Question

the equations of two sides of a square are Y=3X-1 and X+3Y-6 =0. (0,-1) is one of the vertex of the square. Find the coordinates of the other vertices.

Answer 1

I hope this will be helpful
pls.Comment if wrong

Answer 2

The equation will be as follows: y=3x-1 and x+3y-6=0 because the vertices are (0,1)

Point A will be (0,1)

y=3x-1

x+3y-6=0 --> y = (-1/3)x + 2

x + 3(3x-1) = 6

10x = 9

x = 0.9, y = 1.7

= B(0.9,1.7)

This is the distance between the points:

s=Sqroot of diffy^2+diffx^2 whic= sqrt 2.7^2 + 0.9^2 which = sqrt 8.1 or 9sqrt10/10

Now we will find the line thru A(0,1) which is parallel to

x+3y-6=0

y = (-1/3)x - 1

The x-distance from A to D is 2.7, and the y-distance is 0.9

D (-2.7,-0.1)

Now the side CD wil go through D with the slope of 3 this will also be parallel

y + 0.1 = 3(x + 2.7)

y = 3x + 8

now the vertex c is the  intersection of y = 3x + 8 and y = (-1/3)x + 2

3x + 8 = -x/3 + 2

10x/3 = -6

x = -1.8, y = 2.6

C(-1.8,2.6)

point (9/10,17/10), or (0.9,1.7) is one vertex,

xB-Xa=0.9–0=0 and yb-ya=1.7-(-1)=2.7

xE = xb + 2.7 = 0.9 + 2.7 = 3.6

yE = yB - 0.9 = 1.7 - 0.9

= 0.8

xF = xA +2.7

= 0 + 2.7 = 2.7

yF = yA - 0.9

= 1 - 0.9 = - 1.9 ( these are the sides of the square)