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Patterns and Algebra - The Cartesian Coordinate Plane (Four Quadrants)

Grade 6IB

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

🔑Concepts

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The Cartesian Coordinate Plane consists of two perpendicular axes: the horizontal x-axis and the vertical y-axis. Their intersection point is called the Origin, represented by (0,0)(0, 0). The plane is divided into four regions called quadrants, numbered I to IV in a counter-clockwise direction.

Cartesian plane showing four quadrants and the origin.
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Points are located using ordered pairs (x,y)(x, y). The first number (x-coordinate) tells you how far to move left or right from the origin. The second number (y-coordinate) tells you how far to move up or down. For example, (3,−2)(3, -2) moves 3 units right and 2 units down.

A point P located at (3, -2) on the coordinate plane.
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Reflections involve 'flipping' a point over an axis. Reflecting across the x-axis changes the sign of the y-coordinate (x,y)→(x,−y)(x, y) \rightarrow (x, -y). Reflecting across the y-axis changes the sign of the x-coordinate (x,y)→(−x,y)(x, y) \rightarrow (-x, y). This creates a mirror image relative to the chosen axis.

Reflection of point A(2,3) across the y-axis to A'(-2,3).
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Distances between points on the same horizontal or vertical line can be found by calculating the absolute difference between the non-matching coordinates. For a vertical segment, distance =∣y2−y1∣= |y_2 - y_1|.

📐Formulae

Ordered Pair=(x,y)\text{Ordered Pair} = (x, y)

Origin=(0,0)\text{Origin} = (0, 0)

Horizontal Distance=∣x2−x1∣\text{Horizontal Distance} = |x_2 - x_1| (when yy values are equal)

Vertical Distance=∣y2−y1∣\text{Vertical Distance} = |y_2 - y_1| (when xx values are equal)

Reflection across x-axis of (x,y)→(x,−y)\text{Reflection across x-axis of } (x, y) \rightarrow (x, -y)

Reflection across y-axis of (x,y)→(−x,y)\text{Reflection across y-axis of } (x, y) \rightarrow (-x, y)

💡Examples

Problem 1:

Determine the coordinates and the quadrant for a point AA that is located 55 units to the left of the y-axis and 22 units above the x-axis.

Solution:

Step 1: Identify the x-coordinate. '5 units to the left' corresponds to an x-value of −5-5. Step 2: Identify the y-coordinate. '2 units above' corresponds to a y-value of +2+2. Step 3: Combine them into an ordered pair: (−5,2)(-5, 2). Step 4: Determine the quadrant. Since the x-coordinate is negative and the y-coordinate is positive (−,+)(-, +), the point lies in Quadrant II.

Explanation:

We use the directional descriptions to assign signs to the coordinates and then apply the quadrant rules based on those signs.

Problem 2:

Find the distance between point M(−3,4)M(-3, 4) and point N(5,4)N(5, 4).

Solution:

Step 1: Observe the coordinates. Both points have the same y-coordinate, which is 44. This means the points lie on a horizontal line. Step 2: Use the horizontal distance formula: ∣x2−x1∣|x_2 - x_1|. Step 3: Substitute the values: ∣5−(−3)∣|5 - (-3)|. Step 4: Calculate the absolute difference: ∣5+3∣=∣8∣=8|5 + 3| = |8| = 8. The distance is 88 units.

Explanation:

Because the vertical position is identical, we only need to find how many units apart the points are along the horizontal x-axis.

Problem 3:

Point CC is at (−4,−3)(-4, -3). If point DD is the reflection of point CC across the x-axis, find the coordinates of DD and calculate the vertical distance between CC and DD.

Vertical line segment connecting C(-4,-3) and D(-4,3).

Solution:

  1. Reflection of (−4,−3)(-4, -3) across the x-axis keeps the x-coordinate same and negates the y-coordinate: D=(−4,3)D = (-4, 3).
  2. Distance =∣yD−yC∣=∣3−(−3)∣=∣3+3∣=6= |y_D - y_C| = |3 - (-3)| = |3 + 3| = 6 units.

Explanation:

Reflecting over the x-axis moves the point from Quadrant III to Quadrant II. The vertical distance is the total units traveled from y=−3y = -3 up to y=3y = 3.

Problem 4:

A square is drawn on a coordinate plane. Three of its vertices are P(1,1)P(1, 1), Q(1,4)Q(1, 4), and R(4,4)R(4, 4). Determine the coordinates of the fourth vertex SS and the side length of the square.

Square PQRS plotted on a coordinate plane with side length 3.

Solution:

  1. Observing the x-coordinates: PP and QQ are on the line x=1x=1. QQ and RR are on the line y=4y=4.
  2. To complete the square, SS must have the same x-coordinate as RR (which is 44) and the same y-coordinate as PP (which is 11). Thus, S=(4,1)S = (4, 1).
  3. Side length =∣yQ−yP∣=∣4−1∣=3= |y_Q - y_P| = |4 - 1| = 3 units.

Explanation:

In a square, adjacent sides are perpendicular. By matching the coordinates of the existing points, we find the point that closes the shape. The distance between P(1,1)P(1,1) and Q(1,4)Q(1,4) gives the side length of 33.