A vertical line divides the triangle whose vertices are at (0,0), (1,1), and (9,1) into two parts; a triangle and a quadrilateral. If the areas of the triangle and the quadrilateral are equal, what is the equation of the vertical line?
A vertical line divides the triangle whose vertices are at (0,0), (1,1), and (9,1) into two parts; a triangle and a quadrilateral. If the areas of the triangle and the quadrilateral are equal, what is the equation of the vertical line?
for vertical line $\displaystyle x=a$ make the areas of the shapes the same.
You can try to solve $\displaystyle \displaystyle \int_a^9 f(x)~dx = \int_0^a g(x)~dx $
All you have to do is determine $\displaystyle f(x), g(x)$ which are linear combinations of the lines joining the triangle.
Can you find the equations of the 3 lines that form the triangle?
Spoiler:
from the diagram,
$\displaystyle \frac{d}{1}=\frac{(9-a)}{9}$
$\displaystyle d=\frac{(9-a)}{9}$
considering the area of the small triangle,find the area of it
If A is the area of the larger triangle,
$\displaystyle A=(A+A') - A'$
$\displaystyle A1 = \frac{A}{2}$
substitute A, A1, (A+A') and A' with d and a
this is when vertical line meets AB
do the same thing when the vertical line meets CA
Ok, now that you've been shown how, try this similar one:
A vertical line divides the triangle whose vertices are at (0,0), (7,24), and (32,24) into two parts;
a triangle and a quadrilateral. If the areas of the triangle and the quadrilateral are equal,
what is the LENGTH of the vertical line?
You'll find that all side lengths (plus the areas) are integers;
smallest case, I believe, for an all-integer.
SOLUTION (since no response!):
A(0,0), B(7,24), C(32,24)
u=7,v=32,w=24 : so A(0,0), B(u,w), C(v,w)
Vertical line length = wSQRT[(v - u) / (2v)] = 15
Smallest all-integer case is smaller than I thought: A(0,0), B(4,15), C(36,15) ; vertical line = 10
Might be easier to just find the area of the full triangle
$\displaystyle \displaystyle A = \frac{1}{2}\times 8\times \sqrt{2}\times \sin {135}\approx 0.4998$
Take half of that and solve for the smaller triangle section.
$\displaystyle \displaystyle \int_a^9 1-\frac{x}{9}~dx = \frac{0.4998}{2}$