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7.1 Q-8

Question Statement

Given the points B(4,1)B(4, 1), C(βˆ’2,3)C(-2, 3), and D(βˆ’8,0)D(-8, 0), use the vector method to find the coordinates of the following points:

(i) Point AA, if ABCDA B C D is a parallelogram.
(ii) Point EE, if AEBDA E B D is a parallelogram.


Background and Explanation

In this problem, we need to use the vector method to find the coordinates of unknown points in a parallelogram. The key properties of a parallelogram we’ll use are:

  1. Opposite sides are equal and parallel. For example, if ABCDA B C D is a parallelogram, then the vector ABA B is equal to the vector DCD C and vice versa.
  2. Vector addition: The vector sum of two adjacent sides of a parallelogram should result in the vector across the diagonal.

By applying these principles, we can solve for the coordinates of points AA and EE.


Solution

(i) Find the coordinates of point AA, given that ABCDA B C D is a parallelogram

  1. Set up the equation:
    Since ABCDA B C D is a parallelogram, we know that AB=DCA B = D C.
    The vector ABA B is the difference between the coordinates of B(4,1)B(4, 1) and A(x,y)A(x, y), and the vector DCD C is the difference between the coordinates of D(βˆ’8,0)D(-8, 0) and C(βˆ’2,3)C(-2, 3).
    Thus: AB=DCA B = D C
    This translates to the equation: (4,1)βˆ’(x,y)=(βˆ’2,3)βˆ’(βˆ’8,0)(4, 1) - (x, y) = (-2, 3) - (-8, 0)

  2. Simplify the equation:
    Subtract the coordinates:
    (4βˆ’x,1βˆ’y)=(6,3)(4 - x, 1 - y) = (6, 3)

  3. Solve for xx and yy:
    By comparing the components:

    • 4βˆ’x=64 - x = 6 gives x=βˆ’2x = -2
    • 1βˆ’y=31 - y = 3 gives y=βˆ’2y = -2

Thus, the coordinates of point AA are (βˆ’2,βˆ’2)(-2, -2).


(ii) Find the coordinates of point EE, given that AEBDA E B D is a parallelogram

  1. Set up the equation:
    Since AEBDA E B D is a parallelogram, we know that AE=DBA E = D B.
    The vector AEA E is the difference between the coordinates of E(x,y)E(x, y) and A(βˆ’2,βˆ’2)A(-2, -2), and the vector DBD B is the difference between the coordinates of D(βˆ’8,0)D(-8, 0) and B(4,1)B(4, 1).
    Thus: AE=DBA E = D B
    This translates to the equation: (x,y)βˆ’(βˆ’2,βˆ’2)=(4,1)βˆ’(βˆ’8,0)(x, y) - (-2, -2) = (4, 1) - (-8, 0)

  2. Simplify the equation:
    Subtract the coordinates:
    (x+2,y+2)=(12,1)(x + 2, y + 2) = (12, 1)

  3. Solve for xx and yy:
    By comparing the components:

    • x+2=12x + 2 = 12 gives x=10x = 10
    • y+2=1y + 2 = 1 gives y=βˆ’1y = -1

Thus, the coordinates of point EE are (10,βˆ’1)(10, -1).


Key Formulas or Methods Used

  • Vector Addition/Subtraction:
    The vector between two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) is given by:
    PQβ†’=(x2βˆ’x1,y2βˆ’y1)\overrightarrow{P Q} = (x_2 - x_1, y_2 - y_1)

  • Parallelogram Property:
    Opposite sides of a parallelogram are equal and parallel:
    AB→=DC→andAE→=DB→\overrightarrow{A B} = \overrightarrow{D C} \quad \text{and} \quad \overrightarrow{A E} = \overrightarrow{D B}


Summary of Steps

  1. For ABCDA B C D is a parallelogram:

    • Set up the equation AB=DCA B = D C.
    • Solve for the coordinates of point AA using vector subtraction.
  2. For AEBDA E B D is a parallelogram:

    • Set up the equation AE=DBA E = D B.
    • Solve for the coordinates of point EE using vector subtraction.
  3. Simplify the equations to find the unknown coordinates.