(i) A quadrilateral is a parallelogram, if mid points of diagonals AC and BD are same.
(ii) A parallelogram is not a rectangle, if diagonals AC ≠ BD.
(iii) A parallelogram may be a rhombus if AB = BC.
(iv) If in a parallelogram diagonals are equal, then it is rectangle.
In a rectangle if the sides AB = BC, then the rectangle is a square.
For parallelogram with vertices A(2, –2), B(7, 3), C(11, –1), D(6, –6).
mid point of AC = mid point of BD
\Rightarrow\left(\frac{2+11}{2}, \frac{-2-1}{2}\right)=\left(\frac{7+6}{2}, \frac{3-6}{2}\right)\\ \Rightarrow\left(\frac{13}{2}, \frac{-3}{2}\right)=\left(\frac{13}{2}, \frac{-3}{2}\right), \text { which is true. }
Hence, ABCD is a parallelogram.
How, we will check whether AC = BD
or AC2 = BD2
\Rightarrow(11-2)^{2}+(-1+2)^{2}=(6-7)^{2}+(-6-3)^{2}\\ \Rightarrow(9)^{2}+(1)^{2}=(-1)^{2}+(-9)^{2}\\ \Rightarrow 81+1=1+81\\ \Rightarrow 82=82, \text { which is true. }
As the diagonals are equal so it is a rectangle or square. Now, we will check whether adjacent sides AB = BC
\text { or } \mathrm{AB}^{2}=\mathrm{BC}^{2}\\ \Rightarrow(7-2)^{2}+(3+2)^{2}=(11-7)^{2}+(-1-3)^{2}\\ \Rightarrow 5^{2}+5^{2}=(4)^{2}+(-4)^{2}\\ \Rightarrow 25+25=16+16\\ \Rightarrow 50 \neq 32, \text { which is false. }
So, ABCD is not a square. Hence ABCD is a rectangle.
Solution:
Given, the points are A(2, -2) B(7, 3) C(11, -1) and D(6, -6)
We have to find the type of a quadrilateral formed by the given points.
The distance between two points P (x₁ , y₁) and Q (x₂ , y₂) is
√[(x₂ - x₁)² + (y₂ - y₁)²]
Distance between A(2, -2) and B(7, 3) = √[(7 - 2)² + (3 - (-2))²]
= √[(5)² + (5)²]
= √(25 + 25)
= 5√2
Distance between B(7, 3) and C(11, -1) = √[(11 - 7)² + (-1 - 3)²]
= √[(4)² + (-4)²]
= √(16 + 16)
= 4√2
Distance between C(11, -1) and D(6, -6) = √[(6 - 11)² + (-6 - (-1))²]
= √[(-5)² + (-5)²]
= √(25 + 25)
= 5√2
Distance between A(2, -2) and D(6, -6) = √[(6 - 2)² + (-6 - (-2))²]
= √[(4)² + (-4)²]
= √(16 + 16)
= 4√2
Distance between A(2, -2) and C(11, -1) = √[(11 - 2)² + (-1 - (-2))²]
= √[(9)² + (1)²]
= √(81 + 1)
= √82
Distance between B(7, 3) and D(6, -6) = √[(6 - 7)² + (-6 - 3)²]
= √[(-1)² + (-9)²]
= √(81 + 1)
= √82
We observe that AB = CD and BC = AD.
Also, AC = BD
The opposite sides of the quadrilateral are equal.
The diagonals of the quadrilateral are equal.
Therefore, the given points represent a rectangle.
✦ Try This: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:(-1, -2), (1, 0), (-1, 2), (-3, 0)
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7
NCERT Exemplar Class 10 Maths Exercise 7.3 Problem 3
Summary:
The type of a quadrilateral formed by the points A (2, –2), B (7, 3), C (11, –1) and D (6, –6) taken in that order is a rectangle as the opposite sides and diagonals are equal
☛ Related Questions:
Question 3 What type of quadrilateral do the points A 2, 2, B 7,3 C 11, 1 and D 6, 6 taken in that order form?
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1
The points are A(2, –2), B(7, 3), C(11, –1) and D(6, –6)
Using distance formula,
d = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)`
AB = `sqrt((7 - 2)^2 + (3 + 2)^2`
= `sqrt((5)^2 + (5)^2`
= `sqrt(25 + 25)`
= `sqrt(50)`
= 5`sqrt(2)`
BC = `sqrt((11 - 7)^2 + (-1 - 3)^2`
= `sqrt((4)^2 + (-4)^2`
= `sqrt(16 + 16)`
= `sqrt(32)`
= `4sqrt(2)`
CD = `sqrt((6 - 11)^2 + (-6 + 1)^2`
= `sqrt((-5)^2 + (-5)^2`
= `sqrt(25 + 25)`
= `sqrt(50)`
= `5sqrt(2)`
DA = `sqrt((2 - 6)^2 + (-2 + 6)^2`
= `sqrt((-4)^2 + (4)^2`
= `sqrt(16 + 16)`
= `sqrt(32)`
= `4sqrt(2)`
Finding diagonals AC and BD, we get,
AC = `sqrt((11 - 2)^2 + (-1 + 2)^2`
= `sqrt((9)^2 + (1)^2`
= `sqrt(81 + 1)`
= `sqrt(82)`
And BD = `sqrt((6 - 7)^2 + (-6 - 3)^2`
= `sqrt((-1)^2 + (-9)^2`
= `sqrt(1 + 81)`
= `sqrt(82)`
The Quadrilateral formed is rectangle.