AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The give equation of line is,
23x2−48xy+3y2=0
3y2−48xy+23x2=0
Here,
m1+m2=16
mim2=

23

3

∴m1−m2=√(m1+m2)2−4m1m2
On putting the value in above expression.
m1−m2=√(16)2−4×

23

3

m1−m2=

26

√3

On solving.
m1=8+

13

√3

and m2=8−

13

√3

The angle of slope is.
tan‌θ=

m1−m2

1+m1m2

=

( 26 √3 )

1+ 23 3

=√3
θ=60°
The slope of line 2x+3y+4=0 is −

2

3

The angle between line of slope‌8+

13

√3

and −

2

3

is,
tan‌α=

8+ 13 √3 + 2 3

1+(8+ 13 √3 )( 2 3 )

=√3
α=60°
Thus, the lines form an equilateral triangle.