TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

The point of intersection of lines
L1=x−y−7=0 and L2≡x+y−5=0 is (6,−1)
Now, for point A(x1,y1) such that PA=3√2, take the equation of line L1 in symmetric form
L1:‌

x−6

cos‌ π 4

=‌

y+1

sin‌ π 4

=±3√2
⇒‌‌x1=6±3 and y1=−1±3
∵‌‌x1,y1≥0, so A(x1,y1) is (9,2)
Similarly, for point B(x2,y2) such that PB=√2 take the equation of line L2 in symmetric form
L2:‌

x−6

−cos‌ π 4

=‌

y+1

sin‌ π 4

=±√2
⇒‌‌x2=6+1 and y2=−1±1
∵x2,y2≥0, so B(x2,y2) is (5,0)
∴ Slope of line joining origin and point A is m1=‌

2

9

and slope of line joining origin and point B is m2=0
=cos−1(‌

9

√85

)