Question

Find the equation to the straight line which passes throuth the points (3,4) and have intercepts on the axes:such that their sum is 14

Answer 1

Step-by-step explanation:

Let the equation of the line be $\frac{x}{a}+\frac{y}{b}=1$

This passes through $(3,4)$

Therefore,$\frac{3}{a}+\frac{4}{b}=1$

It is given that $a+b=14$

                           $b=14-a$

Putting $b=14-a$ in (2), we get

$⇒\frac{3}{a}+\frac{4}{14-a}=1$

$⇒42-3a+4a=14a-a²⇒a²-13a+42=0$

$⇒(a-7)(a-6)=0$

$∴a=6,7$

$then, b=8,7$

Hence the required equation are,

$\frac{x}{6}+\frac{y}{8}=1$ and $\frac{x}{7}+\frac{y}{7}=1$

$4x+3y=24$ and $x+y=7$

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