Coordinate Geometry — NCERT Solutions, AI Tutor & Practice

(i) Origin: The point where the x-axis and the y-axis intersect. Its coordinates are $(0, 0)$.

(ii) Abscissa: The x-coordinate of a point; it is the perpendicular distance of the point from the y-axis, measured along the x-axis.

(iii) Ordinate: The y-coordinate of a point; it is the perpendicular distance of the point from the x-axis, measured along the y-axis.

(iv) Quadrant: The two axes divide the plane into four parts called quadrants — numbered I, II, III and IV in the anticlockwise direction starting from the upper-right region.

Use the sign rule: I $(+,+)$, II $(-,+)$, III $(-,-)$, IV $(+,-)$.

• $A(3,5)$: both positive → Quadrant I
• $B(-4,2)$: $(-,+)$ → Quadrant II
• $C(-6,-1)$: $(-,-)$ → Quadrant III
• $D(5,-3)$: $(+,-)$ → Quadrant IV
• $E(0,4)$: x-coordinate is $0$ → lies on the y-axis
• $F(-2,0)$: y-coordinate is $0$ → lies on the x-axis

To plot a point $(x, y)$, move $x$ units horizontally (right if positive, left if negative) from the origin, then $y$ units vertically (up if positive, down if negative).

• $P(2,3)$: 2 right, 3 up → Quadrant I
• $Q(-3,2)$: 3 left, 2 up → Quadrant II
• $R(-1,-4)$: 1 left, 4 down → Quadrant III
• $S(4,-2)$: 4 right, 2 down → Quadrant IV

Each point sits in a different quadrant, one in each region of the plane.

(i) On the x-axis, the y-coordinate is $0$. 'Left of origin' means negative x. So the point is $(-5, 0)$.

(ii) On the y-axis, the x-coordinate is $0$. 'Below the origin' means negative y. So the point is $(0, -3)$.

The four vertices are $O(0,0)$, $A(3,0)$, $B(3,4)$, $C(0,4)$.

Plotting them: $O$ and $A$ lie on the x-axis (3 units apart), $O$ and $C$ lie on the y-axis (4 units apart), and all angles are right angles. The figure is a rectangle.

• Length (along x-axis) $= 3$ units
• Breadth (along y-axis) $= 4$ units

$\text{Area} = \text{length} \times \text{breadth} = 3 \times 4 = 12 \text{ square units}$

For the point $(-7, 6)$:

• Abscissa (x-coordinate) $= -7$
• Ordinate (y-coordinate) $= 6$

Reflection in the y-axis: The y-axis reflection keeps the ordinate the same and changes the sign of the abscissa.

$(-7, 6) \;\longrightarrow\; (7, 6)$

So the mirror image is $(7, 6)$.

• $(-5, -8)$: abscissa $= -5$ → negative; ordinate $= -8$ → negative
• $(0, 7)$: abscissa $= 0$ → zero; ordinate $= 7$ → positive
• $(6, 0)$: abscissa $= 6$ → positive; ordinate $= 0$ → zero

Note that $(0,7)$ lies on the y-axis and $(6,0)$ lies on the x-axis.

Take the corner post as the origin $(0,0)$, with 'right' along the positive x-axis and 'up' along the positive y-axis.

• 4 rows to the right → x-coordinate $= 4$
• 6 columns up → y-coordinate $= 6$

So the tree's position is $(4, 6)$.

Both coordinates are positive, so the tree lies in Quadrant I.

Take the centre as origin. Let east be positive x, west negative x; north be positive y.

Position P: 3 m east, 2 m north → $P(3, 2)$.

Position Q: 3 m west, 2 m north → $Q(-3, 2)$.

Observation: The two points have the same ordinate ($2$) but opposite abscissae ($3$ and $-3$). Therefore $P$ and $Q$ are mirror images of each other in the y-axis (they are equidistant from, and on opposite sides of, the y-axis).

(i) School $(2,3)$ and Market $(2,-1)$: they share the same abscissa ($x = 2$), so they lie on a vertical line. The distance is the difference of ordinates: $|3 - (-1)| = |3 + 1| = 4 \text{ units}$ In metres: $4 \times 100 = \mathbf{400\text{ m}}$.

(ii) School $(2,3)$ and Park $(-3,3)$: they share the same ordinate ($y = 3$), so they lie on a horizontal line. The distance is the difference of abscissae: $|2 - (-3)| = |2 + 3| = 5 \text{ units}$ In metres: $5 \times 100 = \mathbf{500\text{ m}}$.

Starting at harbour $O(0,0)$:

• $O(0,0) \to A(4,0)$: only x increases → sails 4 units east.
• $A(4,0) \to B(4,5)$: only y increases → sails 5 units north.
• $B(4,5) \to C(0,5)$: x decreases to 0 → sails 4 units west.
• $C(0,5) \to O(0,0)$: y decreases to 0 → sails 5 units south back to harbour.

The path traces a rectangle of length 4 units and breadth 5 units.

Point $B(4,5)$ has both coordinates positive, so it lies in Quadrant I.