>Ans: m=-27/13

>>I tried using the distance from a point (-7,5) to a line
>>(y=mx) but am at a loss.
>>
>
>Distance along would not help. This will serve as an
>altitude in the triangle in question. To find the area you
>also need the base, so you have to find the intersections of
>y = mx with the two given lines. I do not see how the
>geometry here may simplify the problem. It appears to
>laborious to me.

>As you probably remember the area S of a triangle of which
>the three vertices coordinates are given equal to one half
>of the determinant formed with them as follows:
>
>X1 Y1 1
>X2 Y2 1 = 2*S
>X3 Y3 1
>
>
>Then you already got the coordinate of one the vertex (-7,
>5) . To obtain the other two, remember since the straight
>line must pass through the origin both vertices must comply
>with the equation y = mx besides each given equation, where
>m is the unknown
>
>So by substitution of y = mx in the two original equations
>you get :
>
>X2 = 12/(m-1) and Y2 = 12m/(m-1) for eq. y = x +12
>
>And
>
>X3 = -9/(m+2) and Y2 = -9m/(m+2) for eq. y = -2x -9
>
>
>Plugging these coordinates in the determinant above and
>solving it
>
>After an easy simplification you get a second degree
>equation in m as follows
>
>50 m2 + 71m + 23 = 0 (m2 means the square of m)
>
>Solving it you get the two solution
>
> m = -1/2 and m= -23/25
>
>I let you to have check the answers by graphic and algebra,
>I already have done it