The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are x−y+5=0 and x+2y=0 respectively. If the co-ordinates of A are (1,−2), then the equation of BC is
A
23x+14y−40=0
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B
14x+23y−40=0
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C
23x−14y+40=0
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D
None of these
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Solution
The correct option is B14x+23y−40=0 ∵AB line is perpendicular to EO(perpendicular bisector to side AB) and passes through vertex A, Equation of AB::x+y+1=0
Coordinates of E=(x1+12,y1−22)
Which is also intersection point of AB with x−y+5=0(EO line equation)
So, x1+12=−3;y1−22=2⇒(x1,y1)≡(−7,6)
similarly equation of AC::y=2x−4
and coordinates of F=(x2+12,y2−22)
Which is also intersection point of AC with x+2y=0(F0 line equation)
So, x2+12=85;y2−22=−45⇒(x2,y2)≡(11/5,2/5)
So, the equation of sideBC: y−6=2/5−611/5+7(x+7)⇒14x+23y−40=0