TS EAMCET 19-July-2022 Shift 1 Solved Paper

Given, x+(5λ+1)y+1−3λ=0
⇒(x+y+1)+λ(5y−3)=0
The given family of lines contains two lines i.e. x+v+1=0 and 5v−3=0
5y−3=0⇒y=‌

3

5

x+‌

3

5

+1=0⇒x=−‌

8

5

The point of concurrent of family of lines
x+(5λ+1)y+1−3λ‌ is ‌(‌

−8

5

,‌

3

5

)
‌(5µ+2x−3y+3+6µ=0
‌2x−3y+3+µ(5x+6)=0
The family of lines contain 2x−3y+3=0 and 5x+6=0
x=‌

−6

5

and y=‌

1

5

The point of concurrent of family of lines
(5µ+2)x−3y+3+6µ=0 is (‌

−6

5

,‌

1

5

)
∴ Required distance
‌=√(‌

−8

5

+‌

6

5

)2+(‌

3

5

−‌

1

5

)2
‌=√‌

4

25

+‌

4

25

=√‌

8

25

=‌

2√2

5