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`

Options

A. 14x+23y -40=0
B. 14x-23y+40=0
C. 23x+14y -40=0
D. 23x-14y+40=0

Answer

A

Explanation

Let B (`x_(1),y_(1) ) and C(x_(2) ,y_(2))` be two vertices. Thus
` p ((x_(1)+1)/2, (y_(1) -2)/(2))` les on perpendicular bisector
x -y + 5 =0
`(x_(1) +1)/2 , (y_(1) -2)/2 = -5`
` x_(1) -y_(2) =-13`
Also, PN is perpnedicular to AB.
`(y_(1) +2)/(x_(1) -1) xx 1=-1`
` Rightarrow x_(1) +y_(1) =-1`
On solving Eqs. (i) and (ii) we get
` x_(1) =-7 and y_(1) =6`
Thus the coordinates of B are (-7,6) similarly, the coordinates of C are `(11/5 ,2/5)`
Hence, the equation of BC is ` y-6 = (2/5 -6)/(11/5 +7) = (x+7)`
` Rightarrow y -6 = (-14)/(23) (x+7)`
14 x +23y -40 =0