The points A (–1, 0), B (3, 1), C (2, 2) and D (–2, 1) are the vertices of a parallelogram. Is the following statement true or false
Solution:
We know that
The distance between two points (x₁, y₁) and (x₂, y₂) is
d=\(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)
First let us consider
AB = \(\sqrt{(3+1)^{2}+(1-0)^{2}}=\sqrt{17}\)
BC = \(\sqrt{(2-3)^{2}+(2-1)^{2}}=\sqrt{2}\)
CD = \(\sqrt{(-2-2)^{2}+(1-2)^{2}}=\sqrt{17}\)
AD = \(\sqrt{(-2+1)^{2}+(1-0)^{2}}=\sqrt{2}\)
So we get
AB = CD and BC = AD
Therefore, the points are the vertices of a parallelogram.
✦ Try This: The points A (-2, 0), B (4, 2), C (3, 3) and D (-3, 1) are the vertices of a parallelogram.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7
NCERT Exemplar Class 10 Maths Exercise 7.2 Sample Problem 1
The points A (–1, 0), B (3, 1), C (2, 2) and D (–2, 1) are the vertices of a parallelogram. Is the following statement true or false
Summary:
True, the points A (–1, 0), B (3, 1), C (2, 2) and D (–2, 1) are the vertices of a parallelogram
☛ Related Questions:
- The points (4, 5), (7, 6) and (6, 3) are collinear. Is the following statement true or false
- Point P (0, –7) is the point of intersection of y-axis and perpendicular bisector of line segment jo . . . .
- ∆ ABC with vertices A (–2, 0), B (2, 0) and C (0, 2) is similar to ∆ DEF with vertices D (–4, 0) E ( . . . .
visual curriculum