Find the equations of the line which passes through the point `(3,4)` and the sum of its intercepts on the axes is`14` .

Answer

N/a

Explanation

Let the intercept along the axes be a and b.
Given, `a+b=14rArrb=14-a`
Now the equation of line is `(x)/(a)+(y)/(b)=1`
`rArr (x)/(a)+(y)/(14-a)=1`
Since, the point (3,4)lies on the line .
`therefore (3)/(a)+(4)/(14-a)=1`
`rArr (42-3a+4a)/(a(14-a))=1rArr42+a=14a-a^(2)`
`rArr a^2-13a+42=0rArr a^2-7a-6a+42=0`
`rArr a(a-7)-6(a-7)=-0rArr(a-7)(a-7)(a-6)=0`
`rArr a-7=0`or `a -6=0`
`therefore 1=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)=1rArrx+y=7`
So, the equation of line, when a=6and b=8 is `(x)/(6)+(x)/(8)=1`