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# help

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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
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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   Jan 19, 2019
edited by CPhill  Jan 19, 2019
edited by CPhill  Jan 19, 2019
edited by CPhill  Jan 19, 2019
#2
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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
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i think you might have messed up ni the final steps.

Jan 19, 2019
#4
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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   Jan 19, 2019
edited by CPhill  Jan 19, 2019
#5
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wait so what was the answer?

Jan 19, 2019
#6
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Yours ( and mine)  are correct!!!!   CPhill  Jan 19, 2019
#7
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oh...

Jan 19, 2019