x, y)	(−x, y)
Origin	(x, y)	(−x, −y)
Line y = x	(x, y)	(y, x)

Real-life analogy: Reflection is like looking at your face in a mirror — the mirror is the axis. When you raise your left hand, your reflection raises its right hand!

┌─────────────────────────────────────────────────────────────┐
│         AREA OF TRIANGLE & REFLECTION OF POINTS              │
└─────────────────────────────────────────────────────────────┘

AREA OF TRIANGLE FORMULA:

    A(x₁, y₁)
       /\
      /  \
     /    \
    /      \
   /________\
B(x₂, y₂)   C(x₃, y₃)

Area = ½ |x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

EXAMPLE: A(1,1), B(4,1), C(4,5)

Area = ½ |1(1-5) + 4(5-1) + 4(1-1)|
     = ½ |1(-4) + 4(4) + 4(0)|
     = ½ |-4 + 16 + 0| = ½ × 12 = 6 sq units ✓


REFLECTION OF POINTS IN AXES:

    ORIGINAL POINT P(3, 4)
    
    Reflection in x-axis:
        P'(3, -4)
        
    Reflection in y-axis:
        P''(-3, 4)
        
    Reflection in origin:
        P'''(-3, -4)


REFLECTION VISUAL ON GRID:

          y
          │
     (-3,4)│   (3,4)
         \ │ /
          \│/
    ────────┼──────── x
          /│\
         / │ \
    (-3,-4)│  (3,-4)
          │
          
    Each point's reflection across axes forms a rectangle!


MIDPOINT AS AVERAGE COORDINATES:

    A(2, 3) and B(10, 7)
    
    Midpoint M = ( (2+10)/2, (3+7)/2 ) = (6, 5)
    
    This is the "average" of the two endpoints!

Solved Examples

Worked Example

Find the area of triangle with vertices A(1, 2), B(4, 5), and C(7, 8).

Solution

Step 1: Use area formula: ½ |$x_{1}$($y_{2}-y_{3}$) + $x_{2}$($y_{3}-y_{1}$) + $x_{3}$($y_{1}-y_{2}$)|
Step 2: Substitute: ½ |1(5−8) + 4(8−2) + 7(2−5)|
Step 3: Simplify: ½ |1(−3) + 4(6) + 7(−3)| = ½ |−3 + 24 − 21|
Step 4: ½ |0| = 0

Answer: Area = 0 → points are collinear

Worked Example

Find the reflection of point P(5, −3) in the x-axis and in the y-axis.

Solution

Step 1: Reflection in x-axis: (x, y) → (x, −y)
Step 2: P(5, −3) → $P_{1}$(5, 3)
Step 3: Reflection in y-axis: (x, y) → (−x, y)
Step 4: P(5, −3) → $P_{2}$(−5, −3)

Answer: In x-axis: (5, 3); in y-axis: (−5, −3)

Worked Example

If the midpoint of a line segment joining A(2a, 4) and B(−2, 3b) is C(1, 2a+1), find a and b.

Solution

Step 1: Midpoint formula: ((2a + (−2))/2, (4 + 3b)/2) = (1, 2a+1)
Step 2: x-coordinate: (2a − 2)/2 = 1 → 2a − 2 = 2 → 2a = 4 → a = 2
Step 3: y-coordinate: (4 + 3b)/2 = 2a + 1 = 2(2) + 1 = 5
Step 4: 4 + 3b = 10 → 3b = 6 → b = 2

Answer: a = 2, b = 2

Key Points

Area of triangle = ½ |$x_{1}$($y_{2}-y_{3}$) + $x_{2}$($y_{3}-y_{1}$) + $x_{3}$($y_{1}-y_{2}$)|
If area = 0, points are collinear
Reflection in x-axis: (x, y) → (x, −y)
Reflection in y-axis: (x, y) → (−x, y)
Reflection in origin: (x, y) → (−x, −y)
Midpoint = average of coordinates = (($x_{1}+x_{2}$)/2, ($y_{1}+y_{2}$)/2)

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.The area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is:

Explanation: Coordinate area formula.

Q2.Three points are collinear if the area of the triangle they form is:

Explanation: Zero area means collinear.

Q3.The centroid of a triangle is:

Explanation: Average of the three vertices.

Q4.The reflection of $(3,5)$ in the x-axis is:

Explanation: $y\to-y$.

Practice more MCQs → 📝 Assignment · 35 marks