TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Let rectangle be ABCD. Here line AB||CD
→ slope of line AB= slope of line CD
⇒‌‌−3=‌

−b

2

⇒b=6
Now line AB⟂rBC
⇒ (slope of line AB)×( slope of line BC)=−1
⇒‌‌(−3)(‌

1

a

)=−1⇒a=3
∴ Eq. of BC:x−3y−10=0
Eq. of CD:6x+2y+9=0
∴ Distance between AB and CD
∴ i.e., ‌‌BC=|‌

−4−9∕2

√9+1

|=‌

17

2√10

Now line AD pass through the point (1,2)
⇒ Distance CD= distance of point (1,2) from line BC
=|‌

1−6−10

√1+9

|=‌

15

√10

Now, area of rectangle ABCD=BC×CD
=‌

17

2√10

×‌

15

√0

=‌

17×15

2×10

=‌

51

4