Question

A straight line is drawn through a given point P (1, 4). Determine the least value of the sum of the intercepts on the coordinate axes.

Answer 1

Let us the slope of the line passing through the point P(1,4) be m. 

We know that equation of a straight line passing through the point (x₁,y₁) and having slope m is given by : 

⇒ y - y₁ = m(x - x₁) 

The equation of the straight line is : 

⇒ y - 4 = m(x - 1) 

⇒ mx - y = m - 4

⇒ \(\frac{mx}{m-4}\) - \(\frac{y}{m-4}\) = 1

⇒ \(\frac{x}{\frac{m-4}{m}}\) - \(\frac{y}{4-m}\) = 1

This resembles the standard form \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1,where a is x - intercept and b is y - intercept.

Here,

x - intercept a = \(\frac{m-4}{m}\) and b = 4 - m

According to the problem, we need sum of intercepts to be minimum, 

Let us take the sum of intercepts to be S,

⇒ S = a + b

⇒ S = \(\frac{m-4}{m}\) + (4 - m)

⇒ S = 5 - \(\frac{4}{m}\) - m

Let us assume S is the function of m, 

We know that for maxima and minima,

⇒ 4 - m² = 0 (∵ m²>0) 

⇒ m = ±2 

Differentiating S again,

We have got minima for m = - 2 

Using this value we find the sum of intercepts :

⇒ S_min = 3 + 6 

⇒ S_min = 9 

∴ The least value of sum of intercepts is 9.