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Case Study – 2
Morning assembly is an integral part of every school's schedule. In the assembly, students always stand in rows and columns and this makes a coordinate system.
In a school, there are $200$ students and they all assemble for prayer in $10$ rows. $4$ students are at A, B, C and D with the following positions of the coordinate system :
A $(3, 4)$, B $(6, 7)$, C $(9, 4)$ and D $(6, 1)$.
Based on the above, answer the following questions :
(a) Find the distance between A and B.
(b) Find the distance between C and D.
(c) Show that ABCD forms a parallelogram.
OR
(c) Find the mid-point of the line segments AC and BD.
Morning assembly is an integral part of every school's schedule. In the assembly, students always stand in rows and columns and this makes a coordinate system.
In a school, there are $200$ students and they all assemble for prayer in $10$ rows. $4$ students are at A, B, C and D with the following positions of the coordinate system :
A $(3, 4)$, B $(6, 7)$, C $(9, 4)$ and D $(6, 1)$.
Based on the above, answer the following questions :
(a) Find the distance between A and B.
(b) Find the distance between C and D.
(c) Show that ABCD forms a parallelogram.
OR
(c) Find the mid-point of the line segments AC and BD.
Show SolutionHide Solution↓
(a) Distance between A $(3,4)$ and B $(6,7)$:
$AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
(b) Distance between C $(9,4)$ and D $(6,1)$:
$CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
(c) (i) To show ABCD forms a parallelogram, we can show opposite sides are equal or diagonals bisect each other.
Using distance formula for all sides:
$AB = 3\sqrt{2}$ (from part a)
$CD = 3\sqrt{2}$ (from part b)
Distance between B $(6,7)$ and C $(9,4)$:
$BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
Distance between A $(3,4)$ and D $(6,1)$:
$AD = \sqrt{(6-3)^2 + (1-4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
Since all sides are equal ($AB=BC=CD=DA=3\sqrt{2}$), ABCD is a rhombus, which is a type of parallelogram.
OR
(c) (ii) Mid-point of AC:
A $(3,4)$, C $(9,4)$
Mid-point $= (\frac{3+9}{2}, \frac{4+4}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6,4)$.
Mid-point of BD:
B $(6,7)$, D $(6,1)$
Mid-point $= (\frac{6+6}{2}, \frac{7+1}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6,4)$.
Since the mid-points of AC and BD are the same, the diagonals bisect each other. Therefore, ABCD is a parallelogram.
$AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
(b) Distance between C $(9,4)$ and D $(6,1)$:
$CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
(c) (i) To show ABCD forms a parallelogram, we can show opposite sides are equal or diagonals bisect each other.
Using distance formula for all sides:
$AB = 3\sqrt{2}$ (from part a)
$CD = 3\sqrt{2}$ (from part b)
Distance between B $(6,7)$ and C $(9,4)$:
$BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
Distance between A $(3,4)$ and D $(6,1)$:
$AD = \sqrt{(6-3)^2 + (1-4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units.
Since all sides are equal ($AB=BC=CD=DA=3\sqrt{2}$), ABCD is a rhombus, which is a type of parallelogram.
OR
(c) (ii) Mid-point of AC:
A $(3,4)$, C $(9,4)$
Mid-point $= (\frac{3+9}{2}, \frac{4+4}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6,4)$.
Mid-point of BD:
B $(6,7)$, D $(6,1)$
Mid-point $= (\frac{6+6}{2}, \frac{7+1}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6,4)$.
Since the mid-points of AC and BD are the same, the diagonals bisect each other. Therefore, ABCD is a parallelogram.