Set up an equation of the straight line that goes through the origin and which forms with the lines whose equations are
x-y+12=0 and 2x+y+9=0
a triangle whose area is 1.5 square units.

I graphed the given lines, which intersect at (-7,5). The line I want is y=mx because it goes through the origin and m, from graphing, seems to be negative.

I tried using the distance from a point (-7,5) to a line (y=mx) but am at a loss.

Thank you.

>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.

Ans: m=-27/13

Incorrect. Also, there are two answers. See.

>Ans: m=-27/13

Forgive me, but your reply was unhelpful. See another post from someone who also got the correct answer. Thank you all the same. I also figured it out with help, using the determinant method.

>>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

Thank you - I got this answer with the method you suggested.

>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

This is so easy you have x-y+12=0 and 2x+y+9=0
in my math class I did the same way, so you have (-7,5) and y=mx+b you have to plue it in that you already have.

thank you