TS EAMCET 7 May 2018 Shift 1 Solved Paper

Equation of line passing through the points (3,0) and (1,

4

3

)
y−0=

4 3 −0

1−3

(x−3)
y=−

2

3

(x−3)
3y=−2x+6
2x+3y=6
And the line perpendicular to the above line which passes through (8,1)
y−1=

−1

( −2 3 )

(x−8)
y−1=

3

2

(x−8)
2y−2=3x−24
3x−2y=22
The two equation cuts at (6,-2)
Therefore, the area of the required triangle is
=

1

2

|

6	−2	1
0	2	1
0	−11	1

|
=

1

2

.6(2+11)
=39 sq unit