AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The slope of line BC is given as,
BC=(

0+1

4−2

)=

1

2

The equation of AD is given as.
y−3=

−1

BC

(x+2)
y−3=

−1

1 2

(x+2)
y−3=−2(x+2)
2x+y+1=0
The slope of line AC is given as,
AC=

0−3

4+2

=

−3

6

=

−1

2

The equation of BE is given as,
y+1=

−1

AC

(x−2)
y+1=

−1

(− 1 2 )

(x−2)
y+1=2(x−2)
y−2x+5=0
On solving the above equations we get orthocenter at (1,-3)
x,y=(

−2+2+4

3

,

3−1+0

3

)
x,y=(

4

3

,

2

3

)
The equation of line joining (0,-1) and (

4

3

,

2

3

) is
y+3=

2 3 +3

4 3 −1

(x−1)
y+3=11(x−1)
y+3=11x−11
11x−y−14=0