Question

(c) (-4, -1), (-1, -1), (-1, 2) and (a, b) are
the vertices of a square. Find a, b and
the length of diagonal of a square.[4]​

Answer 1

Given :-

Let us assume the vertices of square as A, B, C, D such that

• A (- 4, - 1)

• B (- 1, - 1)

• C (- 1, 2)

and

• D (a, b)

To Find :-

• Value of a and b

• Length of diagonal

Concept Used :-

• In order to find the values of a and b from the given vertices of A, B, C, D taken in order forms a square, we have to used the concept, midpoint of AC is equals to midpoint of BD as diagonals bisect each other in square.

$\large\underline{\bf{Solution-}}$

Given that vertices of square ABCD are

• A (- 4, - 1)

• B (- 1, - 1)

• C (- 1, 2)

and

• D (a, b)

We know,

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the midpoint of AB, then coordinates of C is

$\boxed{ \quad\sf \:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg) \quad}</p><p>$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: B(x_2,y_2)$

Let us first find midpoint of AC.

• Coordinates of A = ( - 4, - 1)

• Coordinates of C = (- 1, 2)

Using midpoint Formula,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: AC \: = \: \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg)$

Here,

• x₁ = - 4

• x₂ = - 1

• y₁ = - 1

• y₂ = 2

So,

$\rm :\longmapsto\:Midpoint \: of \: AC = \bigg(\dfrac{ - 4 - 1}{2} ,\dfrac{ - 1 + 2}{2} \bigg)$

$\rm :\longmapsto\:Midpoint \: of \: AC = \bigg(\dfrac{ - 5}{2} ,\dfrac{1}{2} \bigg)$

Now,

Let us first find midpoint of BD.

• Coordinates of B = ( - 1, - 1)

• Coordinates of D = (a, b)

Using midpoint Formula,

$\rm :\longmapsto\: \sf \: Midpoint \: of \: BD \: = \: \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg)$

Here,

• x₁ = - 1

• x₂ = a

• y₁ = - 1

• y₂ = b

So,

$\rm :\longmapsto\:Midpoint \: of \: BD = \bigg(\dfrac{ - 1 + a}{2} ,\dfrac{ - 1 + b}{2} \bigg)$

We know,

• In a square diagonals bisect each other,

So,

• Midpoint of AC = Midpoint of BD

$\rm :\longmapsto\:\bigg(\dfrac{ - 5}{2} ,\dfrac{1}{2} \bigg) = \bigg(\dfrac{ - 1 + a}{2} ,\dfrac{ - 1 + b}{2} \bigg)$

So, on comparing we get

$\rm :\implies\: - 5 = - 1 + a \: \: \: and \: \: \: 1 = - 1 + b$

$\red{\bf :\implies\:a = - 4 \: \: \: and \: \: \: b = 2}$

Now to find the length of diagonal of a square.

We know,

• In square, the length of diagonals are equal.

So,

We evaluate, Length of diagonal AC

• Coordinates of A = ( - 4, - 1)

• Coordinates of C = (- 1, 2)

Using Distance Formula,

We know,

${\underline{\boxed{\rm{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

Here,

• x₁ = - 4

• x₂ = - 1

• y₁ = - 1

• y₂ = 2

So,

${\rm{\quad \: AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}$

${\rm{\quad \: AC = \sqrt{( - 1 + 4)^2 + (2 + 1)^2} \quad}}$

${\rm{\quad \: AC = \sqrt{(3)^2 + (3)^2} \quad}}$

${\rm{\quad \: AC = \sqrt{(9) + (9)} \quad}}$

${\rm{\quad \: AC = \sqrt{18} \quad}}$

${\rm{\quad \: AC = \sqrt{3 \times 3 \times 2} \quad}}$

$\purple{{\rm{\quad \: AC = 3\sqrt{2} \quad}} \: \rm \: units}$