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Question

The equation of the prependicular bisectors of the sides AB & AC of a triangle ABC are xy+5=0 & x+2y=0, respectively. If the point A is (1, -2) find the equation of the line BC.

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Solution

Equation of perpendicular bisector of AB is xy+5=0 ----- (1)
Also equation of AB can be written as
y+x+λ=0 (perpendicular to xy+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 y122=2 (since, D is mid-point of AB.)

x1=7 and y1=6
Hence, point of B=(7,6) --- (3)

Now equation AC
2xy4=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 q22=45
Hence,
C(l,m)=(115,25)

Therefore equation of line BC which is passing through B and C is given by,
ym=y1mx1l(xl)
14x+23y=40
Which is required equation of line.

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