What type of a quadrilateral do the points a 1 2 b 13 4 c 6 8 and d 2 4 taken in that order form

What type of a quadrilateral do the points A (2, –2), B (7, 3), C (11, –1) and D (6, –6) taken in that order, form?

Solution:

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

Now using the distance formula,

Now finding the diagonals AC and BD,

As a result, the quadrilateral so formed is 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:

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

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.