Equation of a line passing through (x1,y1) having slope m is given by y−y1=m(x−x1) Since the line PQ is passing through (1,2) therefore its equation is (y−2)=m(x−1) where m is the slope of the line PQ.
Now, point P(x,0) will also satisfy the equation of PQ ‌∴y−2=m(x−1) ‌⇒0−2=m(x−1) ‌⇒−2=m(x−1) ‌⇒x−1=‌
−2
m
‌⇒x=‌
−2
m
+1 Also, OP=√(x−0)2+(0−0)2=x=‌
−2
m
+1 Similarly, point Q(0,y) will satisfy equation of PQ ‌∴y−2=m(x−1) ‌⇒y−2=m(−1)⇒y=2−mb and OQ=y=2−m Area of △POQ=‌
1
2
(OP)(OQ)=‌
1
2
(1−‌
2
m
)(2−m) ‌‌ (As Area of ‌∆=‌
1
2
×‌ base ‌×‌ height) ‌ ‌=‌
1
2
[2−m−‌
4
m
+2]=‌
1
2
[4−(m+‌
4
m
)] ‌=2−‌
m
2
−‌
2
m
Let Area =f(m)=2−‌
m
2
−‌
2
m
Now, f′(m)=‌
−1
2
+‌
2
m2
Put f′(m)=0 ⇒m2=4⇒m=±2 Now, f′′(m)=‌
−4
m3
f′′(m)
‌
‌
|m=2=−‌
1
2
<0 f′′(m)
‌
‌
|m=−2=‌
1
2
>0 Area will be least at m=−2 Hence, slope of PQ is −2.