Find Possible Slopes of Line

A line with a negative slope passes through the point (3,1). The area of the triangle bounded by this line and the axes is 6 square units. Find possible slopes of this line.

y = mx+b
1 = 3m+b
b = 1-3m

y = mx + (1-3m)

x intercept: x = (3m-1)/m
y intercept: y = (1-3m)

Area = xy/2 = -(3m-1)²/2m = 6
(3m-1)² = -12m
9m²-6m+1 = -12m
9m² + 6m + 1 = 0
(3m+1)² = 0
m = -1/3

The slope of the line is -1/3

EDIT: Let's try it with slopes of -1/5 and -5/3, and (3,1) being a point on the line:
Slope = -1/15
y = -(1/15)x + b
1 = -3/15 + b
y intercept: b = 18/15
y = -(1/15)x + 18/15
x intercept:
0 = -x + 18
x = 18

Intercepts are (18,0) and (0, 18/15)
Area = 18*18/30 = 10.8 > 6 : Nope, -1/15 cannot be a correct slope

Slope = -5/3
y = -(5/3)x + b
1 = -5+b
y intercept: b = 6
y = -(5/3)x + 6
x intercept:
0 = -(5/3)x + 6
x = 18/5
A = 6*18/10 = 10.8 > 6 : Nope, -5/3 cannot be a correct slope

There is only one line through (3,1) the meets the stated criteria, and that one line only has one slope.

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Let the equation is x/a + y/b =1............(i) where a and b are intercepts on x axis and y-axis
Since the line passes through the point (3,1)
Therefore 3/a+1/b =1
or 3b + a = ab ...........(ii)
Since the are is 6 square units then 1/2(ab) = 6 or ab = 12 ........(iii)
From (ii) and (iii)
a+3b =12.........(iv)
Slope is b/a < 0, therefore two posibilities are there a is +ve and b is -ve and a is -ve and b is +ve
Let a =15 then b = -1 then slope = b/a = -1/15 ......Ans
Let a = -3 then b =5 then slope = b/a =5/-3 = -(5/3) ....Ans