TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Equation of line L1 passing through (2,1) and (3,‌

5

2

)
y−1=‌

‌ 5 2 −1

3−2

(x−2)
⇒‌‌L1⇒3x−2y=4 . . . (i)
∵‌L2⟂L1
⇒‌‌L2⇒2x+3y
⇒‌‌L2⇒2x+3y+λ=0
It passes through (4,−1).
∴‌‌2(4)+3(−1)+λ=0
⇒‌‌λ=−5
L2⇒2x+3y−5=0 . . . (ii)
Solving Eqs. (i) and (ii), we get
x=‌

22

13

‌ and ‌y=‌

7

13

Point of intersection of L1 and L2 at x=0,
from Eq. (i), we get
y=−2
From Eq. (ii), we get
y=‌

5

3

Area of triangle formed by vertices (0,−2),(0,‌

5

3

)
∆‌‌=‌

1

2

|

‌ 22 13	‌ 7 13	1
0	−2	1
0	‌ 5 3	1

|
‌‌=‌

1

2

⋅‌

22

13

⋅‌

11

3

=‌

121

39

‌ sq. units ‌