Equation of perpendicular bisector of
AB is
x−y+5=0 -----
(1)Also equation of AB can be written as
y+x+λ=0 (perpendicular to x−y+5=0)
Since, passes through A=(1,−2)
Therefore,
−2+1+λ=0
λ=1
And,
y+x+1=0 ---- (2)
From (1) and (2)
The coordinates of D, which is middle point of AB is given by
D=(3,2)
Now,
Let the coordinates of B be (x1,y1)
Then,
x1+12=−3 and y1−22=2 (since, D is mid-point of AB.)
x1=−7 and y1=6
Hence, point of B=(−7,6) --- (3)
Now equation AC
2x−y−4=0 ---- (4)
Solve this equation bey the given equation
x+2y=0
We get, the point E which is the mid-point of AC
hence,
E=(85,−45)
Let the coordinates of C be (l,m)
Then,
l+12=85 and q−22=45
Hence,
C(l,m)=(115,25)
Therefore equation of line BC which is passing through B and C is given by,
y−m=y1−mx1−l(x−l)
14x+23y=40
Which is required equation of line.