can anybody answer 2 one please....
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first check
a1a2+b1b2 sign
i.e (1)(7)+(-1)(1)=6 positive
then {a1x+b1y+c1}/(root (a1^2+b1^2))={a2x+b2y+c2}/(root (a2^2+b2^2)) is obtuse bisector
now apply
{x-y+2}/root2={7x+y+1}/root50
5(x-y+2)=(7x+y+1)
2x+6y-9=0 is obtuse angle bisector
in the Same way we find out acute angle bisector
i.e 12x-4y+11=0
a1a2+b1b2 sign
i.e (1)(7)+(-1)(1)=6 positive
then {a1x+b1y+c1}/(root (a1^2+b1^2))={a2x+b2y+c2}/(root (a2^2+b2^2)) is obtuse bisector
now apply
{x-y+2}/root2={7x+y+1}/root50
5(x-y+2)=(7x+y+1)
2x+6y-9=0 is obtuse angle bisector
in the Same way we find out acute angle bisector
i.e 12x-4y+11=0
abhi178:
(x+2)-2 (y-2)=x-2y+6=0
now you proceed this two equation by angle bisector concept
oj
ok
one more
or slope concept
I solve this by slope concept
both equation you find slope
in the above 1 problem can you see and answer 7 one
is this answer correct
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