Coordinate Geometry — Distance, Section Formula, Area of Triangle

What Is Coordinate Geometry Really About?

Coordinate geometry gives us a way to convert geometric problems into algebraic ones. Every point gets an address $(x, y)$ , and from there we can compute distances, divide line segments, and find areas — all with formulas. For Classes 10 and 11, this chapter is a consistent scoring opportunity in CBSE boards and competitive exams.

The beauty is in precision. No more “roughly equal” or “looks like” — we prove things with exact calculations. Three formulas do most of the heavy lifting here: Distance Formula, Section Formula, and the Area of a Triangle formula.

graph TD
    A[Coordinate Geometry Problem] --> B{What do we need?}
    B -->|Distance between points| C[Distance Formula]
    B -->|Point dividing a segment| D[Section Formula]
    B -->|Area of triangle/polygon| E[Area Formula]
    B -->|Midpoint of segment| F[Midpoint Formula]
    B -->|Collinearity check| G[Area = 0 Test]
    C --> H[√ of sum of squares]
    D --> I[Internal or External?]
    I -->|Internal| J[m₁x₂+m₂x₁ / m₁+m₂]
    I -->|External| K[m₁x₂−m₂x₁ / m₁−m₂]
    E --> L[½|x₁ y₂−y₃ + ...|]
    F --> M[Special case: m=1:1]

Key Terms & Definitions

Cartesian Plane — The $xy$ -plane formed by two perpendicular number lines (axes). The horizontal axis is the $x$ -axis, the vertical is the $y$ -axis, and they meet at the origin $O(0,0)$ .

Coordinates — The ordered pair $(x, y)$ that uniquely identifies a point. The $x$ -value is the abscissa, the $y$ -value is the ordinate.

Quadrants — The four regions of the Cartesian plane. In Quadrant I, both $x$ and $y$ are positive. In Quadrant II, $x$ is negative, $y$ is positive. In Quadrant III, both are negative. In Quadrant IV, $x$ is positive, $y$ is negative.

Collinear Points — Three or more points that lie on the same straight line. We test this by checking if the area of the triangle they form is zero.

The Three Core Formulas

1. Distance Formula

For two points $A(x_1, y_1)$ and $B(x_2, y_2)$ :

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

This comes directly from the Pythagorean theorem — the horizontal and vertical differences form the two legs of a right triangle, and the distance is the hypotenuse.

When to use: Finding lengths of sides, checking if a triangle is equilateral/isosceles/right-angled, verifying if a quadrilateral is a parallelogram or rhombus.

Worked Example (Easy — CBSE):

Find the distance between $P(3, 4)$ and $Q(-1, 1)$ .

d = \sqrt{(-1-3)^2 + (1-4)^2} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5}

2. Section Formula

If point $P$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m:n$ internally:

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

Special case — Midpoint (when $m:n = 1:1$ ):

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

A quick memory trick: in the section formula, the ratio component $m$ multiplies the coordinates of the far point $B$ , and $n$ multiplies the near point $A$ . Think “cross-multiplication” — $m$ goes with $B$ , $n$ goes with $A$ .

Worked Example (Medium — JEE Main):

Find the point that divides the join of $A(2, -3)$ and $B(6, 5)$ in ratio $3:1$ internally.

x = \frac{3(6) + 1(2)}{3+1} = \frac{20}{4} = 5

y = \frac{3(5) + 1(-3)}{3+1} = \frac{12}{4} = 3

The point is $\mathbf{(5, 3)}$ .

3. Area of a Triangle

For a triangle with vertices $A(x_1, y_1)$ , $B(x_2, y_2)$ , $C(x_3, y_3)$ :

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

If the area equals zero, the three points are collinear.

Worked Example (Medium — CBSE):

Find the area of the triangle with vertices $A(1, 2)$ , $B(4, 6)$ , $C(7, 2)$ .

\text{Area} = \frac{1}{2} |1(6-2) + 4(2-2) + 7(2-6)|

= \frac{1}{2} |4 + 0 - 28| = \frac{1}{2} \times 24 = \mathbf{12 \text{ sq. units}}

Solved Examples — Easy to Hard

Example 1 (Easy — CBSE)

Show that $A(1,1)$ , $B(3,3)$ , $C(5,5)$ are collinear.

Area $= \frac{1}{2}|1(3-5) + 3(5-1) + 5(1-3)| = \frac{1}{2}|-2 + 12 - 10| = \frac{1}{2}(0) = 0$

Since area = 0, the points are collinear.

Example 2 (Medium — JEE Main)

Find the ratio in which the line $3x + y - 9 = 0$ divides the segment joining $A(1,3)$ and $B(2,7)$ .

Let the line divide $AB$ in ratio $k:1$ . The dividing point is:

P = \left(\frac{2k+1}{k+1},\ \frac{7k+3}{k+1}\right)

Since $P$ lies on $3x + y - 9 = 0$ :

3\cdot\frac{2k+1}{k+1} + \frac{7k+3}{k+1} - 9 = 0

\frac{6k+3+7k+3}{k+1} = 9 \implies 13k + 6 = 9k + 9 \implies 4k = 3 \implies k = \frac{3}{4}

The ratio is $\mathbf{3:4}$ .

Example 3 (Hard — JEE Advanced)

The vertices of a triangle are $A(at_1^2, 2at_1)$ , $B(at_2^2, 2at_2)$ , $C(at_3^2, 2at_3)$ , where these lie on the parabola $y^2 = 4ax$ . Find the area of the triangle.

\text{Area} = \frac{1}{2}|at_1^2(2at_2 - 2at_3) + at_2^2(2at_3 - 2at_1) + at_3^2(2at_1 - 2at_2)|

= \frac{1}{2} \cdot 2a^2 |t_1^2(t_2 - t_3) + t_2^2(t_3 - t_1) + t_3^2(t_1 - t_2)|

= a^2 |(t_1 - t_2)(t_2 - t_3)(t_3 - t_1)|

This elegant result shows the area depends only on the parameter differences.

Exam-Specific Tips

CBSE Boards: This chapter carries 6-8 marks (Class 10). Questions are very formula-driven — 1 question on distance, 1 on section formula, and 1 on area/collinearity. Practise the standard types and you’ll score full marks.

JEE Main: Coordinate geometry in 2D is a massive chapter for JEE (with straight lines, circles, conics). The basics from Class 10 form the foundation. Section formula questions often appear combined with locus problems.

ICSE: The pattern is nearly identical to CBSE. Additionally, ICSE sometimes asks for the centroid $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ — know this formula cold.

Common Mistakes to Avoid

Mistake 1 — Sign errors in the distance formula. When subtracting coordinates like $(-3) - (5)$ , students write $-3 - 5 = -2$ instead of $-8$ . Since it gets squared, the answer is wrong but not obviously so. Always be careful with negatives.

Mistake 2 — Forgetting the modulus in the area formula. The area formula can give a negative value depending on the order of vertices. That’s why we take the absolute value. Without it, you might report a negative area.

Mistake 3 — Confusing internal and external division. In external division, the formula has a minus sign in the denominator: $(mx_2 - nx_1)/(m - n)$ . Using the internal formula for external division gives a wrong point.

Mistake 4 — Ratio vs. distances. If a point $P$ divides $AB$ in ratio $3:1$ , then $AP:PB = 3:1$ , meaning $P$ is closer to $B$ . Students sometimes reverse which segment is which.

Mistake 5 — Not converting units consistently. In word problems, ensure both coordinates use the same units before applying formulas.

Practice Questions

Q1. Find the distance between $(-2, 3)$ and $(4, -1)$ .

$d = \sqrt{(4-(-2))^2 + (-1-3)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$

Q2. Find the midpoint of the segment joining $(7, -4)$ and $(3, 2)$ .

$M = \left(\frac{7+3}{2}, \frac{-4+2}{2}\right) = (5, -1)$

Q3. In what ratio does the point $(3, 5)$ divide the join of $(1, 3)$ and $(5, 7)$ ?

Let the ratio be $k:1$ . Then $3 = \frac{5k + 1}{k+1}$ , giving $3k + 3 = 5k + 1$ , so $k = 1$ . The ratio is $1:1$ — it’s the midpoint.

Q4. Show that the points $(1, -1)$ , $(5, 2)$ , $(9, 5)$ are collinear.

Area $= \frac{1}{2}|1(2-5) + 5(5-(-1)) + 9(-1-2)| = \frac{1}{2}|-3 + 30 - 27| = 0$ . Since area = 0, the points are collinear.

Q5. Find the area of the quadrilateral with vertices $(1, 1)$ , $(7, -3)$ , $(12, 2)$ , $(7, 21)$ .

Split into two triangles using diagonal from $(1,1)$ to $(12,2)$ . Triangle 1: $(1,1), (7,-3), (12,2)$ . Area $= \frac{1}{2}|1(-3-2) + 7(2-1) + 12(1+3)| = \frac{1}{2}|-5+7+48| = 25$ . Triangle 2: $(1,1), (12,2), (7,21)$ . Area $= \frac{1}{2}|1(2-21) + 12(21-1) + 7(1-2)| = \frac{1}{2}|-19+240-7| = 107$ . Total = $\mathbf{132}$ sq. units.

Q6. Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in ratio $2:3$ .

$x = \frac{2(4) + 3(-1)}{2+3} = \frac{5}{5} = 1$ , $y = \frac{2(-3) + 3(7)}{5} = \frac{15}{5} = 3$ . The point is $(1, 3)$ .

Q7. The centroid of a triangle is $(3, -2)$ . Two vertices are $(5, 1)$ and $(-1, -4)$ . Find the third vertex.

Centroid $= \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ . So $3 = \frac{5+(-1)+x_3}{3}$ gives $x_3 = 5$ . And $-2 = \frac{1+(-4)+y_3}{3}$ gives $y_3 = -3$ . Third vertex is $(5, -3)$ .

Q8. For what value of $k$ are the points $(7, -2)$ , $(5, 1)$ , $(3, k)$ collinear?

Area $= 0$ : $\frac{1}{2}|7(1-k) + 5(k+2) + 3(-2-1)| = 0$ . Simplify: $7 - 7k + 5k + 10 - 9 = 0$ , so $-2k + 8 = 0$ , giving $k = 4$ .

FAQs

What is the difference between distance formula and section formula?

The distance formula tells you how far apart two points are — it gives a length. The section formula tells you where a point is located that divides a segment in a given ratio — it gives coordinates. They answer completely different questions.

How do we check if four points form a parallelogram?

Calculate all four side lengths using the distance formula. If opposite sides are equal ( $AB = CD$ and $BC = DA$ ), it’s a parallelogram. Alternatively, check if the diagonals bisect each other — find midpoints of both diagonals and see if they’re the same point.

Why does area = 0 mean collinear?

Three collinear points lie on a single straight line. A “triangle” formed by three points on a line has zero width — it’s flat. Hence, its area is zero. This is a reliable algebraic test for collinearity.

What is the external section formula?

When a point divides a line segment externally in ratio $m:n$ , the formula uses subtraction: $P = \left(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}\right)$ . The point lies on the line but outside the segment.

Can the distance formula give a negative answer?

No. The formula involves squared differences (always non-negative) under a square root (always non-negative). Distance is always $\geq 0$ , and it equals 0 only when both points are the same.

How is coordinate geometry used in JEE?

In JEE, the basics from Class 10 (distance, section, area formulas) are building blocks for advanced topics like straight lines, circles, parabolas, ellipses, and hyperbolas. Almost every conic sections problem uses these foundational formulas.

What is the formula for the centroid?

The centroid of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ is $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$ . It’s the point where all three medians intersect.

How do we find the area of a polygon using coordinates?

Use the Shoelace formula. For a polygon with $n$ vertices listed in order, the area is $\frac{1}{2}|x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + \ldots + x_n y_1 - x_1 y_n|$ . This generalizes the triangle formula to any polygon.