Question

Find the coordinates of the vertex S, if the other 3 vertices of square PQRS are P(0,-3), Q(0,3) and R(6,3)​

Answer 1

Step-by-step explanation:

Let co-ordinates of P are (x

1

,0) and side of square is 'a'

∴Q(x

1

+a,0)

S(x

1

,a)

R(x

1

+a,a)

Now,

m

AS

=m

AC

⇒

x

1

a

=

2

1

⇒x

1

=2a ....(1)

m

BR

=m

BC

⇒

x

1

+a−3

a

=−1⇒x

1

+2a−3=0 ....(2)

from (1) & (2) a=

4

3

& x

1

=

2

3

Hence co-ordinates of P, Q, R, & S can be determined.

as (

2

3

,0),(

4

9

,0),(

4

9

,

4

3

) and (

2

3

,

4

3

)

Answer 2

Answer:

$\huge\tt\green{❥}\tt\pink{S}\tt\blue{o}\tt\pink{l}\tt\purple{u}\tt\orange{t}\tt\orange{ion}\huge\tt\purple{꧂{}}$

There is a line through PQ. Find this line. The line through SR is parallel to it so it has the same slope as PQ.

Use this slope, and the point slope form of a line with the point R. You know that S must lie on this line.

Next, there is a line through QR. Find this line. It is parallel to the line PS. Use the point slope form with the slope from QR and the point P. You know that S lies on this line.

Find the intersection of the two lines we just discussed. It is S.

There is probably a way to do this with vectors and dot products as well.

Edit:

I think it is easier to notice that the vector PQ is the vector SR. PQ=Q-P=<3,1> So you need vector SR=R-S=<2,-1>-<x,y>=<3,1> which means S=<-1,-2>

Now you can compare your answers!