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7.1 Q-7

Question Statement

Given the points A(2,βˆ’4)A(2, -4), B(4,0)B(4, 0), and C(1,6)C(1, 6), use the vector method to find the coordinates of point DD if:

i. ABCDA B C D is a parallelogram
ii. ADBCA D B C is a parallelogram


Background and Explanation

In this problem, we are asked to find the coordinates of point DD given certain conditions about the geometry of the points. We use the properties of a parallelogram and the vector method to solve the problem.

  • Parallelogram Property: In a parallelogram, opposite sides are equal and parallel. This means for the parallelogram ABCDA B C D, the vector ABβ†’\overrightarrow{A B} is equal to the vector CDβ†’\overrightarrow{C D}, and similarly for the other pair of opposite sides.

  • Vector Addition: To find the unknown point DD, we use the fact that the sum of the vectors along the sides of the parallelogram should lead us back to the same point.


Solution

(i) ABCDA B C D is a parallelogram

  1. Set up the relationship between the vectors:
    Since ABCDA B C D is a parallelogram, the vector ABA B is equal to the vector DCD C. This gives us the equation: AB=DCA B = D C
    So,
    (4,0)βˆ’(2,βˆ’4)=(1,6)βˆ’(x,y)(4, 0) - (2, -4) = (1, 6) - (x, y)

  2. Simplify the equation:
    Subtract the coordinates:
    (4βˆ’2,0+4)=(1βˆ’x,6βˆ’y)(4 - 2, 0 + 4) = (1 - x, 6 - y)
    This simplifies to:
    (2,4)=(1βˆ’x,6βˆ’y)(2, 4) = (1 - x, 6 - y)

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

    • 1βˆ’x=21 - x = 2 gives x=βˆ’1x = -1
    • 6βˆ’y=46 - y = 4 gives y=2y = 2

Thus, the coordinates of point DD are (βˆ’1,2)(-1, 2).


(ii) ADBCA D B C is a parallelogram

  1. Set up the relationship between the vectors:
    Since ADBCA D B C is a parallelogram, the vector ADA D is equal to the vector CBC B. Additionally, the vector AD→\overrightarrow{A D} is parallel to CD→\overrightarrow{C D}.
    AD=CBA D = C B
    This gives us the equation:
    (x,y)βˆ’(2,βˆ’4)=(4,0)βˆ’(1,6)(x, y) - (2, -4) = (4, 0) - (1, 6)

  2. Simplify the equation:
    Subtract the coordinates:
    (xβˆ’2,y+4)=(4βˆ’1,0βˆ’6)(x - 2, y + 4) = (4 - 1, 0 - 6)
    This simplifies to:
    (xβˆ’2,y+4)=(3,βˆ’6)(x - 2, y + 4) = (3, -6)

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

    • xβˆ’2=3x - 2 = 3 gives x=5x = 5
    • y+4=βˆ’6y + 4 = -6 gives y=βˆ’10y = -10

Thus, the coordinates of point DD are (5,βˆ’10)(5, -10).


Key Formulas or Methods Used

  • Vector Addition/Subtraction:
    The vector from one point to another is given by the difference in their coordinates: ABβ†’=(xBβˆ’xA,yBβˆ’yA)\overrightarrow{A B} = (x_B - x_A, y_B - y_A)

  • Parallelogram Property:
    Opposite sides of a parallelogram are equal and parallel: AB→=CD→\overrightarrow{A B} = \overrightarrow{C D}


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 DD using vector subtraction.
  2. For ADBCA D B C is a parallelogram:

    • Set up the equation AD=CBA D = C B.
    • Solve for the coordinates of point DD using vector subtraction.
  3. Simplify and solve the resulting system of equations for xx and yy.