What type of a quadrilateral do the points a 12 b 13 4 c 6 8 and d 2 4 taken in that order form

(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?

Open in App

Suggest Corrections

The points are A(2, –2), B(7, 3), C(11, –1) and D(6, –6)