TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

We have, ax+by+c=0 and 2a+3b=4c
Putting c=‌

2a+3b

4

in ax+by+c=0, we get
ax+by+‌

2a+3b

4

=0
⇒4ax+4by+2a+3b=0
⇒a(4x+2)+b(4y+3)=0
⇒(4x+2)+‌

b

a

(4y+3)=0
This is of the form L1+λL2=0
So, set of lines are
4x+2=0and4y+3=0
These two lines intersect at (‌

−1

2

,‌

−3

4

)
∴‌‌P(l,m)=(‌

−1

2

,‌

−3

4

)Equation of line passing through (‌

−1

2

,‌

−3

4

) and
perpendicular to x+y+1=0 is
y+‌

3

4

=1(x+‌

1

2

)
⇒4y=4x−1⇒4x−4y−1=0
On solving x+y+1=0 and 4x−4y−1=0,
we get Q(−‌

3

8

,−‌

5

8

)