Show that A(4, -1), B(6, 0), C(7, -2) and D(5, -3)are vertices of a square.
15. Show that A(4, -1), B(6, 0), C(7, -2) and D(5, -3)are vertices of a square.
The given points are A(4, –1), B(6, 0), C(7, –2) and D(5, –3).
AB = \[\sqrt{\left( 6 - 4 \right)^2 + \left( 0 + 1 \right)^2} = \sqrt{4 + 1} = \sqrt{5}\]
BC =\[\sqrt{\left( 6 - 7 \right)^2 + \left( 0 + 2 \right)^2} = \sqrt{1 + 4} = \sqrt{5}\]
CD =\[\sqrt{\left( 7 - 5 \right)^2 + \left( - 2 + 3 \right)^2} = \sqrt{4 + 1} = \sqrt{5}\]
AD = \[\sqrt{\left( 5 - 4 \right)^2 + \left( - 3 + 1 \right)^2} = \sqrt{1 + 4} = \sqrt{5}\]
AB = BC = CD = DA
Slope of AB = \[\frac{0 + 1}{6 - 4} = \frac{1}{2}\]
Slope of BC = \[\frac{- 2 - 0}{7 - 6} = - 2\]
Slope of CD = \[\frac{- 3 + 2}{5 - 7} = \frac{1}{2}\]
Slope of AD = \[\frac{- 3 + 1}{5 - 4} = - 2\]
Thus, AB perpendicular to BC and AD. Also, CD perpendicular to AD and AB.
So, all the sides are equal to each other and are perpendicular.
Thus, they form a square.
Explanation:-
The given points are A(4, -1), B(6, 0), C(7, -2), and D(5, -3).
To find the length of each side of the quadrilateral ABCD, we use the distance formula.
AB = √[(6 – 4)^2 + (0 + 1)^2] = √[4 + 1] = √5
BC = √[(7 – 6)^2 + (0 + 2)^2] = √[1 + 4] = √5
CD = √[(7 – 5)^2 + (-2 + 3)^2] = √[4 + 1] = √5
AD = √[(5 – 4)^2 + (-3 + 1)^2] = √[1 + 4] = √5
We observe that all four sides of the quadrilateral are equal to each other and have a length of √5.
To determine if the quadrilateral is a square, we look at the slopes of the sides.
The slope of AB = (0 + 1)/(6 – 4) = 1/2
The slope of BC = (-2 – 0)/(7 – 6) = -2
The slope of CD = (-3 + 2)/(5 – 7) = 1/2
The slope of AD = (-3 + 1)/(5 – 4) = -2
We notice that AB is perpendicular to BC and AD because the product of their slopes is -1. Similarly, CD is perpendicular to AD and AB. Therefore, all four sides of the quadrilateral are perpendicular to each other.
Since all sides of the quadrilateral are equal in length and perpendicular to each other, we can conclude that it is a square.
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)