Coordinate Geometry
NCERT Exercise
7.1 Part 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.