ICSE Class 10 Mathematics — Chapter 12 Master Notes
Coordinate Geometry uses pairs of real numbers (coordinates) to uniquely specify the location of points in a two-dimensional plane.
Definition: When an object point $P$ is placed before a line mirror $l$, its image point $P'$ is formed such that:
Notation: $M_l(P) = P'$ represents the reflection of point $P$ in line $l$.
1. Reflection in the X-axis ($y = 0$):
When a point is reflected in the x-axis, the sign of its ordinate ($y$-coordinate) changes while the abscissa ($x$-coordinate) remains unchanged.
$$M_x(x, y) = (x, -y)$$
2. Reflection in the Y-axis ($x = 0$):
When a point is reflected in the y-axis, the sign of its abscissa ($x$-coordinate) changes while the ordinate ($y$-coordinate) remains unchanged.
$$M_y(x, y) = (-x, y)$$
3. Reflection in the Origin $O(0, 0)$:
When a point is reflected in the origin, the signs of both its abscissa and ordinate change.
$$M_o(x, y) = (-x, -y)$$
The combination (composition) of consecutive reflections obeys fundamental rules:
Definition: Any point that remains unaltered (unchanged) under a given transformation is called an invariant point.
Fundamental Law of Invariant Points: A point is invariant under reflection in a line $l$ if and only if the point lies ON the line $l$ ($M_l(P) = P$).
| Mirror Line / Center | Condition for Point $P(x, y)$ to be Invariant | Examples of Invariant Points |
|---|---|---|
| X-axis ($y = 0$) | $y = 0$ (point lies on x-axis) | $(5, 0), (-3, 0), (a, 0)$ |
| Y-axis ($x = 0$) | $x = 0$ (point lies on y-axis) | $(0, 5), (0, -4), (0, b)$ |
| Origin $O(0, 0)$ | $x = 0$ and $y = 0$ (point is origin) | $(0, 0)$ |
| Line $x = a$ | $x = a$ | $(a, 2), (a, -7)$ |
| Line $y = b$ | $y = b$ | $(3, b), (-4, b)$ |
Reflection Formulas for Parallel Lines:
$$M_{x=a}(x, y) = (2a - x, y)$$
Note: Ordinate $y$ remains unchanged.$$M_{y=b}(x, y) = (x, 2b - y)$$
Note: Abscissa $x$ remains unchanged.Visualizing $\triangle ABC$ ($A(2,3), B(4,5), C(7,2)$) reflected in the origin to $\triangle A'B'C'$ ($-x, -y$).
Example 1 (Successive Reflections): The triangle $A(1, 2)$, $B(4, 4)$, $C(3, 7)$ is first reflected in line $y = 0$ onto $\triangle A'B'C'$ and then $\triangle A'B'C'$ is reflected in the origin onto $\triangle A''B''C''$. Write down coordinates of:
Solution:
(i) Reflection in $y = 0$ (x-axis): $M_x(x, y) = (x, -y)$
$$A' = (1, -2), \quad B' = (4, -4), \quad C' = (3, -7)$$(ii) Reflection in origin: $M_o(x, y) = (-x, -y)$
$$A'' = M_o(1, -2) = (-1, 2)$$ $$B'' = M_o(4, -4) = (-4, 4)$$ $$C'' = M_o(3, -7) = (-3, 7)$$Observation: Notice that $(A'', B'', C'')$ is equivalent to $M_y(A, B, C)$ (reflection in y-axis)!
Example 2 (Finding Object Point from Image): A point $P$ is reflected in the x-axis. The co-ordinates of its image are $(8, -6)$. Find:
Solution:
(i) Since $M_x(x, y) = (x, -y) = (8, -6) \implies x = 8, y = 6$.
$$\therefore P = (8, 6)$$(ii) Reflection of $P(8, 6)$ in y-axis:
$$M_y(8, 6) = (-8, 6)$$Example 3 (Commutativity Verification): Perform operations $M_x \cdot M_y$ and $M_y \cdot M_x$ on point $(3, -4)$. State whether $M_x \cdot M_y = M_y \cdot M_x$.
Solution:
1. $M_x \cdot M_y(3, -4) = M_x[M_y(3, -4)] = M_x(-3, -4) = (-3, 4)$.
2. $M_y \cdot M_x(3, -4) = M_y[M_x(3, -4)] = M_y(3, 4) = (-3, 4)$.
Conclusion: $M_x \cdot M_y = M_y \cdot M_x = M_o$. Yes, reflection in mutually perpendicular axes is always commutative!
Example 4 (Reflection in $x = a$ and $y + 3 = 0$):
Solution:
(i) Line $x = 2$ is parallel to y-axis at distance 2. Distance of $P(-1, 3)$ from $x=2$ is $2 - (-1) = 3$ units to the left. Image $P'$ will be 3 units to the right of $x=2$:
$$x_{P'} = 2 + 3 = 5, \quad y_{P'} = 3 \implies P' = (5, 3)$$Using formula: $M_{x=a}(x, y) = (2a - x, y) = (2(2) - (-1), 3) = (5, 3)$.
(ii) Line $y = -3$ is parallel to x-axis at distance 3 below origin. Using formula $M_{y=b}(x, y) = (x, 2b - y)$:
$$y_{Q'} = 2(-3) - 1 = -7, \quad x_{Q'} = 2 \implies Q' = (2, -7)$$Example 5 (Geometrical Figure PMQN & Area): Points $P(3, 2)$ and $Q(-3, -2)$ are plotted. From $P$ and $Q$, perpendiculars $PM$ and $QN$ are drawn to x-axis.
Solution:
(a) Foot of perpendicular $PM$ on x-axis is $M(3, 0)$. Foot $QN$ on x-axis is $N(-3, 0)$.
(b) $M_o(3, 2) = (-3, -2) = Q$. Thus, $Q$ is the image of $P$ under reflection in origin.
(c) Quad $PMQN$ is a Parallelogram with base $MN = 3 - (-3) = 6$ units and height = distance between parallel sides $= 2 - (-2) = 4$ units (or sum of heights $2 + 2 = 4$).
$$\text{Area of Parallelogram } PMQN = \text{Base} \times \text{Height} = 6 \times 2 = 12 \text{ sq units.}$$| Transformation | Operator | Object Point $P(x, y) \to$ Image Point $P'$ | Invariant Point Condition |
|---|---|---|---|
| Reflection in X-axis | $M_x$ | $(x, -y)$ | $y = 0$ (lies on x-axis) |
| Reflection in Y-axis | $M_y$ | $(-x, y)$ | $x = 0$ (lies on y-axis) |
| Reflection in Origin | $M_o$ | $(-x, -y)$ | $(0, 0)$ (only origin) |
| Reflection in line $x = a$ | $M_{x=a}$ | $(2a - x, y)$ | $x = a$ |
| Reflection in line $y = b$ | $M_{y=b}$ | $(x, 2b - y)$ | $y = b$ |
BOARD Points $A(6, 4)$ and $B(0, 4)$ are reflected in the origin to get images $A'$ and $B'$.
BOARD The point $P(2, -4)$ is reflected about line $x = 0$ to get image $Q$. Point $Q$ is reflected about line $y = 0$ to get image $R$.
BOARD Points $A(2, 3)$, $B(4, 5)$, and $C(7, 2)$ are vertices of $\triangle ABC$. Write down coordinates of $A'', B'', C''$ if $\triangle A''B''C''$ is the image of $\triangle ABC$ reflected in the x-axis. Mention the special name of quadrilateral $BCC''B''$ and find its area.
BOARD $P$ and $Q$ have coordinates $(0, 5)$ and $(-2, 4)$ respectively.
BOARD Quadrilateral $ABCD$ with vertices $A(2, 2)$, $B(5, 1)$, $C(4, 2)$, and $D(2, 2)$ is reflected in the origin onto $A', B', C', D'$. Locate $A', B', C', D'$ and state whether $D, A, A', D'$ are collinear.
BOARD Point $P(3, 4)$ is reflected to $P'$ in the x-axis, and $O'$ is the image of origin $O(0,0)$ when reflected in line $PP'$. Write coordinates of $P'$ and $O'$, find length of $PP'$, and name quadrilateral $POP'O'$.
BOARD Point $P(5, 3)$ is reflected in origin to $P'$. $M$ is foot of perpendicular from $P$ to x-axis, $N$ is foot of perpendicular from $P'$ to x-axis. Name figure $PMP'N$ and find its area.