TS EAMCET 04 May 2019 Shift 1 Solved Paper

Consider the equation,
x2+2√2xy+ky2
Then
tan‌θ‌‌=‌

2√2−k

k+1

tan‌45∘‌‌=‌

2√2−k

k+1

(k+1)2‌‌=4(2−k)
k2+2k+1‌‌=8−4k
Further simplify the above,
k2+6k−7‌‌=0
(k+7)(k−1)‌‌=0
k‌‌=1,k≠−7
Now the equation of the angle bisector of the line is given by,
x2+2√2xy+y2=0
‌

x2−y2

0

=‌

xy

√2

x2−y2=0
The angle of the triangle formed by the lines x2−y2=0 and
x+2y+1=0 is given by
∆‌‌=‌

|1√0+1(1)|

|1(2)2−0−12|

‌‌=‌

1

3