Coordinate Geometry Class 10 Mathematics Notes

7.2 Distance Formula

The distance formula is a fundamental concept in coordinate geometry that allows us to calculate the straight-line distance between two points in a Cartesian plane. It is derived directly from the Pythagorean theorem ( $a^{2} + b^{2} = c^{2}$ ).

Imagine two points, $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ , plotted on a graph. To find the distance between them, we can construct a right-angled triangle.

Draw a line segment from P parallel to the x-axis.
Draw a line segment from Q parallel to the y-axis.
These two lines intersect at a point, let's call it T. The coordinates of T will be $(x_{2}, y_{1})$ .
Now, we have a right-angled triangle PTQ, with the right angle at T.

The length of the side PT is the difference in the x-coordinates: $∣ x_{2} - x_{1} ∣$ . The length of the side QT is the difference in the y-coordinates: $∣ y_{2} - y_{1} ∣$ .

According to the Pythagorean theorem, in the right-angled triangle PTQ:

P Q^{2} = P T^{2} + Q T^{2}

P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Taking the square root of both sides, we get the distance formula:

P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Key Points:

The order of points does not matter because the differences are squared, making the result always positive.
The distance is always a non-negative value.
If one of the points is the origin $(0, 0)$ , the formula simplifies to $O P = \sqrt{x^{2} + y^{2}}$ .

Example 1: Finding the distance between two points

Given: Points P(3, 2) and Q(–2, –3). To find: The distance PQ.

Solution: Using the distance formula: $P Q = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

P Q = \sqrt{(- 2 - 3)^{2} + (- 3 - 2)^{2}}

P Q = \sqrt{(- 5)^{2} + (- 5)^{2}}

P Q = \sqrt{25 + 25}

P Q = \sqrt{50}

P Q = \sqrt{25 \times 2}

P Q = 5 \sqrt{2}

Answer: The distance between the points is $5 \sqrt{2}$ units.

Example 2: Determining if points form a triangle

Given: Points P(3, 2), Q(–2, –3), and R(2, 3). To find: Whether these points form a triangle and what type of triangle it is.

Solution: First, calculate the distances between each pair of points:

P Q = \sqrt{(- 2 - 3)^{2} + (- 3 - 2)^{2}} = \sqrt{(- 5)^{2} + (- 5)^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2}

Q R = \sqrt{(2 - (- 2))^{2} + (3 - (- 3))^{2}} = \sqrt{(2 + 2)^{2} + (3 + 3)^{2}} = \sqrt{4^{2} + 6^{2}} = \sqrt{16 + 36} = \sqrt{52} = 2 \sqrt{13}

P R = \sqrt{(2 - 3)^{2} + (3 - 2)^{2}} = \sqrt{(- 1)^{2} + 1^{2}} = \sqrt{1 + 1} = \sqrt{2}

Check the triangle inequality theorem: The sum of any two sides must be greater than the third side. $P Q + P R = 5 \sqrt{2} + \sqrt{2} = 6 \sqrt{2} \approx 8.48$ . $Q R = \sqrt{52} \approx 7.21$ . $6 \sqrt{2} > \sqrt{52}$ . $P Q + Q R = 5 \sqrt{2} + 2 \sqrt{13} > \sqrt{2}$ . $P R + Q R = \sqrt{2} + 2 \sqrt{13} > 5 \sqrt{2}$ . Since all conditions hold, the points form a triangle.

Now, check for a right angle using the converse of the Pythagorean theorem:

P Q^{2} = (\sqrt{50})^{2} = 50

Q R^{2} = (\sqrt{52})^{2} = 52

P R^{2} = (\sqrt{2})^{2} = 2

We see that $P Q^{2} + P R^{2} = 50 + 2 = 52$ , which is equal to $Q R^{2}$ . Therefore, by the converse of the Pythagorean theorem, the angle at P is $90^{\circ}$ .

Answer: The points P, Q, and R form a right-angled triangle.

Example 3: Checking for collinearity

Given: Points A(1, 5), B(2, 3), and C(–2, –11). To find: Whether the points are collinear.

Solution: Calculate the distances:

A B = \sqrt{(2 - 1)^{2} + (3 - 5)^{2}} = \sqrt{1^{2} + (- 2)^{2}} = \sqrt{1 + 4} = \sqrt{5}

B C = \sqrt{(- 2 - 2)^{2} + (- 11 - 3)^{2}} = \sqrt{(- 4)^{2} + (- 14)^{2}} = \sqrt{16 + 196} = \sqrt{212}

A C = \sqrt{(- 2 - 1)^{2} + (- 11 - 5)^{2}} = \sqrt{(- 3)^{2} + (- 16)^{2}} = \sqrt{9 + 256} = \sqrt{265}

For the points to be collinear, the sum of the lengths of two shorter segments must equal the length of the longest segment. $\sqrt{5} \approx 2.24$ , $\sqrt{212} \approx 14.56$ , $\sqrt{265} \approx 16.28$ . Clearly, $\sqrt{5} + \sqrt{212} \neq \sqrt{265}$ .

Answer: The points (1, 5), (2, 3), and (–2, –11) are not collinear.

Example 4: Finding a relation between x and y for equidistant points

Given: Point P(x, y) is equidistant from A(7, 1) and B(3, 5). To find: A relation between x and y.

Solution: Since P is equidistant from A and B, AP = BP. Squaring both sides, $A P^{2} = B P^{2}$ . Using the distance formula for $A P^{2}$ : $(x - 7)^{2} + (y - 1)^{2}$ Using the distance formula for $B P^{2}$ : $(x - 3)^{2} + (y - 5)^{2}$

Equating them:

(x - 7)^{2} + (y - 1)^{2} = (x - 3)^{2} + (y - 5)^{2}

x^{2} - 14 x + 49 + y^{2} - 2 y + 1 = x^{2} - 6 x + 9 + y^{2} - 10 y + 25

Cancel out $x^{2}$ and $y^{2}$ from both sides:

- 14 x - 2 y + 50 = - 6 x - 10 y + 34

Rearrange terms to group x and y:

- 14 x + 6 x - 2 y + 10 y = 34 - 50

- 8 x + 8 y = - 16

Divide by -8:

x - y = 2

Answer: The relation between x and y is $x - y = 2$ . This equation represents the perpendicular bisector of the line segment AB.

Example 5: Finding a point on an axis equidistant from two given points

Given: Point P on the y-axis is equidistant from A(6, 5) and B(–4, 3). To find: The coordinates of point P.

Solution: A point on the y-axis has coordinates of the form (0, y). Let P be (0, y). Since P is equidistant from A and B, $A P^{2} = B P^{2}$ .

A P^{2} = (6 - 0)^{2} + (5 - y)^{2} = 6^{2} + (5 - y)^{2} = 36 + 25 - 10 y + y^{2} = 61 - 10 y + y^{2}

B P^{2} = (- 4 - 0)^{2} + (3 - y)^{2} = (- 4)^{2} + (3 - y)^{2} = 16 + 9 - 6 y + y^{2} = 25 - 6 y + y^{2}

Equating $A P^{2}$ and $B P^{2}$ :

61 - 10 y + y^{2} = 25 - 6 y + y^{2}

Cancel out $y^{2}$ :

61 - 10 y = 25 - 6 y

Rearrange terms:

61 - 25 = 10 y - 6 y

36 = 4 y

y = \frac{36}{4}

y = 9

So, the coordinates of point P are (0, 9).

Answer: The point on the y-axis equidistant from A(6, 5) and B(–4, 3) is (0, 9).

Remember
Always square the distances before equating them to avoid dealing with square roots, especially when solving for variables.

Exam-smart
Expect questions where you need to find the distance between two towns given their north and east displacements (forming a right triangle), or to prove if given points form a specific shape (square, rectangle, rhombus) by calculating all side lengths and diagonals.

7.3 Section Formula

The section formula helps us find the coordinates of a point that divides a line segment joining two other points in a given ratio. This is crucial for dividing lines into specific proportions, which is common in geometry and applications.

Let a point $P (x, y)$ divide the line segment joining $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ internally in the ratio $m_{1} : m_{2}$ .

To derive the formula, we can use similar triangles, similar to how the distance formula was derived. Draw perpendiculars from A, P, and B to the x-axis, and also draw a line parallel to the x-axis through P. This forms two similar right-angled triangles.

By AA similarity, the triangles are similar, and the ratio of their corresponding sides is equal.

\frac{A P}{P B} = \frac{A Q}{P C} = \frac{P Q}{B C}

Where:

$A P / P B = m_{1} / m_{2}$ (given ratio)
$A Q = x - x_{1}$
$P C = x_{2} - x$
$P Q = y - y_{1}$
$B C = y_{2} - y$

Substituting these into the ratios:

\frac{m_{1}}{m_{2}} = \frac{x - x_{1}}{x_{2} - x}

m_{1} (x_{2} - x) = m_{2} (x - x_{1})

m_{1} x_{2} - m_{1} x = m_{2} x - m_{2} x_{1}

m_{1} x_{2} + m_{2} x_{1} = m_{2} x + m_{1} x

m_{1} x_{2} + m_{2} x_{1} = x (m_{1} + m_{2})

x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}}

Similarly, for the y-coordinate:

\frac{m_{1}}{m_{2}} = \frac{y - y_{1}}{y_{2} - y}

Solving this gives:

y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}}

Thus, the Section Formula for internal division is:

P (x, y) = (\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}})

Special Case: Mid-point Formula If a point divides a line segment in the ratio 1:1 (i.e., it's the mid-point), the formula simplifies: $m_{1} = 1, m_{2} = 1$ .

M i d - p o i n t (x, y) = (\frac{1 \cdot x_{2} + 1 \cdot x_{1}}{1 + 1}, \frac{1 \cdot y_{2} + 1 \cdot y_{1}}{1 + 1}) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})

External Division: While the CBSE syllabus for Class 10 focuses on internal division, it's important to know that a similar formula exists for external division. If a point P divides the line segment AB externally in the ratio $m_{1} : m_{2}$ , then the formula becomes:

P (x, y) = (\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}})

(Note: This is for awareness; focus on the internal division formula for exams.)

Example 6: Finding a point dividing a segment in a given ratio

Given: Points A(4, –3) and B(8, 5). Point P divides AB internally in the ratio 3:1. To find: The coordinates of point P.

Solution: Here, $(x_{1}, y_{1}) = (4, - 3)$ , $(x_{2}, y_{2}) = (8, 5)$ , $m_{1} = 3$ , $m_{2} = 1$ . Using the section formula:

x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} = \frac{3 (8) + 1 (4)}{3 + 1} = \frac{24 + 4}{4} = \frac{28}{4} = 7

y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} = \frac{3 (5) + 1 (- 3)}{3 + 1} = \frac{15 - 3}{4} = \frac{12}{4} = 3

Answer: The coordinates of point P are (7, 3).

Example 7: Finding the ratio in which a point divides a segment

Given: Points A(–6, 10) and B(3, –8). Point P(–4, 6) divides AB. To find: The ratio in which P divides AB.

Solution: Let the ratio be $m_{1} : m_{2}$ . Using the section formula for the x-coordinate:

- 4 = \frac{m_{1} (3) + m_{2} (- 6)}{m_{1} + m_{2}}

- 4 (m_{1} + m_{2}) = 3 m_{1} - 6 m_{2}

- 4 m_{1} - 4 m_{2} = 3 m_{1} - 6 m_{2}

- 4 m_{1} - 3 m_{1} = - 6 m_{2} + 4 m_{2}

- 7 m_{1} = - 2 m_{2}

\frac{m_{1}}{m_{2}} = \frac{- 2}{- 7} = \frac{2}{7}

So, the ratio is $m_{1} : m_{2} = 2 : 7$ .

We should verify this using the y-coordinate:

6 = \frac{m_{1} (- 8) + m_{2} (10)}{m_{1} + m_{2}}

If $m_{1} = 2 k$ and $m_{2} = 7 k$ (where k is a common factor):

6 = \frac{2 k (- 8) + 7 k (10)}{2 k + 7 k} = \frac{- 16 k + 70 k}{9 k} = \frac{54 k}{9 k} = 6

The y-coordinate also matches.

Answer: The point (–4, 6) divides the line segment joining A(–6, 10) and B(3, –8) in the ratio 2 : 7.

Example 8: Points of trisection

Given: Points A(2, –2) and B(–7, 4). To find: The coordinates of the points that divide AB into three equal parts (trisection points).

Solution: Let the points of trisection be P and Q. These points divide the segment AB into three equal parts: AP = PQ = QB.

Point P divides AB in the ratio 1:2.
Point Q divides AB in the ratio 2:1.

For point P (ratio 1:2): $(x_{1}, y_{1}) = (2, - 2)$ , $(x_{2}, y_{2}) = (- 7, 4)$ , $m_{1} = 1$ , $m_{2} = 2$ .

x_{P} = \frac{1 (- 7) + 2 (2)}{1 + 2} = \frac{- 7 + 4}{3} = \frac{- 3}{3} = - 1

y_{P} = \frac{1 (4) + 2 (- 2)}{1 + 2} = \frac{4 - 4}{3} = \frac{0}{3} = 0

So, P is at (–1, 0).

For point Q (ratio 2:1): $(x_{1}, y_{1}) = (2, - 2)$ , $(x_{2}, y_{2}) = (- 7, 4)$ , $m_{1} = 2$ , $m_{2} = 1$ .

x_{Q} = \frac{2 (- 7) + 1 (2)}{2 + 1} = \frac{- 14 + 2}{3} = \frac{- 12}{3} = - 4

y_{Q} = \frac{2 (4) + 1 (- 2)}{2 + 1} = \frac{8 - 2}{3} = \frac{6}{3} = 2

So, Q is at (–4, 2).

Answer: The points of trisection are (–1, 0) and (–4, 2).

Example 9: Ratio where a segment is divided by an axis

Given: Points A(5, –6) and B(–1, –4). The y-axis divides the line segment AB. To find: The ratio in which the y-axis divides AB, and the point of intersection.

Solution: A point on the y-axis has an x-coordinate of 0. Let the ratio be $k : 1$ . Using the section formula for the x-coordinate:

x = \frac{k \cdot x_{2} + 1 \cdot x_{1}}{k + 1}

0 = \frac{k (- 1) + 1 (5)}{k + 1}

0 = - k + 5

k = 5

So, the ratio is $5 : 1$ .

Now, find the y-coordinate using this ratio ( $k = 5$ ):

y = \frac{k \cdot y_{2} + 1 \cdot y_{1}}{k + 1} = \frac{5 (- 4) + 1 (- 6)}{5 + 1} = \frac{- 20 - 6}{6} = \frac{- 26}{6} = - \frac{13}{3}

The point of intersection is $(0, - 13 / 3)$ .

Answer: The y-axis divides the line segment in the ratio 5 : 1. The point of intersection is (0, –13/3).

Example 10: Vertices of a parallelogram

Given: Points A(6, 1), B(8, 2), C(9, 4), and D(p, 3) are vertices of a parallelogram in order. To find: The value of p.

Solution: A key property of a parallelogram is that its diagonals bisect each other. This means the mid-point of diagonal AC is the same as the mid-point of diagonal BD.

Mid-point of AC:

M_{A C} = (\frac{6 + 9}{2}, \frac{1 + 4}{2}) = (\frac{15}{2}, \frac{5}{2})

Mid-point of BD:

M_{B D} = (\frac{8 + p}{2}, \frac{2 + 3}{2}) = (\frac{8 + p}{2}, \frac{5}{2})

Equating the mid-points:

(\frac{15}{2}, \frac{5}{2}) = (\frac{8 + p}{2}, \frac{5}{2})

Equating the x-coordinates:

\frac{15}{2} = \frac{8 + p}{2}

15 = 8 + p

p = 15 - 8

p = 7

The y-coordinates are already equal ( $\frac{5}{2}$ ).

Answer: The value of p is 7.

Remember
When a question involves finding a ratio or a point on an axis, assume the ratio is $k : 1$ and use the section formula. The axis property (e.g., x=0 for y-axis, y=0 for x-axis) will help you solve for $k$ .

Exam-smart
Questions often involve finding a point that divides a segment in a specific ratio (e.g., 2:3, 1:2 for trisection), finding the ratio given the points and the dividing point, or using the mid-point formula for properties of shapes like parallelograms.

Coordinate Geometry — Notes

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