AP EAPCET 23-Aug-2021 Shift 1 Paper

Given equation, 15x2+31xy+14y2=0
⇒15x2+10xy+21xy+14y2=0
⇒5x(3x+2y)+7y(3x+2y)=0
⇒(3x+2y)(5x+7y)=0
∴ Two lines have following equation
3x+2y=0 ...(i)
5x+7y=0 ...(ii)
Perpendicular distance from (2,3) to the line 3x+2y=0 is given as
|

3(2)+2(3)

√9+4

|=

12

√13

Perpendicular distance from (2,3) on line 5x+7y=0 is given as
|

5(2)+7(3)

√25+49

|=

31

√74

Given, p1>p2
⇒p1=

31

√74

and p2=

12

√13

⇒p12+

1

74

−p22+

1

13

=

(31)2

74

+

1

74

−

(12)2

13

+

1

13

=

962

74

−

143

13

=

1924

962

=2