Find the equation of a line Find the equation of the line through (12/5 , 1), forming with the axes a triangle area of 5. 
There are 4 solutions and how can i get it?
 A: The line crossing $A(2.4,1)$ and forming a triangle with the axes and crosses x axis at $x=a$ and y axis at $y=b$ is of the form $$\frac{x}{a} + \frac{y}{b} = 1$$
The triangle may be formed on quarter 1,2 and 4.
In your case,$$|a|\times |b|=2\times5=10$$
$$\frac{2.4}{a} + \frac{1}{b} = 1$$
There are two cases;Solving the corresponding systems yields:
$$\bullet b=\frac{10}{a}\implies 0.1\times a^2 -a+2.4=0$$
$$a=\frac{1+0.2}{0.2},a=\frac{1-0.2}{0.2}\implies a=6, b=\frac53,a=4, b=2.5$$
$$\bullet b=-\frac{10}{a}\implies -0.1\times a^2 -a+2.4=0$$
$$a=\frac{1+1.4}{-0.2},a=\frac{1-1.4}{-0.2}\implies a=-12, b=\frac56,a=2, b=-5 \blacktriangle $$
Here are the plots of the 4 cases:


A: Further to the solution of PooyaM, above:
Note first that the solution depends on solving a quadratic equation.  In this case, there are two distinct solutions.  But if the "pivot point" had been $(2.5,1)$ there would have been only one solution, and if the pivot point had been $(2.5+, 1)$, there would have been no solution.
Two other solutions are found by assuming that $a$ and $b$ have different signs.  This would lead to a different quadratic:$$b=\frac{-10}{a}\implies -0.1\times a^2-a+2.4=0$$This quadratic will always have two solutions.  In this case, the solutions are $$a=-12, b=\frac{5}{6}$$and $$a=2, b=-5$$
