TS EAMCET 04 May 2019 Shift 2 Solved Paper

The equation of the perpendicular bisector of the sides AB and
‌ AC of ‌∆‌ ABC are ‌
‌‌‌‌x−y+5=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I)
And,
x+2y=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II)
Consider the figure shown below.B is the image of A with respect to the line x−y+5=0
Therefore,
‌

x2−1

1

=‌

y2+2

−1

=−2(‌

1+2+5

2

)
‌

x2−1

1

=‌

y2+2

−1

=−8
Hence,
x2=−7
And,
y2=6
So the coordinates of B is (-7,6)
Now, C is the image of A with respect to the line x+2y=0
Therefore,
‌

x3−1

1

=‌

y3+2

2

=−2(‌

1−4

5

)
‌

x3−1

1

=‌

y3+2

2

=‌

6

5

Hence,
x3=‌

11

5

And,
y3=‌

2

5

Solve equation (I) and (II)
x=−‌

10

3

y=‌

5

3

Therefore, the coordinate of point D is
=(‌

−7+ 11 5

2

,‌

6+ 2 5

2

)
=(‌

−12

5

,‌

16

5

)
The equation of perpendicular bisector of side BC.
y−‌

5

3

‌‌=‌

48−25

−36+50

(x+‌

10

3

)
23x−14y+100‌‌=0