# find the equation of the line that passes through (3,1) forming w/ the axes a triangle of area 6.

*print*Print*list*Cite

Let's start with the triangle. The formula for the area of a triangle is:

`A = 1/2 bh` where b would be the base, and h the height. The triangle of interest here has legs that coincide with the axes, and hypotenus passing through the given point. Since it's a right triangle, b and h would simply be the length of the legs which would by **a** and b in this case, where a corresponds to the length of a side with endpoints (0, 0) and (a, 0), and b to the side with endpoindt (0, 0) and (0, b).

Hence, `1/2 ab = 6 \Rightarrow ab = 12` [E1]` `

Next, we know that the line passes through (3,1). This point is contained by the hypotenuse. Also contained by the hypotenuse are points (a, 0) and (0, b) -- in fact, they are the endpoints of the hypotenuse. Hence, these three points lie on the same line. Since they lie on the same line, the slope must be the same. Slope is given by:

`m = (y_2 - y_1)/(x_2 - x_1)`

Using any pair of points -- say, (3, 1) and (a, 0), and (3, 1) and (0, b) -- we can compute for the slope in different ways:

`m = (1 - 0)/(3-a) = 1/(3-a)`

`m = (1 - b)/(3 - 0) = (1-b)/3`

Since all of these points are on the same line, the two values must be equal:

`1/(3-a) = (1-b)/3` [E2]

Now, we have two variables a and b, and two equations. We can solve this system.

[E1] `ab = 12`

[E2] `1/(3-a) = (1-b)/3`

[E2] can be re-written as `3 = (1-b)(3-a) \Rightarrow 3 = (3 - a - 3b + ab)` .

Using [E1], we can express either a or b in terms of the other. Let's use `a = 12/b` .

Substituting this to the transformed, and also using the fact that `ab = 12` [E2]:

`3 = (3 - 12/b - 3b + 12)`

`3 = 15 - 12/b - 3b`

`-12 = -12/b - 3b`

`-12b = -12 - 3b^2`

`3b^2 - 12b + 12 = 0`

`b^2 -4b + 4 = 0`

`(b - 2)^2 = 0`

`b - 2 = 0`

`b = 2`

Since `ab = 12` , we know that `a = 6` .` `

Using either value, we can calculate the slope of the line:

`m = 1/(3-a) = 1/(3-6) = 1/-3 = -1/3`

Hence, the equation of the line is (using the point slope form):

`y - 1 = -1/3 (x - 3)`

`-3y + 3 = x - 3`

`x + 3y = 6`

Equation is x + 3y = 6 or y = (6 - x)/3