I need help solving this question using integration

Question

Consider the triangle whose vertices are (0, 0), (1, 0), and (0, 1)). Let Lm be the line with slope m that passes through the point (2, 0). Find the value of m ∈ R which divides the triangle into two halves, each of equal area.

Answer 1

Find the equation for the line with slope m that goes through (2,0)

Find the expression (in m) of the x where the two lines intersect. (2m+1)/(m+1) is what I got.

Integrate the top triangle (The area between the two lines from x = 0 to point of intersection)

Your result needs to be one half of the original area of the triangle. Solve for m.

Answer 2

The line that eats (0,1) and (1,0) is Line 1

The line that eats (2,0) is Line 2

The equation of Line 1 is y = -x + 1.

y = mx + b

b = y - mx is the intercept equation

Line 2 has intercept n = 0 - m(2) = -2m

The intercept of Line 2 is (0 , -2m)

the equation of Line 2 is y = mx - 2m

It is easier to integrate with resepct to y,

so that line is always on top. Therefore, we need

the inverses.

Line #1 is y = 1-x

Line #2 is y =(1/m) x + 2

Let t be the intersection, 0 < t < 1

Then

1 - t = (1/m) t + 2

1 - 2 = (1/m) t + t

-1 = [(1/m) + 1] t

-1 = [ (m+1)/m] t

t = -m/(m+1)

Solving for m:

t(m+1) = -m

mt + t = -m

mt+m = -t

m(t+1) = -t

m = -t/(t+1) <--- please label this equation DELTA

integral [1 - x] = x - (1/2)x^2

Area under the first curve: x=(0,t)

limit as x--> t is t - (1/2)t^2

limit as x--> 0 is 0

So the area of the first curve is t - (1/2)t^2

Area under the second curve: x=(t,1)

limit as x-->1 : 1 - (1/2) = 1/2

limit as x-->t: t - (1/2)t^2

subtracting them:

1/2 - t + (1/2)t^2

t - (1/2)t^2 = 1/2 - t + (1/2)t^2

moving everything from left to right:

0 = 1/2 - 2t + t^2

0 = 1 - 4t 2t^2

2t^2 - 4t + 1 = 0

t = [4 +or- sqrt( 16 - 4(2))]/4

= [ 4 +or- sqrt(16 - 8)]/ 4

= [ 4 +or- sqrt(8)] / 4

= [ 4 +or- 2*sqrt(2)] / 4

= [ 2 +or- sqrt(2)]/ 2

the negative branch is between zer0 and 1,

so t = (2 - sqrt(2))/2

Per equation DELTA: m = -t/(t+1)

t+1 = (2 - sqrt(2))/2 + 2/2 = (4 - sqrt(2))/2

so the slope is forced to be per equation DELTA:

(sqrt(2)-2)/2 divided by (4 - sqrt(2))/2

= (sqrt(2)-2)/2 * 2/(4 - sqrt(2))

=(sqrt(2)-2))/(4 - sqrt(2))

= (sqrt(2)-2))(4 + sqrt(2)) / (4-sqrt(2))(4+sqrt(2)) <--- rationalizes denomiantor

= (sqrt(2)-2))(4 + sqrt(2)) / (16 - 2)

= (sqrt(2)-2))(4+sqrt(2))/14

which is approximately -0.22654

This will get you started