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 \(k\).

Guest Jan 19, 2019

#1**+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

CPhill Jan 19, 2019

#2**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??!

asdf335 Jan 19, 2019