TS EAMCET 4 May 2018 Shift 1 Solved Paper

The equation of the line is,
x+2y+c=0
According to given information.

2|C−3|

√5

=

|8−C|

√5

2(C−3)=8−C
3C=14
C=

14

3

Thus, the equation becomes,
x+2y+

14

3

=0
The normal form of the equation is given as,
x‌cos‌α+y‌sin‌α=p
Therefore,
cos‌α‌−1=

sin‌α

−2

=

p

14 3

So,

cos‌α

−1

=

sin‌α

−2

tan‌α=2
α=tan−12
And,
p=

14

3√5

=

14

√45

Hence, the required equation is,
x‌cos‌α+y‌sin‌α=

14

√45

And,
α=π+tan−12