Question

A straight line with negative slope passing through the point \( (1,4) \) meets the coordinate axes at A and B. The minimum value of \( OA + OB \) equals to

Answer 1

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.