Step 1: Find the intercepts of the line.
Let, the slope of the line be −m
And, the y-intercept of the line be c.
∴ The equation of the line : y = −mx + c ....(1)
The line passes through (1, 4).
Putting in eq(1)
⇒ 4 = −m + c
⇒ c = m + 4 .....(2)
Putting the value of c in eq(1)
⇒ y = −mx + m + 4
⇒ y + mx = m + 4 ......(3)
Now, for point A, putting y = 0 in eq(3)
⇒ \(x = \frac{m + 4}m\)a
\(\therefore OA = \frac {m + 4}m\) ......(4)
Again, for point B, putting x = 0 in eq(3)
⇒ y = m + 4
∴ OB = m + 4 .....(5)
Step 2: Apply relevant condition to find the desired value.
Now,
⇒ \(OA + OB = \frac{m + 4}m + m + 4\)
⇒ \(OA + OB = 1+ \frac{ 4}m + m + 4\)
⇒ \(OA + OB = 5+ \frac{ 4}m + m \)
Let,
⇒ \(D = 5 + \frac 4m + m\) ......(6)
Differentiating both sides with respect to m,
⇒ \(\frac{dD}{dm} = - \frac 4{m^2 + 1}\)
Putting,
⇒ \(\frac{dD}{dm} = 0\)
⇒ \( - \frac 4{m^2 + 1}=0\)
⇒ \(m = \pm 2\)
Now, putting m = 2 in eq(6)
⇒ D = 5 + 2 + 2 = 9
And, putting m = −2 in eq(6)
⇒ D = 5 - 2 - 2 = 1
Hence, the minimum value of OA + OB is 1.