Question

Three vertices of a square PQRS are (-4,0), Q(1,0), R(1,-5). Plot on points. Also find the co-ordinates of the missing vertex S.

Answer 1

HELLO DEAR,

[figure is in the attachment]

now,we know the properties of Square
all sides of a square is equal
and all angels are of 90°

GIVEN:- Three vertices of a square PQRS are P(-4,0) ,Q(1,0),R(1,-5),S(x , y).

let the vertices of S(x , y).

so, mid point of QS = PR

mid point formula is $\sf{\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2}}$

$\sf{\frac{1 + x}{2} , \frac{0 + y}{2} = \frac{1 - 4}{2} , \frac{0 - 5}{2}}$

$\sf{1 + x = -3}$ , $\sf{y = -5}$

⇒x = -4 , y = -5

$\sf{\boxed{HENCE,\;\; point s(-4 , -5)}}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

we know that,

the properties of square all sides of a square us equal and all angles of 90°.

let vertices of S (x,y)

Given:- three vertices of square PQRS are

P (-4,0)

Q (1,0)

R (1,-5)

and vertice of S (x,y)

mid point of QS = PR

mid point formula is

$\frac{x1 + x2}{2} = \frac{y1 + y2}{2}$

$\frac{1 + x}{2} and \: \frac{0 + y}{2} = \frac{1 - 4}{2} and \: \frac{0 - 5}{2}$

1 + x = 3 , y = - 5

x = - 4 , y = - 5

Hence, points of S (-4, -5)