Prove that the lines (a+b)x + (a-b)y+d=o and
(a²-b²) x + (a-b)²y + k=0 are parallel to each other.
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Answer:
Given lines
(a+b)x+(a−b)y−2ab=0−−−−(1),(a−b)x+(a+b)y−2ab=0−−−−(2) and x+y=0−−−−−(3)
On comparing above eq (1) (2) and (3) with y=mx+c, we get
m
1
=−
a−b
a+b
m
2
=−
a+b
a−b
and m
3
=−1
Angle between (1) and (3) by formula
tanα=
∣
∣
∣
∣
∣
1+m
1
m
3
m
1
−m
3
∣
∣
∣
∣
∣
tanα=
∣
∣
∣
∣
∣
∣
∣
∣
1+
a−b
a+b
−
a−b
a+b
+1
∣
∣
∣
∣
∣
∣
∣
∣
tanα=
∣
∣
∣
∣
∣
a−b+a+b
−a−b+a−b
∣
∣
∣
∣
∣
tanα=
∣
∣
∣
∣
∣
2a
−2b
∣
∣
∣
∣
∣
tanα=
a
b
Angle between (2) and (3) by formula
tanβ=
∣
∣
∣
∣
∣
1+m
2
m
3
m
2
−m
3
∣
∣
∣
∣
∣
tanβ=
∣
∣
∣
∣
∣
∣
∣
∣
1+
a+b
a−b
−
a+b
a−b
+1
∣
∣
∣
∣
∣
∣
∣
∣
tanβ=
∣
∣
∣
∣
∣
a+b+a−b
−a+b+a+b
∣
∣
∣
∣
∣
tanβ=
∣
∣
∣
∣
∣
2a
2b
∣
∣
∣
∣
∣
tanβ=
a
b
Here tanβ=tanα
Hence traingle is isosceles triangle and the vertical angle is π−2tan
−1
a
b
=2tan
−1
b
a
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