Question

Show that p(a,b),q(a+3,b+4),r(a-1,b+7),s(a-4,b+3) are the vertices of a square. What is the area of the square?

Answer 1

W know that a quadrilateral is a square if all sides are equal and diagonal are equal to each other

Here PQRS is quadrilateral, to prove it is a square we need to prove

PQ = QR = RS = SP and PR = QS

For which we will used Distance formula to find desired distance between given coordinates

We know that Distance between two co-ordinates =

$\sqrt{(x_2 - x_1)^2 +(y_2-y_1)^2}$

∴ Length of PQ

$\sqrt{(a+3-a)^2+(b+4-b)^2}  = \sqrt{3^2 + 4^2 } = \sqrt{25} = 5 units$

QR

$\sqrt{(a-1-a-3)^2+(b+7-b-4)^2}  = \sqrt{(-4)^2 + 3^2 } = \sqrt{25} = 5 units$

RS

$\sqrt{(a-4-a+1)^2+(b+3-b-7)^2}  = \sqrt{(-3)^2 + (-4)^2 } = \sqrt{25} = 5 units$

SP

$\sqrt{(a-4-a)^2+(b+3-b)^2}  = \sqrt{4^2 + 3^2 } = \sqrt{25} = 5 units$

All sides are equal (PQ = QR = RS = SP = 5 units

But it can be a Rhombus too (all sides equal), so to prove it is a square we need to prove its diagonals are equal to each other (PE = QS)

PR

$\sqrt{(a-1-a)^2+(b+7-b)^2}  = \sqrt{1^2 + 7^2 } = \sqrt{50} = 5\sqrt2 units$

QS

$\sqrt{(a-4-a-3)^2+(b+3-b-4)^2}  = \sqrt{(-7)^2 + (-1)^2 } = \sqrt{50} = 5\sqrt2 units$

So here All sides of given quadrilateral are equal ( 5 units each) and both diagonals are equal ( 5√2 units), we can say PQRS is a Square

Area of Square PQRS = Side² = 5² =25 square units

Answer 2

Refer the picture

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