The equations of the perpendicular bisectors of the sides `A Ba n dA C` of triangle `A B C` are `x-y+5=0` and `x+2y=0` , respectively. If the point `A` is `(1,-2)` , then find the equation of the line `B Cdot`

Explanation

Let the coordinates of `B` and `C` be `(x_(1),y_(1))` and `(x_(2),y_(2))` respectively. Let `m_(1)` and `m_(2)` be the slopes of `AB` and `AC`, respectively. Then,

`m_(1)="slope of"AB=(y_(1)+2)/(x_(1)-1)`
and `m_(2)="slope of"AC=(y_(2)+2)/(x_(2)-1)`
Let `F` and `E` be the mid point of `AB` and `AC` , respectively.
Then, the coordinates of `E` and `F` are
`E((x_(1)+1)/(2),(y_(2)-2)/(2))` and `F ((x_(1)+1)/(2),(y_(2)-2)/(2))`, respectively.
Now, `F` lies on `x-y+5=0`
`implies(x_(1)+1)/(2)-(y_(1)-2)/(2)=-5`
`impliesx_(1)-y_(1)+13=0` .........`(i)`
Since, `AB` is perpendicular to `x-y+5=0`
`:.("slope of"AB)*("slope of"x-y+5=0)=-1`
`implies(y_(1)+2)/(x_(1)-1)*(1)=-1`
`impliesy_(1)+2=-x_(1)+1`
`impliesx_(1)+y_(1)+1=0`........`(ii)`
On solving Eqs. `(i)` and `(ii)`, we get
`x_(1)=-7`, `y_(1)=6`
So, the coordinates of `B` are `(-7,6)`.
Now, `E` lies on `x+2y=0`
`:. (x_(2)+1)/(2)+2((y_(2)-2)/(2))=0`
`impliesx_(2)+2y_(2)-3=0` ........`(iii)`
Since, `AC` is perpendicular to `x+2y=0`
`:. ("slope of"AC)*(slope of"x+2y=0)=-1`
`implies(y_(2)+2)/(x_(2)-1)*(-(1)/(2))=-1`
`implies2x_(2)-y_(2)=4`....`(iv)`
On solving Eqs. `(iii)` and `(iv)`, we get
`x_(2)=(11)/(5)` and `y_(2)=(2)/(5)`
So, the coordinates of `C` are `((11)/(5),(2)/(5))`
Thus, the equation of `BC` is
`y-6=(2//5-6)/(11//5+7)(x+7)`
`implies-23(y-6)=14(x+7)`
`implies14x+23y-40=0`