Application of Derivatives Part 1

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.