Coordinate Geometry: Distance Formula, Section Formula, and Midpoint Formula

Coordinate geometry (also known as analytical geometry) bridges the gap between algebra and geometry. Developed by the French philosopher and mathematician René Descartes in the 17th century, coordinate geometry allows us to locate points on a two-dimensional plane using ordered pairs of real numbers and analyze geometric figures using algebraic equations.

In CBSE Class 10 Mathematics, coordinate geometry forms a crucial segment of the curriculum. It enables us to calculate distances between spatial locations, determine points that divide lines in given ratios, and identify structural midpoints—tools essential for advanced mathematics, physics, engineering, computer graphics, and geographic information systems (GIS).

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1. In-Depth Conceptual Breakdown

1.1 The Cartesian Coordinate System Refresher

A Cartesian plane is formed by two mutually perpendicular number lines intersecting at their zero points:

• Horizontal axis ($x$-axis): Values to the right of the origin are positive; values to the left are negative.
• Vertical axis ($y$-axis): Values above the origin are positive; values below are negative.
• Origin ($O$): The point of intersection, denoted by coordinates $(0, 0)$.

Any point $P$ in the plane is represented as an ordered pair $(x, y)$, where:

• The abscissa ($x$-coordinate) represents the perpendicular distance of the point from the $y$-axis.
• The ordinate ($y$-coordinate) represents the perpendicular distance of the point from the $x$-axis.

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1.2 The Distance Formula

The distance formula allows us to find the straight-line distance between any two points in a coordinate plane without physical measurement.

Derivation of the Distance Formula

Consider two points $A(x_1, y_1)$ and $B(x_2, y_2)$ on a Cartesian plane.

       Y |
         |                  B(x₂, y₂)
         |                  /|
         |                 / |
         |                /  | |y₂ - y₁|
         |   A(x₁, y₁)   /   |
         |      +-------+----+ C(x₂, y₁)
         |      | |x₂ - x₁|
         |------+-------------------- X
         O |

1. Draw perpendiculars $AP$ and $BQ$ from $A$ and $B$ to the $x$-axis.
2. Draw a perpendicular $AC$ from $A$ to line $BQ$.
3. Notice that $\triangle ACB$ is a right-angled triangle, right-angled at $C$.
4. The horizontal distance $AC = |x_2 - x_1|$.
5. The vertical distance $BC = |y_2 - y_1|$.

By applying the Pythagoras Theorem in $\triangle ACB$:

$AB^2 = AC^2 + BC^2$

$AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Taking the positive square root (since distance is always non-negative):

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

Special Case: Distance from the Origin

The distance of a point $P(x, y)$ from the origin $O(0, 0)$ is:

$OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$

Geometric Applications of the Distance Formula

In the CBSE board examinations, the distance formula is frequently used to verify geometric properties:

1. Collinearity of Three Points: Points $A$, $B$, and $C$ are collinear if the sum of the lengths of any two segments equals the length of the third segment (e.g., $AB + BC = AC$).
2. Types of Triangles:
   • Equilateral Triangle: $AB = BC = CA$
   • Isosceles Triangle: Any two sides are equal ($AB = BC \neq CA$)
   • Scalene Triangle: $AB \neq BC \neq CA$
   • Right-Angled Triangle: Pythagoras theorem holds ($AB^2 + BC^2 = AC^2$)
3. Types of Quadrilaterals:
   • Parallelogram: Opposite sides are equal ($AB = CD$ and $BC = DA$).
   • Rectangle: Opposite sides are equal ($AB = CD$, $BC = DA$) AND diagonals are equal ($AC = BD$).
   • Rhombus: All four sides are equal ($AB = BC = CD = DA$).
   • Square: All four sides are equal ($AB = BC = CD = DA$) AND diagonals are equal ($AC = BD$).

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1.3 The Section Formula

The section formula determines the coordinates of a point $P$ that divides a line segment joining two given points $A$ and $B$ internally in a given ratio $m_1 : m_2$.

Internal Division Formula

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two given points. Let $P(x, y)$ be the point dividing the segment $AB$ internally in the ratio $m_1 : m_2$, such that:

$\frac{AP}{PB} = \frac{m_1}{m_2}$

The coordinates of point $P(x, y)$ are given by:

$P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$

The $k : 1$ Ratio Trick

When a problem asks to find the ratio in which a point divides a line segment, assume the ratio to be $k : 1$ (where $k = \frac{m_1}{m_2}$).

Substituting $m_1 = k$ and $m_2 = 1$ into the section formula yields:

$P(x, y) = \left( \frac{k x_2 + x_1}{k + 1}, \frac{k y_2 + y_1}{k + 1} \right)$

• If $P$ lies on the $x$-axis, its ordinate is zero ($y = 0$). Set the $y$-expression equal to 0 to solve for $k$.
• If $P$ lies on the $y$-axis, its abscissa is zero ($x = 0$). Set the $x$-expression equal to 0 to solve for $k$.

Points of Trisection

Trisection means dividing a line segment into three equal parts. This requires two interior points, say $P$ and $Q$.

• Point $P$ divides $AB$ internally in the ratio $1 : 2$.
• Point $Q$ divides $AB$ internally in the ratio $2 : 1$ (or $Q$ is the midpoint of $PB$).

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1.4 The Midpoint Formula

The midpoint formula is a special case of the section formula where the dividing point $M(x, y)$ bisects the line segment $AB$ into two equal halves.

Here, $m_1 : m_2 = 1 : 1$.

Substituting $m_1 = 1$ and $m_2 = 1$ into the section formula:

$M(x, y) = \left( \frac{1 \cdot x_2 + 1 \cdot x_1}{1 + 1}, \frac{1 \cdot y_2 + 1 \cdot y_1}{1 + 1} \right)$

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

Key Applications of the Midpoint Formula

1. Diagonals of Parallelograms / Rectangles / Rhombuses / Squares: The diagonals of these quadrilaterals bisect each other. Therefore: $\text{Midpoint of Diagonal } AC = \text{Midpoint of Diagonal } BD$
2. Center of a Circle: The center of a circle is the midpoint of any diameter $AB$.

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Formula Comparison Table

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2. Real-World Applications

Application 1: GPS Navigation and Mapping Systems

Modern digital maps (such as Google Maps) overlay coordinate systems onto geographic surfaces.

• The distance formula calculates the straight-line ("as the crow flies") displacement between your current position $(x_1, y_1)$ and your target destination $(x_2, y_2)$.
• The section formula helps navigation engines display progress along a path. If you have completed $70\%$ of your journey along a linear highway segment, the algorithm computes your current location using a ratio of $7 : 3$.

Application 2: Computer Graphics and Game Development

In 2D video game development, screen pixels operate on a Cartesian plane.

• Collision Detection: Game engines use the distance formula to compute distances between character hitboxes and enemy objects to trigger hits or interactions.
• Animation Interpolation: When a character moves smoothly from point $A$ to point $B$, developers use the midpoint and section formulas to calculate intermediate frame positions frame-by-frame.

Application 3: Structural Engineering and Civil Planning

Urban planners and civil engineers rely on coordinate geometry when placing infrastructure.

• To place a power supply station equi-distant between two towns, engineers calculate the midpoint of the linear grid connecting them.
• To locate a logistics hub that services two factories, where Factory A generates double the supply volume of Factory B, engineers use the section formula with ratio $1:2$ to minimize overall transport costs.

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3. Step-by-Step Solved Textbook Examples

Example 1: Verifying a Geometric Shape (Distance Formula)

Problem: Show that the points $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$, and $D(-4, 4)$ are the vertices of a square.

Solution:

To prove $ABCD$ is a square, we must show:

1. All four sides are equal in length ($AB = BC = CD = DA$).
2. Both diagonals are equal in length ($AC = BD$).

Step 1: Calculate the lengths of all four sides using $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

• Side $AB$: $AB = \sqrt{(4 - 1)^2 + (2 - 7)^2} = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$

• Side $BC$: $BC = \sqrt{(-1 - 4)^2 + (-1 - 2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$

• Side $CD$: $CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-4 + 1)^2 + (4 + 1)^2} = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34}$

• Side $DA$: $DA = \sqrt{(1 - (-4))^2 + (7 - 4)^2} = \sqrt{(1 + 4)^2 + (3)^2} = \sqrt{(5)^2 + 9} = \sqrt{25 + 9} = \sqrt{34}$

Since $AB = BC = CD = DA = \sqrt{34}$, all four sides are equal. This confirms the shape is at least a rhombus.

Step 2: Calculate the lengths of the diagonals.

• Diagonal $AC$: $AC = \sqrt{(-1 - 1)^2 + (-1 - 7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68}$

• Diagonal $BD$: $BD = \sqrt{(-4 - 4)^2 + (4 - 2)^2} = \sqrt{(-8)^2 + (2)^2} = \sqrt{64 + 4} = \sqrt{68}$

Since $AC = BD = \sqrt{68}$, the diagonals are equal.

Conclusion: Since all four sides are equal and both diagonals are equal, $ABCD$ is a square.

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Example 2: Finding a Ratio and Point of Intersection (Section Formula)

Problem: Find the ratio in which the $y$-axis divides the line segment joining the points $A(5, -6)$ and $B(-1, -4)$. Also, find the coordinates of the point of intersection.

Solution:

Step 1: Set up the ratio using $k : 1$. Let the required ratio be $k : 1$. Let the point of intersection on the $y$-axis be $P(0, y)$ (since any point on the $y$-axis has an $x$-coordinate of $0$).

Step 2: Apply the Section Formula. Here, $(x_1, y_1) = (5, -6)$ and $(x_2, y_2) = (-1, -4)$.

$P(x, y) = \left( \frac{k(-1) + 1(5)}{k + 1}, \frac{k(-4) + 1(-6)}{k + 1} \right)$

$P(x, y) = \left( \frac{-k + 5}{k + 1}, \frac{-4k - 6}{k + 1} \right)$

Step 3: Solve for $k$ using the known $x$-coordinate ($x = 0$). $\frac{-k + 5}{k + 1} = 0$

$-k + 5 = 0 \implies k = 5$

Therefore, the ratio $k : 1$ is $5 : 1$.

Step 4: Find the $y$-coordinate by substituting $k = 5$. $y = \frac{-4(5) - 6}{5 + 1} = \frac{-20 - 6}{6} = \frac{-26}{6} = -\frac{13}{3}$

Final Answer:

• The ratio in which the $y$-axis divides the segment is $5 : 1$.
• The point of intersection is $\left( 0, -\frac{13}{3} \right)$.

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Example 3: Finding Points of Trisection

Problem: Find the coordinates of the points of trisection of the line segment joining $A(2, -2)$ and $B(-7, 4)$.

Solution:

Trisection divides segment $AB$ into three equal parts $AP = PQ = QB$.

A(2, -2) ------ P(x, y) ------ Q(x', y') ------ B(-7, 4)

Step 1: Find Point $P$. Point $P$ divides $AB$ internally in the ratio $1 : 2$. Here, $m_1 = 1$, $m_2 = 2$, $(x_1, y_1) = (2, -2)$, $(x_2, y_2) = (-7, 4)$.

$P(x, y) = \left( \frac{1(-7) + 2(2)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2} \right)$

$P(x, y) = \left( \frac{-7 + 4}{3}, \frac{4 - 4}{3} \right) = \left( \frac{-3}{3}, \frac{0}{3} \right) = (-1, 0)$

Step 2: Find Point $Q$. Point $Q$ divides $AB$ internally in the ratio $2 : 1$. Here, $m_1 = 2$, $m_2 = 1$.

$Q(x', y') = \left( \frac{2(-7) + 1(2)}{2 + 1}, \frac{2(4) + 1(-2)}{2 + 1} \right)$

$Q(x', y') = \left( \frac{-14 + 2}{3}, \frac{8 - 2}{3} \right) = \left( \frac{-12}{3}, \frac{6}{3} \right) = (-4, 2)$

(Alternative method for $Q$: $Q$ is the midpoint of $PB$, where $P = (-1, 0)$ and $B = (-7, 4)$). $Q = \left( \frac{-1 + (-7)}{2}, \frac{0 + 4}{2} \right) = (-4, 2)$

Final Answer: The coordinates of the points of trisection are $P(-1, 0)$ and $Q(-4, 2)$.

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Example 4: Parallelogram Property (Midpoint Formula)

Problem: If $(1, 2)$, $(4, y)$, $(x, 6)$, and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

Solution:

Let the vertices be $A(1, 2)$, $B(4, y)$, $C(x, 6)$, and $D(3, 5)$.

Key Property: The diagonals of a parallelogram bisect each other. Therefore, the midpoint of diagonal $AC$ is identical to the midpoint of diagonal $BD$.

Step 1: Find the midpoint of $AC$. Using the midpoint formula with $A(1, 2)$ and $C(x, 6)$:

$\text{Midpoint of } AC = \left( \frac{1 + x}{2}, \frac{2 + 6}{2} \right) = \left( \frac{1 + x}{2}, 4 \right)$

Step 2: Find the midpoint of $BD$. Using the midpoint formula with $B(4, y)$ and $D(3, 5)$:

$\text{Midpoint of } BD = \left( \frac{4 + 3}{2}, \frac{y + 5}{2} \right) = \left( \frac{7}{2}, \frac{y + 5}{2} \right)$

Step 3: Equate corresponding coordinates.

• Equating $x$-coordinates: $\frac{1 + x}{2} = \frac{7}{2}$ $1 + x = 7 \implies x = 6$

• Equating $y$-coordinates: $4 = \frac{y + 5}{2}$ $8 = y + 5 \implies y = 3$

Final Answer: The unknown values are $x = 6$ and $y = 3$.

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4. Common Student Mistakes to Avoid

Error 1: Sign Errors with Negative Coordinates

• Mistake: Calculating $x_2 - x_1$ as $3 - 5 = 2$ instead of $-2$, or handling double negatives incorrectly like $-4 - (-2) = -6$.
• Correction: Always use parentheses when substituting negative numbers into formulas: $x_2 - x_1 = -4 - (-2) = -4 + 2 = -2$

Error 2: Swapping $m_1$ and $m_2$ in the Section Formula

• Mistake: Multiplying $m_1$ with $x_1$ and $m_2$ with $x_2$: $\text{WRONG: } x = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
• Correction: Remember the "cross-multiplication" rule: $m_1$ multiplies the coordinates of the second point $(x_2, y_2)$, while $m_2$ multiplies the coordinates of the first point $(x_1, y_1)$: $\text{CORRECT: } x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$

Error 3: Stopping Short when Proving a Square or Rectangle

• Mistake: Proving that all four sides of a quadrilateral are equal and concluding it is a square.
• Correction: Four equal sides only prove a rhombus. You must also calculate and show that the two diagonals are equal ($AC = BD$) to definitively prove it is a square. Similarly, opposite equal sides only prove a parallelogram; equal diagonals are required to prove a rectangle.

Error 4: Reversing Order of Vertices

• Mistake: Plotting or naming vertices out of order (e.g., matching $A$ to $C$ instead of sequential order $A \to B \to C \to D$).
• Correction: When a question states "vertices taken in order", draw a simple continuous perimeter sketch ($A \to B \to C \to D$) to correctly identify which lines form sides and which form diagonals ($AC$ and $BD$).

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5. Practice Questions for Self-Assessment

Question 1

Find the point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.

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Solution:

1. Any point on the $x$-axis has coordinates $P(x, 0)$.
2. Let $A = (2, -5)$ and $B = (-2, 9)$. Given $PA = PB \implies PA^2 = PB^2$.
3. Apply distance formula: $PA^2 = (x - 2)^2 + (0 - (-5))^2 = (x - 2)^2 + 25$ $PB^2 = (x - (-2))^2 + (0 - 9)^2 = (x + 2)^2 + 81$
4. Equate $PA^2$ and $PB^2$: $(x - 2)^2 + 25 = (x + 2)^2 + 81$ $(x^2 - 4x + 4) + 25 = (x^2 + 4x + 4) + 81$
5. Cancel $x^2 + 4$ from both sides: $-4x + 25 = 4x + 81$ $-8x = 56 \implies x = -7$

Answer: The required point on the $x$-axis is $(-7, 0)$.

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Question 2

Find the ratio in which the line segment joining $A(1, -5)$ and $B(-4, 5)$ is divided by the $x$-axis. Also find the coordinates of the point of division.

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Solution:

1. Let the ratio be $k : 1$. Let the point on the $x$-axis be $P(x, 0)$.
2. By Section Formula: $P(x, 0) = \left( \frac{k(-4) + 1(1)}{k + 1}, \frac{k(5) + 1(-5)}{k + 1} \right)$
3. Equate $y$-coordinate to 0: $\frac{5k - 5}{k + 1} = 0 \implies 5k - 5 = 0 \implies k = 1$ The ratio is $1 : 1$ (which means $P$ is the midpoint).
4. Substitute $k = 1$ to find $x$-coordinate: $x = \frac{-4(1) + 1}{1 + 1} = \frac{-3}{2}$

Answer: The ratio is $1 : 1$, and the point of division is $\left( -\frac{3}{2}, 0 \right)$.

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Question 3

If $A(-5, 7)$, $B(-4, -5)$, $C(-1, -6)$, and $D(4, 5)$ are the vertices of a quadrilateral, find the length of its diagonals $AC$ and $BD$.

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Solution:

1. Length of diagonal $AC$ (between $A(-5, 7)$ and $C(-1, -6)$): $AC = \sqrt{(-1 - (-5))^2 + (-6 - 7)^2} = \sqrt{(-1 + 5)^2 + (-13)^2}$ $AC = \sqrt{4^2 + 169} = \sqrt{16 + 169} = \sqrt{185} \text{ units}$

2. Length of diagonal $BD$ (between $B(-4, -5)$ and $D(4, 5)$): $BD = \sqrt{(4 - (-4))^2 + (5 - (-5))^2} = \sqrt{(4 + 4)^2 + (5 + 5)^2}$ $BD = \sqrt{8^2 + 10^2} = \sqrt{64 + 100} = \sqrt{164} = 2\sqrt{41} \text{ units}$

Answer: Diagonal $AC = \sqrt{185}$ units; Diagonal $BD = 2\sqrt{41}$ units.

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Question 4

Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose center is $(2, -3)$ and $B$ is $(1, 4)$.

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Solution:

1. Let the coordinates of point $A$ be $(x_1, y_1)$.
2. The center $C(2, -3)$ is the midpoint of the diameter $AB$, where $B = (1, 4)$.
3. Using the midpoint formula: $\left( \frac{x_1 + 1}{2}, \frac{y_1 + 4}{2} \right) = (2, -3)$
4. Equating coordinates: $\frac{x_1 + 1}{2} = 2 \implies x_1 + 1 = 4 \implies x_1 = 3$ $\frac{y_1 + 4}{2} = -3 \implies y_1 + 4 = -6 \implies y_1 = -10$

Answer: The coordinates of point $A$ are $(3, -10)$.

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6. Exam Revision & FAQs

Q1: How do I prove that three points $A$, $B$, and $C$ are collinear using coordinate geometry?

Answer: Calculate the three segment lengths $AB$, $BC$, and $AC$ using the distance formula. If the sum of the two smaller lengths equals the largest length (e.g., $AB + BC = AC$), then the three points lie on the same straight line and are collinear.

Q2: Why do we use the ratio $k : 1$ instead of $m_1 : m_2$ when finding an unknown ratio?

Answer: Using $m_1 : m_2$ introduces two unknown variables, requiring a system of simultaneous equations. Using $k : 1$ (where $k = \frac{m_1}{m_2}$) reduces the problem to a single unknown variable $k$, making the algebraic linear equation significantly faster and less prone to calculation errors.

Q3: What is the shortcut method to check if a quadrilateral is a parallelogram?

Answer: Rather than calculating all four side lengths using the distance formula, calculate the midpoints of both diagonals $AC$ and $BD$. If:

$\text{Midpoint}(AC) = \text{Midpoint}(BD)$

then $ABCD$ is guaranteed to be a parallelogram. This method uses simple addition and division instead of square roots.

Q4: Is external division included in the CBSE Class 10 syllabus?

Answer: No. The CBSE Class 10 Mathematics syllabus strictly covers internal division using the section formula. You do not need to study external division formulas for board examinations.