ICSE Class 10 Mathematics — Chapter 13 Master Notes
In Coordinate Geometry, for any two known points $A(x_1, y_1)$ and $B(x_2, y_2)$ in a Cartesian plane, we use specialized algebraic formulas to find:
To find the co-ordinates of a point $P(x, y)$ which divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in a given ratio $m_1 : m_2$ (i.e. $AP : PB = m_1 : m_2$).
Complete Geometric Proof / Derivation:
Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be the given points and $P(x, y)$ be the point dividing $AB$ internally in the ratio $m_1 : m_2$.
Draw perpendiculars $AL$, $PM$, and $BN$ onto the x-axis from points $A$, $P$, and $B$ respectively. Thus, $AL \parallel PM \parallel BN$.
From point $A$, draw $AR \perp PM$, and from $P$, draw $PS \perp BN$.
Since $\triangle APR$ and $\triangle PBS$ are similar triangles ($\angle APR = \angle PBS$ and $\angle ARP = \angle PSB = 90^\circ$):
$$\frac{AR}{PS} = \frac{PR}{BS} = \frac{AP}{PB} = \frac{m_1}{m_2}$$Equating X-coordinates ratio:
$$\frac{x - x_1}{x_2 - x} = \frac{m_1}{m_2}$$ $$m_2(x - x_1) = m_1(x_2 - x) \implies m_2 x - m_2 x_1 = m_1 x_2 - m_1 x$$ $$m_1 x + m_2 x = m_1 x_2 + m_2 x_1 \implies x(m_1 + m_2) = m_1 x_2 + m_2 x_1$$ $$\therefore x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}$$Equating Y-coordinates ratio:
$$\frac{y - y_1}{y_2 - y} = \frac{m_1}{m_2}$$ $$m_2(y - y_1) = m_1(y_2 - y) \implies y(m_1 + m_2) = m_1 y_2 + m_2 y_1$$ $$\therefore y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$$Section Formula (Internal Division):
The coordinates of point $P(x, y)$ dividing line joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in ratio $m_1 : m_2$ are:
$$P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$$
Alternative $k : 1$ Method for Finding Ratio:
When finding an unknown ratio in which a point divides a line segment, assume the ratio to be $k : 1$ (where $k = \frac{m_1}{m_2}$).
The coordinates of $P$ become:
$$x = \frac{k x_2 + x_1}{k + 1}, \quad y = \frac{k y_2 + y_1}{k + 1}$$In ICSE Board Exams, questions frequently ask for the ratio in which a line segment is divided by the x-axis, y-axis, or a given line:
Master Tip for Board Exams:
Always draw a rough sketch showing line $AB$ and point $P$. Label $A(x_1, y_1)$ as the starting point and $B(x_2, y_2)$ as the ending point so that $m_1$ multiplies $x_2, y_2$ and $m_2$ multiplies $x_1, y_1$ without cross-over confusion!
If points $P$ and $Q$ lie on line segment $AB$ such that $AP = PQ = QB$, then $P$ and $Q$ are called the points of trisection of line segment $AB$.
Trisection Coordinates Formulas:
For $A(x_1, y_1)$ and $B(x_2, y_2)$:
$$P = \left( \frac{1 \cdot x_2 + 2 \cdot x_1}{1 + 2}, \frac{1 \cdot y_2 + 2 \cdot y_1}{1 + 2} \right) = \left( \frac{x_2 + 2x_1}{3}, \frac{y_2 + 2y_1}{3} \right)$$ $$Q = \left( \frac{2 \cdot x_2 + 1 \cdot x_1}{2 + 1}, \frac{2 \cdot y_2 + 1 \cdot y_1}{2 + 1} \right) = \left( \frac{2x_2 + x_1}{3}, \frac{2y_2 + y_1}{3} \right)$$The mid-point $M(x, y)$ of line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ divides $AB$ in equal ratio $m_1 : m_2 = 1 : 1$.
Substituting $m_1 = 1$ and $m_2 = 1$ in the Section Formula:
$$x = \frac{1 \cdot x_2 + 1 \cdot x_1}{1 + 1} = \frac{x_1 + x_2}{2}$$ $$y = \frac{1 \cdot y_2 + 1 \cdot y_1}{1 + 1} = \frac{y_1 + y_2}{2}$$Mid-Point Formula:
$$M(x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$
Key Geometrical Applications of Mid-Point Formula:
Definition: The point of intersection of the three medians of a triangle is called its Centroid ($G$).
Property: The centroid divides each median internally in the ratio $2 : 1$ measured from the vertex to the midpoint of the opposite side.
Derivation of Centroid Formula:
Let $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ be vertices of $\triangle ABC$.
Let $D$ be the midpoint of side $BC$. By Mid-Point Formula:
$$D = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right)$$The centroid $G(x, y)$ lies on median $AD$ and divides $AD$ in ratio $AG : GD = 2 : 1$.
Applying Section Formula on segment $AD$ with ratio $2 : 1$:
$$x = \frac{2 \left(\frac{x_2 + x_3}{2}\right) + 1 \cdot x_1}{2 + 1} = \frac{x_2 + x_3 + x_1}{3} = \frac{x_1 + x_2 + x_3}{3}$$ $$y = \frac{2 \left(\frac{y_2 + y_3}{2}\right) + 1 \cdot y_1}{2 + 1} = \frac{y_2 + y_3 + y_1}{3} = \frac{y_1 + y_2 + y_3}{3}$$Centroid Formula:
$$G(x, y) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$$
Important Theorem to Remember:
The centroid of the triangle formed by joining the midpoints of the sides of $\triangle ABC$ is identical to the centroid of $\triangle ABC$.
Visualizing line segment joining $A(2, 1)$ and $B(7, 6)$ divided by $P(5, 4)$ in ratio $3 : 2$.
Example 1 (Finding Dividing Point): Find the co-ordinates of point $P$ which divides the line joining $A(4, -5)$ and $B(6, 3)$ in the ratio $2 : 5$.
Solution:
Here $(x_1, y_1) = (4, -5)$, $(x_2, y_2) = (6, 3)$, and $m_1 : m_2 = 2 : 5$.
Using Section Formula:
$$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{2 \times 6 + 5 \times 4}{2 + 5} = \frac{12 + 20}{7} = \frac{32}{7}$$ $$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{2 \times 3 + 5 \times (-5)}{2 + 5} = \frac{6 - 25}{7} = \frac{-19}{7}$$Answer: $P = \left( \frac{32}{7}, \frac{-19}{7} \right)$
Example 2 (Finding Ratio): Find the ratio in which the point $(5, 4)$ divides the line joining points $(2, 1)$ and $(7, 6)$.
Solution:
Let the required ratio be $m_1 : m_2$.
Using section formula for x-coordinate:
$$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} \implies 5 = \frac{m_1 \times 7 + m_2 \times 2}{m_1 + m_2}$$ $$5(m_1 + m_2) = 7m_1 + 2m_2 \implies 5m_1 + 5m_2 = 7m_1 + 2m_2$$ $$3m_2 = 2m_1 \implies \frac{m_1}{m_2} = \frac{3}{2}$$Answer: The required ratio is $3 : 2$.
Example 3 (Division by X-axis): In what ratio is the line joining the points $(4, 2)$ and $(3, -5)$ divided by the x-axis? Also, find the co-ordinates of the point of intersection.
Solution:
Let the ratio be $k : 1$. Since the point lies on x-axis, its coordinates are $P(x, 0)$.
Here $(x_1, y_1) = (4, 2)$ and $(x_2, y_2) = (3, -5)$.
Using section formula for y-coordinate:
$$y = \frac{k y_2 + y_1}{k + 1} \implies 0 = \frac{k(-5) + 2}{k + 1} \implies -5k + 2 = 0 \implies k = \frac{2}{5}$$Thus, ratio $m_1 : m_2 = 2 : 5$.
Now finding x-coordinate:
$$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{2(3) + 5(4)}{2 + 5} = \frac{6 + 20}{7} = \frac{26}{7}$$Answer: Ratio $= 2 : 5$, Point of intersection $= \left( \frac{26}{7}, 0 \right)$.
Example 4 (Division by horizontal line $y=3$): Calculate the ratio in which the line joining $(4, 6)$ and $(-5, -4)$ is divided by the line $y = 3$. Also, find the co-ordinates of the point of intersection.
Solution:
Every point on line $y = 3$ is of the form $(x, 3)$.
Using section formula for y-coordinate with ratio $m_1 : m_2$:
$$3 = \frac{m_1(-4) + m_2(6)}{m_1 + m_2} \implies 3m_1 + 3m_2 = -4m_1 + 6m_2$$ $$7m_1 = 3m_2 \implies \frac{m_1}{m_2} = \frac{3}{7}$$Ratio is $3 : 7$. Now calculating x-coordinate:
$$x = \frac{3(-5) + 7(4)}{3 + 7} = \frac{-15 + 28}{10} = \frac{13}{10}$$Answer: Ratio $= 3 : 7$, Point $= \left(\frac{13}{10}, 3\right)$.
Example 5 (Trisection Calculation): Find the co-ordinates of the points of trisection of the line segment joining $A(6, -2)$ and $B(-8, 10)$.
Solution:
Let $P$ and $Q$ be points of trisection ($AP = PQ = QB$).
For point $P$ ($1 : 2$ ratio):
$$x_P = \frac{1(-8) + 2(6)}{1 + 2} = \frac{-8 + 12}{3} = \frac{4}{3}$$ $$y_P = \frac{1(10) + 2(-2)}{1 + 2} = \frac{10 - 4}{3} = 2$$ $$\implies P = \left( \frac{4}{3}, 2 \right)$$For point $Q$ ($2 : 1$ ratio):
$$x_Q = \frac{2(-8) + 1(6)}{2 + 1} = \frac{-16 + 6}{3} = -\frac{10}{3}$$ $$y_Q = \frac{2(10) + 1(-2)}{2 + 1} = \frac{20 - 2}{3} = 6$$ $$\implies Q = \left( -\frac{10}{3}, 6 \right)$$Example 6 (Mid-Point on Axes Intercept): The mid-point of line segment $AB$ shown in a diagram is $(-3, 5)$ where $A$ lies on the x-axis and $B$ lies on the y-axis. Find the coordinates of $A$ and $B$.
Solution:
Let $A = (x, 0)$ and $B = (0, y)$.
Mid-point of $AB = \left( \frac{x + 0}{2}, \frac{0 + y}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right)$.
Given mid-point $= (-3, 5)$:
$$\frac{x}{2} = -3 \implies x = -6$$ $$\frac{y}{2} = 5 \implies y = 10$$Answer: $A = (-6, 0)$ and $B = (0, 10)$.
Example 7 (Missing Fourth Vertex of Parallelogram): $A(14, -2)$, $B(6, -2)$, and $D(8, 2)$ are three vertices of a parallelogram $ABCD$. Find the co-ordinates of the fourth vertex $C(x, y)$.
Solution:
Diagonals of parallelogram $ABCD$ are $AC$ and $BD$. They bisect each other, so:
$$\text{Mid-point of } AC = \text{Mid-point of } BD$$ $$\left( \frac{14 + x}{2}, \frac{-2 + y}{2} \right) = \left( \frac{6 + 8}{2}, \frac{-2 + 2}{2} \right)$$ $$\left( \frac{14 + x}{2}, \frac{-2 + y}{2} \right) = \left( \frac{14}{2}, \frac{0}{2} \right) = (7, 0)$$Equating components:
$$\frac{14 + x}{2} = 7 \implies 14 + x = 14 \implies x = 0$$ $$\frac{-2 + y}{2} = 0 \implies -2 + y = 0 \implies y = 2$$Answer: Vertex $C = (0, 2)$.
Example 8 (Reconstructing Triangle Vertices from Side Midpoints): In $\triangle ABC$, $P(-2, 5)$ is mid-point of $AB$, $Q(2, 4)$ is mid-point of $BC$, and $R(-1, 2)$ is mid-point of $AC$. Calculate vertices $A, B, C$.
Solution:
Let $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$.
Forming x-coordinate equations:
$$\frac{x_1 + x_2}{2} = -2 \implies x_1 + x_2 = -4 \quad \text{--- (i)}$$ $$\frac{x_2 + x_3}{2} = 2 \implies x_2 + x_3 = 4 \quad \text{--- (ii)}$$ $$\frac{x_1 + x_3}{2} = -1 \implies x_1 + x_3 = -2 \quad \text{--- (iii)}$$Adding (i), (ii), (iii):
$$2(x_1 + x_2 + x_3) = -4 + 4 - 2 = -2 \implies x_1 + x_2 + x_3 = -1 \quad \text{--- (iv)}$$Subtracting (ii) from (iv): $x_1 = -1 - 4 = -5$.
Subtracting (iii) from (iv): $x_2 = -1 - (-2) = 1$.
Subtracting (i) from (iv): $x_3 = -1 - (-4) = 3$.
Forming y-coordinate equations similarly:
$$y_1 + y_2 = 10, \quad y_2 + y_3 = 8, \quad y_1 + y_3 = 4$$ $$2(y_1 + y_2 + y_3) = 22 \implies y_1 + y_2 + y_3 = 11$$ $$y_1 = 11 - 8 = 3, \quad y_2 = 11 - 4 = 7, \quad y_3 = 11 - 10 = 1$$Answer: $A(-5, 3)$, $B(1, 7)$, and $C(3, 1)$.
Example 9 (Centroid Calculation & Side Length): $\triangle ABC$ has centroid $G(4, 3)$. If $A(1, 3)$, $B(4, b)$, and $C(a, 1)$, find $a$ and $b$, and calculate side length $BC$.
Solution:
Using Centroid formula $G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$:
$$\frac{1 + 4 + a}{3} = 4 \implies 5 + a = 12 \implies a = 7$$ $$\frac{3 + b + 1}{3} = 3 \implies 4 + b = 9 \implies b = 5$$Thus, $B = (4, 5)$ and $C = (7, 1)$.
Now calculating side length $BC$ using Distance formula:
$$BC = \sqrt{(7 - 4)^2 + (1 - 5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units.}$$| Concept | Given Parameters | Standard Formula / Condition |
|---|---|---|
| Section Formula (Internal) | $A(x_1,y_1), B(x_2,y_2)$, Ratio $m_1:m_2$ | $P = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)$ |
| Ratio Method ($k:1$) | Point $P(x,y)$, Vertices $A, B$ | $x = \frac{k x_2 + x_1}{k+1}, \quad y = \frac{k y_2 + y_1}{k+1}$ |
| Mid-Point Formula | Ratio $1:1$, Vertices $A, B$ | $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ |
| Trisection Points | Ratio $1:2$ for P, $2:1$ for Q | $P = \left(\frac{x_2+2x_1}{3}, \frac{y_2+2y_1}{3}\right), \quad Q = \left(\frac{2x_2+x_1}{3}, \frac{2y_2+y_1}{3}\right)$ |
| Parallelogram Vertex | Vertices $A, B, C, D$ | $\text{Midpoint of } AC = \text{Midpoint of } BD$ |
| Centroid of Triangle | Vertices $A, B, C$ | $G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$ |
BOARD The line segment joining $A(-5, 8)$ and $B(10, -4)$ is trisected by the coordinate axes. Find the coordinates of the points of trisection.
BOARD Calculate the ratio in which the line joining $A(-4, 2)$ and $B(3, 6)$ is divided by point $P(x, 3)$. Also, find $x$ and the length of $AP$.
BOARD $A(-4, 2)$ and $B(8, -3)$ are given points. $P$ is a point on $AB$ such that $AP : PB = 1 : 2$. Find the coordinates of $A$ and $B$ if $P$ is on the x-axis.
BOARD Given a line segment $AB$ joining $A(-4, 6)$ and $B(8, -3)$. Find:
BOARD Triangle $ABC$ has centroid $G(4, 3)$. If $A(1, 3)$, $B(4, b)$, and $C(a, 1)$, find $a$ and $b$, and the length of side $BC$.
BOARD Find the ratio in which the line segment joining $A(-4, 3)$ and $B(8, -6)$ is divided by the x-axis.
BOARD The line segment joining $A(2, 3)$ and $B(6, -5)$ is intercepted by the x-axis at point $K$. Write down the ordinate of point $K$. Hence, find the ratio in which $K$ divides $AB$ and the coordinates of $K$.
BOARD $P(-4, 5)$ and $Q(3, 2)$ intersect the y-axis at point $R$. $PM$ and $QN$ are perpendiculars from $P$ and $Q$ onto the x-axis. Find: