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  • Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

Answers (1)

Equation of line in intercept form = \frac{x}{a}+\frac{y}{b}=1

Given that   a+b=14    b=14-a

 So equation of line \frac{x}{a}+\frac{y}{14-a}=1

 \frac{x\left ( 14-a \right )+ay}{a\left ( 14-a \right )} =1

14x-ax+ay=14a-a^{2}........................(1)

  If equation 1 passes through (3,4) then 14 * 3 - a*3+a*4=14a-a^{2}

   42-3a+4a-14a+a2=0  

  a2-13a+42=0 

  a2-7a-6a+42=0   

 (a-6)(a-7)=0  

  a=6 or a=7  

 If a=6 then 6+b=14  b=8   

If a=7 the 7+b=14 b=7  

 If a=6 and b=8 then equation of line is \frac{x}{6}+\frac{y}{8}=1

 4x+3y=24 

  If a=7 and b=7 then equation of line \frac{x}{7}+\frac{x}{7}=1

x+y=7

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