Show that points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are vertices of a rhombus ABCD
Answer:-
The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
Distance between two points [= sqrt{((x_2 – x_1)^2 + (y_2 – y_1)^2)}]
According to the distance formula,
[AB =sqrt{([-1-(-4)]^2 +[2-(-7)]^2}]
[∴ AB = sqrt{(3^2+9^2)}]
[∴ AB =sqrt{(9+81)} ]
[∴ AB = sqrt{90} …(1)]
[BC =sqrt{([8-(-1)]^2+(5-2)^2)}]
[∴ BC =sqrt{(9^2+3^2)}]
[∴ BC = sqrt{(81+9)}]
[∴ BC = sqrt90 …(2)]
[CD = sqrt{((5-8)^2 +(-4-5)^2)}]
[∴ CD =sqrt((-3)^2 +(-9)^2)]
[ ∴ CD =sqrt{(9+81)}]
[∴ CD = sqrt90 ……. (3)]
[AD = sqrt{([5-(-4)]^2+ [-4-(-7)]^2)}]
[∴ AD =sqrt{(9^2+3^2)}]
[∴ AD =sqrt{(81+9)}]
[∴ AD = sqrt90 ……. (4)]
From (1), (2), (3), and (4)
AB = BC = CD = AD
Thus, all sides are equal.
In a quadrilateral, if all the sides are equal, then it is a rhombus.
Chapter 5. Co-ordinate Geometry – Practice Set 5.1 (Page 107)
