Coordinate geometry formulas — distance, midpoint, section, area in one page

Question

Summarise all coordinate geometry formulas — distance, midpoint, section formula, and area of triangle — and show when to use each. Given $A(2, 3)$ and $B(8, 11)$ , find the distance, midpoint, and the point dividing $AB$ in ratio $3:1$ .

(CBSE 10-11 Board + competitive exams)

Solution — Step by Step

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

For $A(2, 3)$ and $B(8, 11)$ :

AB = \sqrt{(8-2)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}

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

M = \left(\frac{2 + 8}{2}, \frac{3 + 11}{2}\right) = \mathbf{(5, 7)}

Point dividing $(x_1, y_1)$ and $(x_2, y_2)$ in ratio $m:n$ :

P = \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right)

For ratio $3:1$ :

P = \left(\frac{3(8) + 1(2)}{4}, \frac{3(11) + 1(3)}{4}\right) = \left(\frac{26}{4}, \frac{36}{4}\right) = \mathbf{(6.5, 9)}

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

Use this when three vertices are given. If the area equals 0, the three points are collinear.

flowchart TD
    A["Coordinate Geometry Problem"] --> B{"What is asked?"}
    B -- "Length/Distance" --> C["Distance Formula"]
    B -- "Middle point" --> D["Midpoint Formula"]
    B -- "Point in given ratio" --> E["Section Formula"]
    B -- "Area of triangle" --> F["Area Formula"]
    B -- "Check collinearity" --> G["Area = 0?"]
    E --> H{"Internal or External?"}
    H -- Internal --> I["Use + in denominator: m + n"]
    H -- External --> J["Use - in denominator: m - n"]

Why This Works

All these formulas come from the Pythagorean theorem applied to the coordinate plane. The distance formula is literally Pythagoras — the horizontal difference is one leg, the vertical difference is the other, and the distance is the hypotenuse.

The midpoint is the average of coordinates. The section formula generalises this — instead of equal weights (average), we use weighted average based on the ratio. The area formula comes from the cross-product interpretation of vectors formed by the vertices.

Alternative Method

For the area of a triangle, you can also use the shoelace formula (same formula, different arrangement):

Write the vertices in a column, repeat the first at the bottom:

\frac{1}{2}|x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_1 - x_1 y_3|

Cross-multiply diagonally: products going down-right are positive, products going down-left are negative. Take half the absolute difference.

The midpoint is just the section formula with ratio $1:1$ . If you remember the section formula well, you never need to memorise the midpoint formula separately. Similarly, for external division, just change the sign: replace $m + n$ with $m - n$ in the denominator.

Common Mistake

In the section formula, students swap $m$ and $n$ — applying the ratio weight to the wrong point. The rule: the weight $m$ goes with the FARTHER point and $n$ goes with the NEARER point. If $P$ divides $AB$ in ratio $m:n$ , then $m$ multiplies coordinates of $B$ (the point farther from $A$ ), and $n$ multiplies coordinates of $A$ . Drawing a quick sketch with the ratio marked helps avoid this swap.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next