Question

the equations of perpendicular bisectors of the side AB and AC of the triangle ABC are x-y+5=0 and x+2y=0 respectively . if the point A is (1,-2) ,the equation of the line BC is ???

Answer 1

Answer:

equation for AB is y+2=-1(x-1)

y+x+1=0 (1)

equation of AC is y+2=2(x-1)

y-2x+4=0 (2)

the other bisectors line is x-y+5=0&x+2y=0 (3) & (4)

putting values from(4)→(2)

we get 5y=-4

$y = - \frac{4}{5} and \: x = \frac{8}{5}$

putting values from (3) to (1)

x=−3&y=2x=-3&y=2

B point by plotting we get(−7,6)

$c \: point \: is \: (\frac{11}{5} \frac{2}{5} )$

$y - 6 = \frac{ \frac{28}{5} }{ - \frac{46}{5} } .(x + 7)$

$y - 6 = - \frac{14}{23} (x + 7)$

$23y - 138 = - 14x - 98$

So, eqn of line BC is 23y +14x-40=0

Answer 2

Answer:

23Y+14X+40=0

HOPE IT HELPS

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