In the given figure, the equal sides ABand AC
of an osceles triangle ABC are produced
to Nandu, respectively such that ANCA
Prove that MB =Nc
Answers
Answer:
the base BC be taken along axis of x and its mid-point be chosen as origin so that the points B and C are (−a,0) and (a,0) respectively. The third vertex A will lie on y−axis at (0,b) say. By given condition
AB
BP
=
CQ
AB
=λsayor
AB
BP
=
CQ
AC
=λ∴AB=AC
B divides PA in the ratio λ:1 and C divides AQ in the ratio λ:1 as shown in the figure. Hence by ratio formula the co-ordinate of P are
[−a(λ+1),−λb] and Q is [a
λ
λ+1
,
λ
b
]
Slop of PQ by =
x
2
−x
1
y
2
−y
1
is
a
b
λ+1
λ−1
Hence equation of PQ is
y+λb=
a
b
λ+1
λ−1
[x+a(λ+1)]
Or a(λ+1)y+abλ(λ+1)=b(λ−1)x+ab(λ
2
−1)
Cancel abλ
2
and collect the terms of λ
(ay+bx+ab)+λ(bx−ay−ab)=0
Above is of the form u+λv=0 which represents a family of straight lines passing through the intersection of u=0 i.e. the point the points (0,−b) which is a fixed point.