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**

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