The equation of perpendicular bisector of two side AB and AC of a triangle ABC are x+y+1=0 And x-y+1=0,respectively .if circumradius of triangle ABC is 2 units , then focus of vertex A is
Answers
equation for AB is y+2=-1(x-1)
equation for AB is y+2=-1(x-1)y+x+1=0 (1)
equation for AB is y+2=-1(x-1)y+x+1=0 (1)equation of AC is y+2=2(x-1)
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)
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)
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)
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
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=-4y = -
and \: x =
y=−
y=− 5
y=− 54
y=− 54
y=− 54 andx=
y=− 54 andx= 5
y=− 54 andx= 58
y=− 54 andx= 58
y=− 54 andx= 58
y=− 54 andx= 58 putting values from (3) to (1)
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 5
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52 )
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52 )y - 6 = 28/5- 46/5.(x + 7)y−6=
.(x + 7)y−6= −
.(x + 7)y−6= − 5
.(x + 7)y−6= − 546
.(x + 7)y−6= − 546
.(x + 7)y−6= − 546
.(x + 7)y−6= − 546 5
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528 .(x+7)
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=−
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 23
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314 (x+7)
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314 (x+7)23y - 138 = - 14x - 9823y−138=−14x−98
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314 (x+7)23y - 138 = - 14x - 9823y−138=−14x−98So, eqn of line BC is 23y +14x-40=0
Answer:
answer:
equation for AB is y+2=-1(x-1)
equation for AB is y+2=-1(x-1)y+x+1=0 (1)
equation for AB is y+2=-1(x-1)y+x+1=0 (1)equation of AC is y+2=2(x-1)
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)
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)
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)
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
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=-4y = -
\frac{4}{5}
5
4
and \: x =
\frac{8}{5}
5
8
y=−
y=− 5
y=− 54
y=− 54
y=− 54 andx=
y=− 54 andx= 5
y=− 54 andx= 58
y=− 54 andx= 58
y=− 54 andx= 58
y=− 54 andx= 58 putting values from (3) to (1)
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 5
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52 )
y=− 54 andx= 58 putting values from (3) to (1)x=−3&y=2x=-3&y=2B point by plotting we get(−7,6)11 52 )y - 6 = 28/5- 46/5.(x + 7)y−6=
.(x + 7)y−6= −
.(x + 7)y−6= − 5
.(x + 7)y−6= − 546
.(x + 7)y−6= − 546
.(x + 7)y−6= − 546
.(x + 7)y−6= − 546 5
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528
.(x + 7)y−6= − 546 528 .(x+7)
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=−
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 23
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314 (x+7)
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314 (x+7)23y - 138 = - 14x - 9823y−138=−14x−98
.(x + 7)y−6= − 546 528 .(x+7)y - 6 = - \frac{14}{23} (x + 7)y−6=− 2314 (x+7)23y - 138 = - 14x - 9823y−138=−14x−98So, eqn of line BC is 23y +14x-40=0
Step-by-step explanation: