Could someone help me with this "Linear function problem"?

y = 2x + 4

y = -x/3 + 4

2x + 4 = -x/3 + 4

6x + 12 = -x + 12

7x = 0

x = 0

y = 2(0) + 4

y = 4

Therefore other vertex = (0, 4)

Distance from, (0, 4) to (12, 0) = √(4² + 12²) = √160 = 4√10

3rd vertex on y = 2x + 4

(h - 0)² + (k - 4)² = 160

h² + k² = 160

h² + (2h + 4 - 4)² = 160

h² + 4h² = 160

5h² = 160

h² = 32

h = 4√2

k = 2(4√2) + 4 = 8√2 + 4

Therefore third vertex = (h, k) = (4√2, 8√2 + 4)

Fourth vertex = (12 + (4√2 - 0), 0 + (8√2 + 4- 4))

= (12 + (4√2, 8√2)

So all the vertexes are

A(12, 0),

(0, 4),

(4√2, 8√2 + 4),

(12 + (4√2, 8√2)

the 2 lines intersect at (0,4)

find the distance between the 2 points(0,4),(12,0)

and it is sqrt160

draw out the graph.The other 2 points are on the right side of the graph

slope of one of the side is y=2x+b

use (12,0) as one of the point , find the line is y=2x-24

Use (a,b) as another vertex on this line

b=2a-24

160=(12-a)^2+b^2

solve and get (12+2sqrt3,4sqrt3)

and the other point is (2sqrt3,4+4sqrt3)

hi, i'm little confused with "find the co-ordinates of the other 3 sides", are you asking the equation of the sides? or the vertices?