∵‌ Area of ‌△PAB‌‌=2×‌ Area of ‌△ABC ‌‌=2×1=2 [AB‌‌=√5‌ by using pythagoras in ‌△ABC] Let P(h,k) be the point. ‌ Area of ‌△PAB=‌
1
2
AB×PQ ‌2=(‌
1
2
)√5×PQ ‌PQ=‌
4
√5
‌ Equation of ‌AB=(y−0)(0−1)=(2−0)(x−1) ‌‌−y=2x−2 ⇒‌‌2x+y−2=0 Perpendicular distance from P is ‌