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A triangle is formed with edges along the line \(y=\frac{2}{3}x+5\), the \(x\)-axis, and the line \(x=k\). If the area of the triangle is less than 20, find the sum of all possible integral values of \(k\).

 Jan 19, 2019
 #1
avatar+106539 
+2

The line intersects the x axis at x = -7.5

 

The base length of the triangle will be   k - (-7.5) =  k + 7.5

 

The height of the triangle will be    (2/3)k +  5

 

So...we are trying to solve this

 

(1/2) base * height = 20

 

(1/2) ( k + 7.5) [ (2/3)k + 5]  = 20

 

(k + 7.5) (2/3k + 5)  =  40

 

(2/3)k^2 + 5k + 5k + 37.5 = 40

 

(2/3)k^2 + 10k - 2.5 = 0             multiply through by 3

 

2k^2 + 30k - 7.5   =  0           multiply through by 10

 

20k^2 + 300k - 75  = 0

 

4k^2 - 60k - 15 = 0

 

The soutions to this are   k ≈ -15.246      and k ≈ .245   

 

So....the integer values for k that make this triangle have an area < 20    are  the integers  -15 to 0 inclusive

 

And the sum of these =  - ( 15) (16) / 2  =  -120 

 

 

 

 

 

cool cool cool

 Jan 19, 2019
edited by CPhill  Jan 19, 2019
edited by CPhill  Jan 19, 2019
edited by CPhill  Jan 19, 2019
 #2
avatar+533 
0

first find the ax intercept of the line. this is just 5*3/2 or 7.5, and since the slope is 2/3, then the height is 2/3 of the width of the triangle.

 

the width of the triangle is |k+7.5|, so the height is 2/3|k+7.5|. multiplying these, we can take away the absolute value because a square is always non negative, so you get ((k+7.5)^2)/3 because you have to divide by 2 because of a triangle. this is less than 20, so (k+7.5)^2 is less than 60, so k can be 0,-1,-2,-3,-4,-5,-6,-7,-8,-9,-10,-11,-12,-13,-14, and -15. adding these up you get -120.

 

HOPE THIS HELPED!

 

 

also cphill, i think you might have forgotten that a triangle is width*height/2.

 

edit: nevermind, but you are wrong, i am right. i dont know where you made a mistake though, i graphed my answers on desmos, they work.

 

confused??!

 Jan 19, 2019
edited by asdf335  Jan 19, 2019
 #3
avatar+533 
0

i think you might have messed up ni the final steps.

 Jan 19, 2019
 #4
avatar+106539 
+1

Actually, asdf.....we both are correct....we just need to "fuse" our final answers!!!

 

I had forgotten that the triangle could also lie "below" the x axis

 

 

cool cool cool

 Jan 19, 2019
edited by CPhill  Jan 19, 2019
 #5
avatar+533 
0

wait so what was the answer?

 Jan 19, 2019
 #6
avatar+106539 
0

Yours ( and mine)  are correct!!!!

 

 

cool cool cool

CPhill  Jan 19, 2019
 #7
avatar+533 
0

oh...

 Jan 19, 2019

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