find the equation of the sides of an isosceles right angled triangle whose vertex is (-2,-3) and the base is on the line x=0​

AC:y=x-1  CB:y=-x-5 AB:x=0

Step-by-step 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

40 minutes

step-by-step explanation:

add the time and multiply by 2

17.08

correct me if i am wrong.