NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry — Complete Guide with All Exercises

Mnemonic: Starting from Quadrant I and going anti-clockwise, the signs are $(+,+)$, $(-,+)$, $(-,-)$, $(+,-)$. Some students remember this as: "I is all positive, II has sine (y) positive, III has tangent positive, IV has cosine (x) positive" — though the trig mnemonics are more useful in Class 10.

Points on the Axes:
- Any point on the x-axis has $y = 0$. Its coordinates are of the form $(a, 0)$.
- Any point on the y-axis has $x = 0$. Its coordinates are of the form $(0, b)$.
- The origin lies on both axes: $(0, 0)$.

For a point $P(x, y)$:

• Abscissa = $x$-coordinate = the perpendicular distance of the point from the $y$-axis (measured along the $x$-direction).
  - Ordinate = $y$-coordinate = the perpendicular distance of the point from the $x$-axis (measured along the $y$-direction).

For the point $(7, -3)$:
- Abscissa = $7$
- Ordinate = $-3$

Common Exam Question: "The abscissa of a point is $5$ and the ordinate is $-2$. Write its coordinates." Answer: $(5, -2)$.

Another Common Question: "The ordinate of a point on the $x$-axis is always _____." Answer: $0$.

Question: How will you describe the position of a table lamp on your study table to another person?

Solution:
To describe the position of any object on a flat surface, we need two perpendicular reference lines (axes) and distances from each.

Choose two perpendicular edges of the table as reference axes. Measure:
- The distance of the lamp from the left edge (this acts as the $x$-coordinate).
- The distance of the lamp from the bottom edge (this acts as the $y$-coordinate).

For example, if the lamp is 30 cm from the left edge and 20 cm from the bottom edge, its position is $(30, 20)$ with respect to the chosen edges.

Key Insight: This is exactly the principle behind the Cartesian coordinate system — any position in a plane can be uniquely described by two numbers measured from two perpendicular reference lines.

Question: (Street Map) A city has two main roads crossing at the centre. You are standing at a point that is 5 blocks east and 3 blocks north of the centre. What are your coordinates?

Solution:
Taking the centre as the origin, east as positive $x$, and north as positive $y$:

Your position = $(5, 3)$, which lies in Quadrant I.

If you were 5 blocks west and 3 blocks south, your position would be $(-5, -3)$, in Quadrant III.

Key Insight: The choice of positive direction matters. Convention: right/east = positive $x$, up/north = positive $y$.

Question: In which quadrant or on which axis do the following points lie?
(i) $(-2, 4)$
(ii) $(3, -1)$
(iii) $(-1, -1)$
(iv) $(4, 0)$
(v) $(0, -5)$
(vi) $(-3, 0)$
(vii) $(0, 0)$
(viii) $(2, 7)$

Solution:

Key Rule: If either coordinate is zero, the point lies on an axis, not in a quadrant.

Question: Write the answer to each of the following:
(i) What is the name of the horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.

Solution:

(i) The horizontal line is called the **$x$-axis (or abscissa axis). The vertical line is called the $y$-axis (or ordinate axis). Together, they are called the coordinate axes**.

(ii) Each part of the plane formed by the coordinate axes is called a quadrant. There are four quadrants, numbered I, II, III, IV in anti-clockwise order starting from the top-right.

(iii) The point of intersection of the coordinate axes is called the origin, denoted by $O$. Its coordinates are $(0, 0)$.

Question: See the figure, and write the following:
(i) The coordinates of $B$.
(ii) The coordinates of $C$.
(iii) The point identified by the coordinates $(-3, -5)$.
(iv) The point identified by the coordinates $(2, -4)$.
(v) The abscissa of the point $D$.
(vi) The ordinate of the point $H$.
(vii) The coordinates of the point $L$.
(viii) The coordinates of the point $M$.

Solution (based on typical NCERT figure):

(i) $B = (-5, 2)$ — Read the $x$-coordinate first (horizontal), then the $y$-coordinate (vertical).

(ii) $C = (5, -5)$

(iii) The point at $(-3, -5)$ is $E$ (in the third quadrant).

(iv) The point at $(2, -4)$ is $G$ (in the fourth quadrant).

(v) If $D = (6, 1)$, then the abscissa of $D$ is $6$.

(vi) If $H = (-1, -3)$, then the ordinate of $H$ is $-3$.

(vii) $L = (0, 5)$ — on the positive $y$-axis.

(viii) $M = (-3, 0)$ — on the negative $x$-axis.

Key Technique: When reading coordinates from a graph, always read the $x$-value first (go left/right from the origin) and then the $y$-value (go up/down).

Question: Plot the points $(x, y)$ given by the following table:

Solution:
Plot the points: $(-2, 8)$, $(-1, 1)$, $(0, 0)$, $(1, 1)$, $(2, 8)$.

Observations:
1. The points are symmetric about the $y$-axis: the points $(-2, 8)$ and $(2, 8)$ are mirror images, as are $(-1, 1)$ and $(1, 1)$.
2. When connected by a smooth curve, these points trace a parabola — the graph of $y = x^2 \cdot 2$ (or more precisely, these points lie on $y = x^4/x^2$... actually $y = 2x^2$ gives $(-2,8), (-1,2)$, which doesn't match exactly. The actual function here is $y = 2x^2$ for $(-2,8),(2,8)$ but $(−1,1)$ doesn't fit. These points actually lie on $y = x^2$ for $(\pm 1, 1)$ and $y = 2x^2$ doesn't work. The given table simply represents a set of points — they can be plotted and joined, forming a U-shaped curve.)

Key Insight: Plotting tables of values is the first step toward understanding graphs of functions, which you will study extensively in Chapter 4 (Linear Equations) and in later classes.

Question: Plot the point $P(3, 4)$ on the Cartesian plane. Draw perpendiculars $PM$ and $PN$ from $P$ to the $x$-axis and $y$-axis respectively. Write the coordinates of $M$ and $N$.

Solution:

$P = (3, 4)$

$PM$ is the perpendicular from $P$ to the $x$-axis. The foot $M$ has the same $x$-coordinate as $P$ and $y = 0$:

$PN$ is the perpendicular from $P$ to the $y$-axis. The foot $N$ has $x = 0$ and the same $y$-coordinate as $P$:

Additional Information:
- $PM = |y|= 4$ units (the ordinate gives the distance from the $x$-axis).
- $PN = |x| = 3$ units (the abscissa gives the distance from the $y$-axis).
- The rectangle $OMPN$ has area $= 3 \times 4 = 12$ square units.

Question: Plot the points $A(4, 4)$, $B(-4, 4)$, $C(-4, -4)$, $D(4, -4)$. Name the figure $ABCD$. Find its area.

Solution:

Plotting these four points:
- $A(4, 4)$ is in Quadrant I
- $B(-4, 4)$ is in Quadrant II
- $C(-4, -4)$ is in Quadrant III
- $D(4, -4)$ is in Quadrant IV

Connecting $A \to B \to C \to D \to A$ gives a square.

Side length: $AB$ is horizontal (same $y$-coordinate = 4), from $x = -4$ to $x = 4$.

Similarly, $BC = CD = DA = 8$ units, and all angles are $90°$.

Perimeter $= 4 \times 8 = 32$ units.

Question: Plot the points $P(1, 1)$, $Q(-2, 1)$, $R(-2, -3)$, $S(1, -3)$. What figure does $PQRS$ form? Find its area and perimeter.

Solution:

• $PQ$: from $(1,1)$ to $(-2,1)$ — horizontal line, length $= |1-(-2)| = 3$ units.
  - $QR$: from $(-2,1)$ to $(-2,-3)$ — vertical line, length $= |1-(-3)| = 4$ units.
  - $RS$: from $(-2,-3)$ to $(1,-3)$ — horizontal line, length $= 3$ units.
  - $SP$: from $(1,-3)$ to $(1,1)$ — vertical line, length $= 4$ units.

Opposite sides are equal ($PQ = RS = 3$, $QR = SP = 4$) and all angles are $90°$. So $PQRS$ is a rectangle.

Question: Plot $A(0, 0)$, $B(4, 0)$, $C(4, 3)$. What figure does $ABC$ form? Find its area.

Solution:

• $A(0,0)$ is the origin.
  - $B(4,0)$ is on the $x$-axis.
  - $C(4,3)$ is in Quadrant I.

$AB$ lies along the $x$-axis with length $4$ units.
$BC$ is vertical (same $x = 4$) with length $3$ units.
$\angle B = 90°$.

So $\triangle ABC$ is a right triangle with legs $AB = 4$ and $BC = 3$.

The hypotenuse $AC = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ units (Pythagoras theorem).

Question: The point $P(3, 5)$ is reflected in the $x$-axis to get $P'$, and $P'$ is then reflected in the $y$-axis to get $P''$. Find the coordinates of $P'$ and $P''$.

Solution:

**Reflection in $x$-axis:** $(3, 5) \to P' = (3, -5)$

**Reflection in $y$-axis:** $(3, -5) \to P'' = (-3, -5)$

Note: $P''(-3, -5)$ is the same as the reflection of $P(3, 5)$ in the origin. This makes sense: reflecting in the $x$-axis and then in the $y$-axis (or vice versa) is the same as reflecting in the origin.

Question: Point $A(2, 3)$ is mapped to $A'(2, -3)$. In which axis was $A$ reflected?

Solution:
The $x$-coordinate stays the same ($2$), but the $y$-coordinate changes sign ($3 \to -3$). This is a reflection in the **$x$-axis**.

Question: Point $B(-4, 7)$ is mapped to $B'(4, 7)$. In which axis was $B$ reflected?

Solution:
The $y$-coordinate stays the same ($7$), but the $x$-coordinate changes sign ($-4 \to 4$). This is a reflection in the **$y$-axis**.

Question: Find the distance between $A(1, 2)$ and $B(4, 6)$.

Solution:

Question: Find the distance of the point $P(3, 4)$ from the origin.

Solution:
The origin is $O(0, 0)$.

Notice this is the classic 3-4-5 right triangle!

Question: What is the distance of the point $(5, -7)$ from the $x$-axis? From the $y$-axis?

Solution:

**Distance from the $x$-axis** = $|y|$ = $|-7|$ = $7$ units.

**Distance from the $y$-axis** = $|x|$ = $|5|$ = $5$ units.

Key Rule: The distance of a point $(a, b)$ from the $x$-axis is $|b|$ (the absolute value of the ordinate), and from the $y$-axis is $|a|$ (the absolute value of the abscissa).

Question: Which of the following points lie on the line $y = 2x + 1$?
(i) $(0, 1)$ (ii) $(1, 2)$ (iii) $(2, 5)$ (iv) $(-1, -1)$

Solution:
Substitute each point into $y = 2x + 1$:

(i) $(0, 1)$: $1 = 2(0) + 1 = 1$ ✓ Yes.

(ii) $(1, 2)$: $2 = 2(1) + 1 = 3$. $2 \neq 3$. No.

(iii) $(2, 5)$: $5 = 2(2) + 1 = 5$ ✓ Yes.

(iv) $(-1, -1)$: $-1 = 2(-1) + 1 = -1$ ✓ Yes.

So the points $(0, 1)$, $(2, 5)$, and $(-1, -1)$ lie on the line $y = 2x + 1$.

Question: Write the coordinates of a point that lies on both axes simultaneously.

Solution:
A point on the $x$-axis has $y = 0$. A point on the $y$-axis has $x = 0$. For a point to lie on both axes, we need $x = 0$ AND $y = 0$.

The only such point is the origin $(0, 0)$.

Question: A point is equidistant from both axes. In which quadrant(s) could it lie? Give an example.

Solution:
The distance from the $x$-axis is $|y|$ and the distance from the $y$-axis is $|x|$.

For equidistance: $|x| = |y|$, which means $y = x$ or $y = -x$.

Such points can lie in any quadrant:
- Quadrant I: $(3, 3)$ lies on $y = x$
- Quadrant II: $(-5, 5)$ lies on $y = -x$
- Quadrant III: $(-2, -2)$ lies on $y = x$
- Quadrant IV: $(4, -4)$ lies on $y = -x$
- On the axes: the origin $(0, 0)$

Question: Three points $A(1, 2)$, $B(3, 6)$, and $C(5, 10)$ are plotted on the Cartesian plane. Are they collinear (lying on the same straight line)?

Solution:
Check if the slope between consecutive pairs is the same:

Slope of $AB = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$

Slope of $BC = \frac{10 - 6}{5 - 3} = \frac{4}{2} = 2$

Since the slopes are equal, $A$, $B$, $C$ are collinear — they all lie on the line $y = 2x$.

Alternative Check: Verify that all three points satisfy $y = 2x$:
- $A(1, 2)$: $2 = 2(1)$ ✓
- $B(3, 6)$: $6 = 2(3)$ ✓
- $C(5, 10)$: $10 = 2(5)$ ✓

Question: Plot the points $A(0, 0)$, $B(3, 0)$, $C(3, 4)$. Find the perimeter of $\triangle ABC$.

Solution:
$AB$: horizontal from $(0,0)$ to $(3,0)$, length $= 3$.
$BC$: vertical from $(3,0)$ to $(3,4)$, length $= 4$.
$AC$: from $(0,0)$ to $(3,4)$, length $= \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

Perimeter $= 3 + 4 + 5 = 12$ units.

This is a 3-4-5 right triangle with the right angle at $B$.

Question: Find the coordinates of the midpoint of $AC$.

Solution:
Midpoint $= \left(\frac{0+3}{2}, \frac{0+4}{2}\right) = \left(\frac{3}{2}, 2\right) = (1.5, 2)$.

Question: The vertices of a triangle are $A(2, 3)$, $B(5, 3)$, $C(3, 7)$. Find the vertices of the triangle formed by reflecting $\triangle ABC$ in the $x$-axis.

Solution:
Reflecting each vertex in the $x$-axis (change sign of $y$):

$A(2, 3) \to A'(2, -3)$
$B(5, 3) \to B'(5, -3)$
$C(3, 7) \to C'(3, -7)$

The reflected triangle $A'B'C'$ has the same shape and size as $\triangle ABC$ (reflection preserves distances and angles).