A≡(−5,−2),B≡(4,−2)
Let C → mid-point of AB
C≡(−5+42,−2−22)≡(−12,−2)
∴C≡(−12,−2)
Perpendicular drawn from C to x-axis touches x-axis at Q.
Co-ordinates of Q≡(−12,0)
For ΔABQ,AB=√(4+5)2+(−2+2)2=9
AQ=√(−5+12)2+(−2−0)2=4.924
√(4+12)2+(−2+0)2=4.924
AQ = BQ
∴ΔABQ is isosceles triangle.