AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Given equation of pair is
3x2−5xy+2y2=0
(3x−2y)(x−y)=0
The equation of the two sides of the triangle is
3x−2y=0
x−y=0
The line 2x+3y+k=0 passes through the point (2,1)
4+3+k=0
k=−7
Therefore,
2x+3y−7=0
The point of intersection of line y−x=0 and 2x+3y−7=0 is
x=

7

5

,y=

7

5

The line x+y+k=0 passes through the point (2,1).
2+1+k−0
k=−3
The point of intersection of line x+y−3 and 3x−2y=0 isx=

6

5

,y=

9

5

The equation of the third side of the triangle is
y−

7

5

=

9 5 − 7 5

6 5 − 7 5

(x−

7

5

)
y−

7

5

=

2

−1

(x−

7

5

)

5y−7

5

=

−10x+14

5

10x+5y=21