the equation of the perpendicular bisectors of sides AB and AC of a triangle ABC are x-y+5=0 and x+2y=0respectively. if the point is A (1,-2), find the equation of line BC.
Equation of the perpendicular bisector of AB is x-y+5=0.........(i)
Also, Equation of AB (is perpendicular to x-y+5=0) can be written as y +x+λ=0, but passes through A = (1,-2)
⇒-2+1+λ=0
⇒λ=1
∴AB = y+x+1=0..............(ii)
From (i) and (ii) , we get,
The coordinates of D, the middle point of AB
⇒D=(3,2)
⇒x=-3,y=2
Now, D is the middle point of AB,
Let the coordinates of B be (α,β)
Then,
Now, equation of AC,
2x-y-4=0.................(iv)
Solve this equation by x+2y=0, we get, the coordinates of E , which is the middle point of AC,
Let the coordinate of C be (p, q) , then
Thus, the equation of BC , which is passing through B and C , is given by
y-q=
⇒14x+23y=40