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Orienting Yourself: The Use of Coordinates - Settling In

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Cartesian Coordinate System: A system used to specify the position of points on a plane using a pair of numerical coordinates. It consists of two perpendicular number lines: the horizontal xx-axis and the vertical yy-axis. Their intersection point is called the Origin O(0,0)O(0, 0).

A Cartesian plane showing the X and Y axes intersecting at the origin.
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Quadrants: The two axes divide the plane into four regions called quadrants, numbered I to IV in counter-clockwise order. In Quadrant I, both xx and yy are positive (+,++, +). In Quadrant II, xx is negative and yy is positive (−,+-, +). In Quadrant III, both are negative (−,−-, -). In Quadrant IV, xx is positive and yy is negative (+,−+, -).

The four quadrants of a Cartesian plane with their respective signs.
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Coordinates (Abscissa and Ordinate): For a point P(x,y)P(x, y), the xx-coordinate is called the abscissa (distance from the yy-axis) and the yy-coordinate is called the ordinate (distance from the xx-axis).

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Location of points on Axes: Any point lying on the xx-axis has an ordinate of 00, i.e., its form is (x,0)(x, 0). Any point lying on the yy-axis has an abscissa of 00, i.e., its form is (0,y)(0, y).

📐Formulae

Coordinates of Origin=(0,0)\text{Coordinates of Origin} = (0, 0) Paradigms

Point on X-axis  ⟹  (x,0)\text{Point on X-axis} \implies (x, 0)

Point on Y-axis  ⟹  (0,y)\text{Point on Y-axis} \implies (0, y)

Signs in Quadrants: QI(+,+),QII(−,+),QIII(−,−),QIV(+,−)\text{Signs in Quadrants: } Q_I(+,+), Q_{II}(-,+), Q_{III}(-,-), Q_{IV}(+,-)

💡Examples

Problem 1:

Determine the quadrant or axis where the following points lie: P(3,−5)P(3, -5), Q(−2,−2)Q(-2, -2), and R(0,4)R(0, 4).

Solution:

P(3,−5)P(3, -5) lies in Quadrant IV. Q(−2,−2)Q(-2, -2) lies in Quadrant III. R(0,4)R(0, 4) lies on the YY-axis.

Explanation:

For P(3,−5)P(3, -5), x>0x > 0 and y<0y < 0, which corresponds to Quadrant IV. For Q(−2,−2)Q(-2, -2), both xx and yy are negative, placing it in Quadrant III. For R(0,4)R(0, 4), the xx-coordinate is 00, meaning the point must lie on the YY-axis.

Problem 2:

Find the coordinates of a point MM which is 55 units away from the xx-axis and 33 units away from the yy-axis, given that it lies in the second quadrant.

Solution:

The coordinates of point MM are (−3,5)(-3, 5).

Explanation:

The distance from the yy-axis is the absolute value of the xx-coordinate (abscissa), so ∣x∣=3|x| = 3. The distance from the xx-axis is the absolute value of the yy-coordinate (ordinate), so ∣y∣=5|y| = 5. Since the point is in the second quadrant, xx must be negative and yy must be positive. Thus, x=−3x = -3 and y=5y = 5.

Problem 3:

Calculate the area of a triangle whose vertices are O(0,0)O(0, 0), A(6,0)A(6, 0), and B(0,8)B(0, 8).

Solution:

Area=12×base×height=12×6×8=24 square units\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 8 = 24 \text{ square units}

Explanation:

The point A(6,0)A(6, 0) lies on the xx-axis, so the base OAOA has a length of 66 units. The point B(0,8)B(0, 8) lies on the yy-axis, so the height OBOB has a length of 88 units. Since the xx and yy axes are perpendicular, the triangle is right-angled at the origin.

Problem 4:

Plot the points A(4,3)A(4, 3), B(−4,3)B(-4, 3), C(−4,−3)C(-4, -3), and D(4,−3)D(4, -3) on a Cartesian plane. Join them in order. What shape is formed?

A rectangle ABCD plotted on the coordinate plane.

Solution:

A(4,3)→Quadrant IA(4, 3) \rightarrow \text{Quadrant I} B(−4,3)→Quadrant IIB(-4, 3) \rightarrow \text{Quadrant II} C(−4,−3)→Quadrant IIIC(-4, -3) \rightarrow \text{Quadrant III} D(4,−3)→Quadrant IVD(4, -3) \rightarrow \text{Quadrant IV} Joining A−B−C−D−AA-B-C-D-A forms a rectangle. Length AB=∣4−(−4)∣=8AB = |4 - (-4)| = 8 units. Breadth BC=∣3−(−3)∣=6BC = |3 - (-3)| = 6 units. The shape is a rectangle.

Explanation:

Points are plotted based on their signs. Since the lengths of opposite sides are equal (88 units and 66 units) and the axes are perpendicular, the resulting figure is a rectangle.

Problem 5:

Find the area of a square whose opposite vertices are P(−2,2)P(-2, 2) and R(2,−2)R(2, -2).

A square with vertices at (-2,2), (2,2), (2,-2), and (-2,-2).

Solution:

The coordinates of the four vertices of the square would be P(−2,2)P(-2, 2), Q(2,2)Q(2, 2), R(2,−2)R(2, -2), and S(−2,−2)S(-2, -2). Length of side PQ=∣2−(−2)∣=4PQ = |2 - (-2)| = 4 units. Area of square = side2=42=16\text{side}^2 = 4^2 = 16 sq units.

Explanation:

By plotting opposite vertices PP and RR, we can identify the other two vertices QQ and SS to complete the square. The side length is calculated by the horizontal or vertical distance between adjacent vertices.