X
13. Transform the equation x+y=
+ 1 into the
a b
normal form when a>0 and b>0. If the
perpendicular distance of the straight line
from the origin is p, deduce that
1 1 1
(***)
p
a?
b2
11
+
2
2
III.
Answers
Answer:
Step-by-step explanation:
a
x
+
b
y
=1.......(i),a>0, b>0
a
2
1
+
b
2
1
=
a
2
b
2
a
2
+b
2
=
ab
a
2
+b
2
Divide (i) throughout by
ab
a
2
+b
2
a
ab
a
2
+b
2
x
+
b
ab
a
2
+b
2
y
=
ab
a
2
+b
2
1
x
a
2
+b
2
b
+y
a
2
+b
2
a
=
a
2
+b
2
ab
...........(ii)
Slope of (ii) is given by
tanα=−
Coefficient of y
Coefficient of x
=−
b
1
a
1
=−
a
b
which is negative, hence α is obtuse
tanα=−
a
b
=
Base
Perpendicular
Hypotenuse=
Base
2
+Perpendicular
2
=
a
2
+b
2
For obtuse angles sin is positive and cos is negative
sinα=
a
2
+b
2
b
cosα=−
a
2
+b
2
b
Hence (ii) will become
x(−cosα)+y(sinα)=
a
2
+b
2
ab
We know that
sin(A)=sin(π−A)
and cos(A)=−cos(π−A)
⇒xcos(π−α)+ysin(π−α)=
a
2
+b
2
ab
Which is the required normal form where tanα is slope of given line
Hence right hand side must give perpendicular distance of the straight line from the orogin.
⇒P=
a
2
+b
2
ab
⇒P
2
=
a
2
+b
2
a
2
b
2
⇒
P
2
1
=
a
2
b
2
a
2
+b
2
⇒
P
2
1
=
a
2
b
2
a
2
+
a
2
b
2
b
2
⇒
P
2
1
=
a
2
1
+
b
2
1
Step-by-step explanation:
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