Answer:
AC:y=x-1 CB:y=-x-5 AB:x=0
Explanation:
Consider the triangle. The base AB is on the line x=0, the vertex C is (-2,-3)
The side AC is equal to BC. The angle ACB is 90 degrees. If the base is on the line x=o, it is on the axis Y.Explore the distance from the point C to the AB
c(-2,-3), the distance to the axis Y is equal to the modul of the coordinate x (-2), it is 2. The coordinates of point projected by the point C to the axis Y is N(0,-3). The modul of the height is 2, the height of the isosceles triangle to the base is the bisectrix, so the angle BCA is 90/2=45degrees, CBA is 180-90-45=45 degrees too
the heigt CN is equal to side NB, NB=2
Suppose B is (0,y) (x=0 because the base is on this line)
THe modul of the vector NB is equal to sqrt ((0-0)^2+(y+3)^2)= 2
modul (y+3)= 2
y=-1 or y=-5
(0,-1), (0,-5) - two points, one of them (suppose B) is (0,-5) when A is (0,-1) (A is remote from the point N on the same distance with B, because AB is the median too)
Find CB and AC
Use the equation for AC
(x-0)/(-2-0)= (y+1)/(-3+1)
x/-2= (y+1)/-2
x=y+1
y=x-1
For CB
(x-0)/ (-2-0)= (y+5)/ (-3-(-5))
x/-2= (y+5)/2
-x=y+5
y=-x-5