The vertices of a quadrilateral ABCD are A(6, – 2), B(9, 2), C(5,

Q) The vertices of a quadrilateral ABCD are A(6, – 2), B(9, 2), C(5, – 1) and D(2, – 5). Prove that ABCD is a rhombus, and not a square.

Ans: Let’s quickly understand the comparisons of a Square and a Rhombus:

Key Similarities:
1. Both are polygons with all four sides equal in length
2. both have opposite sides parallel
3. Both shapes’ diagonals bisect each other at 90°.

Key Differences:
1. In a Square, all 4 angles are right angles while in a Rhombus, opposite angles are equal, but may not be right angles.
2. In the Rhombus, one set of opposite angles may be acute, and the other set is obtuse.
3. The square has all equal length diagonals. The Rhombus has unequal diagonals.

Next, we can solve this question by 2 methods:

Solution 1: By coordinate geometry (plotting on a graph):

Let’s first plot all vertices on a graph.

we can clearly that all the four sides are equal in length.

Next, let’s connect diagonals.

We can clearly see that the diagonals are of unequal length.

Hence ABCD shape is a Rhombus, not a Square.

Hence Proved !

Solution 2: By coordinate geometry (without plotting on a graph):

Step 1: Calculate Length of Sides:

$AB = \Sqrt [(6 - 9)^2 + (-2 -2)^2]$
$= \Sqrt [(-3)^2 + (-4)^2] = \Sqrt [9 + 16]$
$= \Sqrt (25) = 5$

$BC = \Sqrt [(9 - 5)^2 + (2 - (-1))^2]$
$= \Sqrt [(4)^2 + (3)^2] = \Sqrt [16 + 9]$
$= \Sqrt (25) = 5$

$CD = \Sqrt [(5-2)^2 + ((-1) - (-5))^2]$
$= \Sqrt [(3)^2 + (-4)^2] = \Sqrt [9 + 16]$
$= \Sqrt (25) = 5$

$DA = \Sqrt [(2 - 6)^2 + ((-5) - (-2))^2]$
$= \Sqrt [(-4)^2 + (-3)^2] = \Sqrt [16 + 9]$
$= \Sqrt (25) = 5$

Since all the four sides have equal length, hence the shape is either Rhombus or a Square

Step 2: Diagonals Length:

$AC = \Sqrt [(6 - 5)^2 + (-2 - (-1))^2]$
$= \Sqrt [(1)^2 + (-1)^2] = \Sqrt [1 + 1]$
$= \Sqrt (2) = \sqrt2$

$BD = \Sqrt [(9 - 2)^2 + (2 - (-5))^2]$
$= \Sqrt [(7)^2 + (7)^2] = \Sqrt [49 + 49]$
$= \Sqrt (49 \times 2)) = 7\sqrt2$

Since the diagonals are unequal in length, hence the shape is a Rhombus, not a square. (Before conclusion, we must check if the diagonals are bisecting at 90° or not.)

Step 3: Diagonals intersection angle:

(i) In Δ AOB, We need to check if AO² + BO² = AB²
AO = half of AC = $\frac{\sqrt 2}{2} = \frac{1}{\sqrt2}$
BO = half of BD = $\frac{7 \sqrt 2}{2} = \frac{7}{\sqrt2}$
AO² + BO² = $(\frac{1}{\sqrt2})^2 + (\frac{7}{\sqrt2})^2 = (\frac{1}{2}) + (\frac{49}{2}) =\frac{50}{2} = 25$
AB² = 5² = 25

Hence, AO² + BO² = AB² and hence, ∠ AOB is 90°.

(ii) Next, No need to check in Δ COD because ∠ COD is also 90° (by opposite angle property).
Hence, we should check in Δ BOC, if BO² + CO² = BC²
BO = half of BD = $\frac{7 \sqrt 2}{2} = \frac{7}{\sqrt2}$
CO = half of AC = $\frac{\sqrt 2}{2} = \frac{1}{\sqrt2}$
BO² + CO² = $(\frac{7}{\sqrt2})^2 + (\frac{1}{\sqrt2})^2 = (\frac{49}{2}) + (\frac{1}{2}) =\frac{50}{2} = 25$
BC² = 52 = 25
Hence, BO² + CO² = BC² and hence, ∠ BOC is also 90°.