Coordinate Geometry NCERT exercise 7.1 solution part 4 Class Ten Mathematics

Coordinate Geometry

Exercise 7.1Part 4

Question 6: Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.

(i) (-1, -2), (1, 0), (-1, 2), (-3,0)

Solution: Let A = (-1, -2), B = (1, 0), C = (-1, 2) and D = (-3,0)

`AB =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = -1, x2 = 1, y1 = -2 and y2 = 0

So, `AB=sqrt((1+1)^2+(0+2)^2)`

`=sqrt(2^2+2^2)=sqrt(4+4)`

`=sqrt(8)=2sqrt2` units

`BC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 1, x2 = -1, y1 = 0 and y2 = 2

So, `BC=sqrt((-1-1)^2+(2-0)^2)`

`=sqrt((-2)^2+2^2)=sqrt(4+4)`

`=sqrt8=2sqrt2` units


`CD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = -1, x2 = -3, y1 = 2 and y2 = 0

So, `CD=sqrt((-3+1)^2+(0-2)^2)`

`=sqrt((-2)^2+(-2)^2)=sqrt(4+4)`
`=sqrt8=2sqrt2` units

`AD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = -1, x2 = -3, y1 = -2 and y2 = 0

So, `AD=sqrt((-3+1)^2+(0+2)^2)`

`=sqrt((-2)^2+2^2)=sqrt(4+4)`
`=sqrt8=2sqrt2` units

Now, let us calculate the diagonals.

`AC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = -1, x2 = -1, y1 = -2 and y2 = 2

So, `AC=sqrt((-1+1)^2+(2+2)^2)`

`=sqrt(0^2+4^2)=sqrt(16)=4` units

`BD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 1, x2 = -3, y1 = 0 and y2 = 0

So, `BD=sqrt((-3-1)^2+(0-0)^2)`

`=sqrt((-4)^2+0^2)=sqrt16=4` units

Here, `AB=BC=CD=AD` and `AC=BD`
Since all sides are equal and diagonals are equal, so given quadrilateral is a square.


(ii) (-3, 5), (3, 1), (0, 3), (-1,-4)

Solution: Let A = (-3, 5), B = (3, 1), C = (0, 3) and = (-1,-4)

`AB =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = -3, x2 = 3, y1 = 5 and y2 = 1

So, `AB=sqrt((3+3)^2+(1-5)^2)`

`=sqrt(6^2+(-4)^2)`

`=sqrt(36+16)=sqrt52``=2sqrt(13)` unit

`BC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 3, x2 = 0, y1 = 1 and y2 = 3

So, `BC=sqrt((0-3)^2+(3-1)^2)`

`=sqrt((-3)^2+2^2)=sqrt(9+4)`

`=sqrt(13)` unit

`CD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 0, x2 = -1, y1 = 3 and y2 = -4

So, `CD=sqrt((-1-0)^2+(-4-3)^2)`

`=sqrt((-1)^2+(-7)^2)=sqrt(1+49)`

`=sqrt(50)=5sqrt2` unit

`AD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = -3, x2 = -1, y1 = 5 and y2 = -4

So, `AD=sqrt((-1+3)^2+(-4-5)^2)`

`=sqrt(2^2+(-9)^2)=sqrt(4+81)`

`=sqrt(85)` unit

Now, let us calculate the diagonal AC:

`AC = sqrt((0+3)^2+(3-5)^2)`

`=sqrt(3^2+(-2)^2)`

`=sqrt(9+4)=sqrt(13)`

So, we get `AC+BC=sqrt(13)+sqrt(13)`

`=2sqrt(13)=AB`

It means that points A, B and C are collinear; with point C coming between points A and B. So, points A, B, C and D cannot make a quadrilateral.


(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution: Let A = (4, 5), B = (7, 6), C = (4, 3) and D = (1, 2)

`AB =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = -3, x2 = 4, y1 = 5 and y2 = 6

So, `AB=sqrt((7-4)^2+(6-5)^2)`
`=sqrt(3^2+1^2)=sqrt(9+1)=sqrt(10)` unit

`BC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 7, x2 = 4, y1 = 6 and y2 = 3

So, `BC=sqrt((4-7)^2+(3-6)^2)`

`=sqrt((-3)^2+(-3)^2)=sqrt(9+9)`

`=sqrt(18)=3sqrt2` unit

`CD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 4, x2 = 1, y1 = 3 and y2 = 2

So, `CD=sqrt((1-4)^2+(2-3)^2)`

`=sqrt((-3)^2+(-1)^2)=sqrt(9+1)`

`=sqrt(10)` unit

`AD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 4, x2 = 1, y1 = 5 and y2 = 2

So, `AD=sqrt((1-4)^2+(2-5)^2)`

`=sqrt((-3)^2+(-3)^2)=sqrt(9+9)`

`=sqrt(18)=3sqrt2` unit

Now let us calculate the diagonals

`AC =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 4, x2 = 4, y1 = 5 and y2 = 3

So, `AC=sqrt((4-4)^2+(3-5)^2)`

`=sqrt(0^2+(-2)^2)=sqrt4=2` unit

`BD =sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Here, x1 = 7, x2 = 1, y1 = 6 and y2 = 2

So, `BD=sqrt((1-7)^2+(2-6)^2)`

`=sqrt((-6)^2+(-4)^2)=sqrt(36+16)`

`=sqrt(52)=2sqrt(13)` unit

Here, `AB=CD=sqrt(10)` unit and `BC=AD=3sqrt2` unit.

Moreover, `AC=2` unit and `BD=2sqrt13` unit

Here opposite sides are equal but diagonals are not equal, thus with the given points a parallelogram is formed.



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