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{(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 = sqrt{90 …(2)}]
[CD = sqrt{((5-8)^2 +(-4-5)^2)}]
[∴ CD =sqrt{((-3)^2 +(-9)^2)}]
[ ∴ CD =sqrt{(9+81)}]
[∴ CD = sqrt{90 ……. (3)}]
[ AD = sqrt{([5-(-4)]^2+ [-4-(-7)]^2)}]
[∴ AD =sqrt{(9^2+3^2)}]
[∴ AD =sqrt{(81+9)}]
[∴ AD = sqrt{90 ……. (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.
∴ square ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.
Chapter 5. Co-ordinate Geometry – Practice Set 5.1 (Page 108)