Introduction to Coordinate Geometry and The Distance Formula for CBSE Class 10

In pure Euclidean geometry, we study geometric figures through axioms, postulates, and visual constructions. However, when French philosopher and mathematician René Descartes introduced the Cartesian coordinate system in the 17th century, he revolutionized the subject by bridging algebra and geometry into a unified discipline: Coordinate Geometry (or Analytical Geometry).

In CBSE Class 10 Mathematics, Chapter 7 (Coordinate Geometry) provides the algebraic tools to locate positions on a grid, measure exact straight-line distances, divide line segments into proportional ratios, and compute enclosed planar areas.

What You Will Learn

• Quick review of the Cartesian coordinate plane: axes, origin, and coordinates $(x, y)$
• Step-by-step derivation of the Distance Formula using the Pythagoras Theorem
• Distance of any point from the origin: $d = \sqrt{x^2 + y^2}$
• Testing collinearity of three points using the distance formula
• Finding coordinates of an unknown point equidistant from two given points
• Solved CBSE board examination problems and common algebraic traps

1. The Cartesian Coordinate System (Review)

A point in a two-dimensional plane is uniquely located by an ordered pair of real numbers $(x, y)$:

• The horizontal reference axis is the $x$-axis; the vertical reference axis is the $y$-axis.
• Their point of intersection is the origin, denoted by $O(0, 0)$.
• The $x$-coordinate is called the abscissa (perpendicular distance from the $y$-axis).
• The $y$-coordinate is called the ordinate (perpendicular distance from the $x$-axis).
• Any point on the $x$-axis has coordinates of the form $(x, 0)$.
• Any point on the $y$-axis has coordinates of the form $(0, y)$.

2. Derivation of the Distance Formula

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be any two points in the Cartesian plane.

                     y
                     |              Q(x2, y2)
                     |             /|
                     |            / |
                     |  P(x1, y1)/  | (y2 - y1)
                     |          +---+
                     |            R(x2, y1)
                     |        (x2 - x1)
                     +----------------------- x
                    O

Geometric Construction:

1. Draw perpendiculars $PM$ and $QN$ from points $P$ and $Q$ to the $x$-axis.
2. Draw a perpendicular $PR$ from point $P$ to line segment $QN$.
3. This forms a right-angled triangle $\Delta PRQ$, right-angled at $R$.

Determining Side Lengths:

• Horizontal distance: $PR = MN = ON - OM = x_2 - x_1$
• Vertical distance: $QR = QN - RN = QN - PM = y_2 - y_1$

Applying the Pythagoras Theorem in $\Delta PRQ$:

$PQ^2 = PR^2 + QR^2$ $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

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

The Distance Formula

The straight-line distance between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by: $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\text{Distance} = \sqrt{(\text{Difference of } x\text{-coordinates})^2 + (\text{Difference of } y\text{-coordinates})^2}$

Important: <u>Because squaring eliminates negative signs, $(x_2 - x_1)^2 = (x_1 - x_2)^2$. Therefore, you can subtract coordinates in either order, provided you do not mix up corresponding $x$ and $y$ values!</u>

3. Distance of a Point from the Origin

If one of the points is the origin $O(0, 0)$ and the other point is $P(x, y)$: $OP = \sqrt{(x - 0)^2 + (y - 0)^2} = \mathbf{\sqrt{x^2 + y^2}}$

Quick Example:

The distance of point $P(-3, 4)$ from the origin is: $OP = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ units}$

4. Solved CBSE Board Examination Problems

Solved Example 1: Standard Distance Calculation

Problem: Find the distance between the points $P(-5, 7)$ and $Q(-1, 3)$.

Solution:

1. Here $x_1 = -5, y_1 = 7$ and $x_2 = -1, y_2 = 3$.
2. Apply the distance formula: $PQ = \sqrt{[-1 - (-5)]^2 + (3 - 7)^2}$ $PQ = \sqrt{(-1 + 5)^2 + (-4)^2} = \sqrt{4^2 + 16} = \sqrt{16 + 16} = \sqrt{32}$
3. Simplify the radical: $PQ = \sqrt{16 \times 2} = 4\sqrt{2}\text{ units}$
4. Therefore, <u>the distance is $4\sqrt{2}\text{ units}$</u>.

Solved Example 2: Testing for Collinearity Using Distance

Problem: Determine if the points $A(1, 5)$, $B(2, 3)$, and $C(-2, -11)$ are collinear.

Solution: Three points $A, B,$ and $C$ are collinear (lie on a single straight line) if and only if the sum of the lengths of any two segments equals the length of the third segment (e.g., $AB + BC = AC$).

1. Calculate $AB$: $AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$
2. Calculate $BC$: $BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53}$
3. Calculate $AC$: $AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}$
4. Notice that: $AB + BC = \sqrt{5} + 2\sqrt{53} \ne \sqrt{265} = AC$
5. Since no sum of two distances equals the third distance, <u>the points $A, B,$ and $C$ are NOT collinear</u>.

Solved Example 3: Equidistant Point on the x-axis (CBSE High-Yield)

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

Solution:

1. Any point on the $x$-axis has its $y$-coordinate equal to $0$. Let the required point be $P(x, 0)$.
2. Let $A(2, -5)$ and $B(-2, 9)$. We are given that $PA = PB$, which implies: $PA^2 = PB^2$
3. Using the squared distance formula: $(x - 2)^2 + [0 - (-5)]^2 = [x - (-2)]^2 + (0 - 9)^2$ $(x - 2)^2 + 5^2 = (x + 2)^2 + (-9)^2$
4. Expand both sides: $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$
5. Subtract $x^2 + 4$ from both sides: $-4x + 25 = 4x + 81$ $-4x - 4x = 81 - 25$ $-8x = 56 \implies x = \frac{56}{-8} = -7$
6. Therefore, <u>the required point on the $x$-axis is $(-7, 0)$</u>.

5. Summary and Examination Tips

Exam Tip: When solving equidistant problems ($PA = PB$), always square both sides ($PA^2 = PB^2$) right at the beginning. This eliminates the square root radicals immediately and prevents algebraic errors!

Common Mistake: Forgetting double negatives when coordinates are negative. In $[-1 - (-5)]$, remember that minus a negative is a positive: $-1 + 5 = +4$, not $-6$!