Question

Find the equation of the straight line which passes through the point (-3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7

Answer 1

Concept Used:

The equation of the line with intercepts a and b is  \(\frac{x}{a}+\frac{y}{b}=1\) 

Given: 

Here a + b = 7, b = 7 – a 

Explanation: 

The line is passing through (-3, 8).

 \(\frac{-3}{a}+\frac{8}{b}=1\) 

Substituting b =7 – a, we get

 \(\frac{x}{a}+\frac{y}{7-b}=1\)  

⇒ -3(7 - a) + = 7a - a²

⇒ a² + 4a – 21 = 0 

⇒ (a – 3)(a + 7)= 0

⇒ a = 3 ( since, a can only be positive) 

Substituting a = 3 in equation (i) we get, b = 7 – 3 = 4 

Hence, the equation of the line is  \(\frac{x}{3}+\frac{y}{4}=1\)