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−‌