Question

find the equation of the lines which passes through the point (3, 4) and cuts off intercept from the coordinate axes such that their sun is 14

Answer 1

Hi

Here is your answer,

Let the intercept along the axes be a and b 

Given,                     a  + b = 14  →  b = 14 - a

Now, the equation of line is x/a + y/b = 1

                                        → x/a + y/14-a = 1

Since, the point (3,4) lies on the line.

                                 3/a + 4/14-a = 1 

   So, now 42 - 3a + 4a / a (14-a) = 1→ 42 + a = 14a - a²

     → a² - 13a + 42 = 0 → a²-7a - 6a + 42 = 0 

     → a (a - 7) - 6 (a-7) =0 

     → (a  - 7) ( a - 6) = 0 

      → a - 7 = 0 or a - 6 = 0 

         → a =7  or a =6 
       
  When,        → a = 7 then  b = 7

 When,       → a = 6 , then b = 8


 Therefore, the equation of line when, a = 7 and b = 7 is 

                                    x/7 + y/7 = 1 → x + y = 7


So, the equation of line , when a = 6 and b = 8 is x/6 + y/8 = 1 



Hope it helps you !

Answer 2

Equation of a Line L with x and y intercepts = a & b respectively,

         x/a + y/b = 1     or,    a y + b x = ab       --- (1)

Given    a + b = 14    ---(2)
So equation (1) becomes:
    a y + (14 -a) x = a (14 - a) = 14 a - a²   ---(3)
                 
Since L passes through Point  P(3,4),
   a * 4 + (14 - a) 3 = 14 a - a²

=> a² - 13 a + 42 = 0
     (a - 6) (a - 7) = 0
=>  a = 6 or 7.
=>  b = 8 or 7.

Equation of lines are:   6 y + 8 x = 48  or, 3 y + 4 x = 24
           and                    7 x + 7 y = 49  or,  x + y = 7

There are two lines possible.