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Coordinate Geometry - Let Us Explore Four Quadrants-advanced

Grade 9CBSE

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

🔑Concepts

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The Cartesian plane is divided into four quadrants by the intersection of the horizontal xx-axis and vertical yy-axis at the origin O(0,0)O(0, 0). 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; and in Quadrant IV, xx is positive and yy is negative.

Cartesian plane showing four quadrants and their sign conventions.
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The location of any point P(x,y)P(x, y) is defined by its signed distances from the axes. The xx-coordinate (abscissa) represents the perpendicular distance from the yy-axis, while the yy-coordinate (ordinate) represents the perpendicular distance from the xx-axis.

A point P in the first quadrant showing its x and y distances from the axes.
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Points lying on the axes do not belong to any quadrant. Any point on the xx-axis has an ordinate of 00 (form (x,0)(x, 0)), and any point on the yy-axis has an abscissa of 00 (form (0,y)(0, y)).

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Reflection of a point P(x,y)P(x, y) in the origin results in a point P′(−x,−y)P'(-x, -y). This transformation is equivalent to a rotation of 180∘180^\circ around the origin.

📐Formulae

Distance from y-axis=∣x∣Distance \text{ from } y\text{-axis} = |x|

Distance from x-axis=∣y∣Distance \text{ from } x\text{-axis} = |y|

Area of Rectangle=∣x2−x1∣×∣y2−y1∣\text{Area of Rectangle} = |x_2 - x_1| \times |y_2 - y_1|

Midpoint of line joining (x1,y) and (x2,y)=(x1+x22,y)\text{Midpoint of line joining } (x_1, y) \text{ and } (x_2, y) = \left( \frac{x_1 + x_2}{2}, y \right)

💡Examples

Problem 1:

A point MM is at a distance of 55 units from the xx-axis and 77 units from the yy-axis. If the point lies in the third quadrant, find its coordinates.

Solution:

  1. In the third quadrant, both xx and yy coordinates are negative.
  2. The distance from the yy-axis represents the absolute value of the xx-coordinate. So, ∣x∣=7  ⟹  x=−7|x| = 7 \implies x = -7 (since it is in the 3rd quadrant).
  3. The distance from the xx-axis represents the absolute value of the yy-coordinate. So, ∣y∣=5  ⟹  y=−5|y| = 5 \implies y = -5.
  4. Therefore, the coordinates of point MM are (−7,−5)(-7, -5).

Explanation:

Distance from axes translates to coordinate values, and the quadrant determines the signs.

Problem 2:

Find the area of the triangle formed by the points A(0,4)A(0, 4), B(0,0)B(0, 0), and C(6,0)C(6, 0).

Solution:

  1. Plotting the points: B(0,0)B(0, 0) is the origin. A(0,4)A(0, 4) lies on the yy-axis. C(6,0)C(6, 0) lies on the xx-axis.
  2. This forms a right-angled triangle where the base is the segment BCBC and the height is the segment ABAB.
  3. Base BC=∣6−0∣=6BC = |6 - 0| = 6 units.
  4. Height AB=∣4−0∣=4AB = |4 - 0| = 4 units.
  5. Area=12×base×heightArea = \frac{1}{2} \times \text{base} \times \text{height}
  6. Area=12×6×4=12 sq. unitsArea = \frac{1}{2} \times 6 \times 4 = 12 \text{ sq. units}

Explanation:

When vertices lie on the axes, the lengths of the base and height are simply the non-zero coordinate values.

Problem 3:

If the point P(3,4)P(3, 4) is reflected in the xx-axis to get P′P' and then P′P' is reflected in the yy-axis to get P′′P'', find the coordinates of P′′P''.

Solution:

  1. Reflection of P(3,4)P(3, 4) in the xx-axis: The xx-coordinate stays the same, and the yy-coordinate changes sign. So, P′=(3,−4)P' = (3, -4).
  2. Reflection of P′(3,−4)P'(3, -4) in the yy-axis: The yy-coordinate stays the same, and the xx-coordinate changes sign. So, P′′=(−3,−4)P'' = (-3, -4).
  3. Note: This is equivalent to a single reflection of the original point PP through the origin.

Explanation:

Reflecting across the xx-axis changes the sign of yy; reflecting across the yy-axis changes the sign of xx.

Problem 4:

A square ABCDABCD has its center at the origin O(0,0)O(0, 0) and its sides are parallel to the axes. If the coordinates of vertex AA are (3,3)(3, 3), find the coordinates of the other vertices and calculate the area of the square.

Square ABCD centered at the origin with vertices at (3,3), (-3,3), (-3,-3), and (3,-3).

Solution:

  1. Since the sides are parallel to the axes and the center is (0,0)(0,0), the vertices must be symmetric about the axes.
  2. Given A(3,3)A(3, 3) is in Quadrant I.
  3. BB (reflecting AA across the yy-axis) is (−3,3)(-3, 3).
  4. CC (reflecting BB across the xx-axis) is (−3,−3)(-3, -3).
  5. DD (reflecting AA across the xx-axis) is (3,−3)(3, -3).
  6. The side length s=∣3−(−3)∣=6s = |3 - (-3)| = 6 units.
  7. Area=s2=6×6=36\text{Area} = s^2 = 6 \times 6 = 36 sq. units.

Explanation:

Because the square is centered at the origin with sides parallel to the axes, the distance from the origin to each side along the axes is equal. The vertices are (±3,±3)(\pm 3, \pm 3).

Problem 5:

Plot the points X(−4,0)X(-4, 0), Y(4,0)Y(4, 0), and Z(0,5)Z(0, 5). Identify the shape formed by joining these points and find its area.

Isosceles triangle XYZ with base on the x-axis and vertex Z on the y-axis.

Solution:

  1. Plot X(−4,0)X(-4, 0) and Y(4,0)Y(4, 0) on the xx-axis.
  2. Plot Z(0,5)Z(0, 5) on the yy-axis.
  3. Joining X,Y,ZX, Y, Z forms a triangle.
  4. Base XY=∣4−(−4)∣=8\text{Base } XY = |4 - (-4)| = 8 units.
  5. Height OZ=∣5−0∣=5\text{Height } OZ = |5 - 0| = 5 units.
  6. Area=12×base×height=12×8×5=20\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 5 = 20 sq. units.

Explanation:

The base of the triangle lies on the xx-axis, and the vertex ZZ lies on the yy-axis, making the segment OZOZ the perpendicular height.