TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the figure shown below.Let the equation of lines passing through the origin and forming the triangle be
3x+4y−5=0
y=m1x
y=m2x
since, the slope of line is

−3

4

Therefore,
tan‌60°=|

m1+ 3 4

1− 3 4 m1

|
±√3=

4m1+3

4−3m1

4√3−3√3m1=4m1+3
m1=

4√3−3

4+3√3

And,
−√3=

4m1+3

4−3m1

−4√3+3√3m1=4m1+3
m1=

3+4√3

3√3−4

Also,
m2=

3+4√3

3√3−4

Or
m2=

4√3−3

4+3√3

The required equation of pair of lines is calculated as,
(y−m1x)(y−m2x)=0
(y−(

4√3−3

4+3√3

)x) (y−(

3+4√3

3√3−4

)x)=0
((4+3√3)y−(4√3−3)x) ((3√3−4)y−(3+4√3)x)=0
[(3√3+4)(3√3−4)y2 −xy((4+3√3)(3+4√3)) +(4√3−2)((3√3−4)+(4√3−3) (4√3+3)x2)]=0
Solve further,
(27−16)y2−xy(96) +(48−9)x2=0
11y2−96xy+39x2=0