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dovid31
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May1207, 07:44 AM (EST) 

"Geometry"

Set up an equation of the straight line that goes through the origin and which forms with the lines whose equations are xy+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. 

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dovid31
guest

May2107, 01:43 PM (EST) 

7. "RE: Geometry"
In response to message #1

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.


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mpdlc
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May1907, 12:02 PM (EST) 

3. "RE: Geometry"
In response to message #0

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/(m1) and Y2 = 12m/(m1) 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 

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dovid31
guest

May2107, 01:43 PM (EST) 

5. "RE: Geometry"
In response to message #3

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/(m1) and Y2 = 12m/(m1) 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


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algerya moye
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May2107, 01:25 PM (EST) 

4. "RE: Geometry"
In response to message #0

This is so easy you have xy+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 

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