TS EAMCET 4 May 2018 Shift 1 Solved Paper

The coordinates of the triangle is,
A(3,4),B(−3,6) and C(−5,1)
Let the coordinates of the point D be D(h,k)
ΔABC=

1

2

||

3	4	1
−3	6	1
−5	1	1

||
=

1

2

|[15−8+27]|
=

34

2

=17
And,
ΔADC=

1

2

|

3	4	1
h	k	1
−5	1	1

|
=

1

2

[3k−3−20−4h+h+5k]
=

1

2

[8k−3h−23]
=17
Now, take locus of points (h,k),
|8x−3y−23|=34
Therefore,
3x−8y=11
(3x−8y−11)=0
Or,
3x−8y=−84
(3x−8y+57)=0
Therefore,
(3x−8y−11)(3x−8y+57)=0