Area between three lines

What's the area bound by the lines x + y = 5, x + 2y = 6 and the y-axis? I present one way to solve this problem. #shorts

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[MottBotMinecraft]

My calc-oriented brain jumped right to integrating and forgot that geometry exists lol

[titancodm]

"Shut up and integrate" ~leibniz and newton

[alberteinstein]

I immediately defaulted to integration lol

[barteqw]

I used integrals.
First x+2y=6 => y=-1/2x+3 and x+y=5 => y=-x+5.
Then we take integral from 0 to 4 substracting lower function from upper.
Int [-x+5-(-1/2x+3)] dx = Int [-1/2x+2] dx = -x^2/4+2x = F(x)
F(4)-F(0)=-16/4+2*4-0=-4+8=4.
Here is your answer.

[sixtysix]

Double integration, setting
0<x<4 and 3-0, 5x < y < 5-x

[Icenflamesrush]

A nice way to freshen up some basic calculus.
I did all the first steps but solved for x by setting the equations equal to each other. Then I used integrals to solve for the area.

[alwaysjustanindividual]

You can also take the integral from 0 to 4 of x + y = 5 (after setting both eqns to slope intercept form) minus the integral from 0 to 4 of x + 2y = 6 but i think this is more for like pre algebra or alg1 students 😅

[onalbetul]

When you solved harder versions of this question at high school...

[Skank_and_Gutterboy]

I did it by determining the intersection point of the two lines, which is the point (4, 1). Then you take the definite integrals of both lines for x=0 to 4. The area that were looking for is the integral from 0 to 4 of function f(x) = -x + 5 minus the integral from 0 to 4 of function g(x) = -x/2 +3. That yields an area of 4.

[Inspirator_AG]

x + y = 5 ➡️ y = 5 - x
x+ 2y = 6 ➡️ 2y = 6 - x ➡️ y = 3 - 0.5x

*Intersection Formula:*
Given two lines: Ax + B, Cx + D
x = (B - D) ÷ (C - A) = (D - B) ÷ (A - C)

In this case, A = -1, B = 5, C = -0.5, and D = 3.

(5 - 3) ÷ (-0.5 + 1) = 2 ÷ 0.5 = 4

It is a triangle, so you can take the difference of the Y-intercepts, 5 - 3, (since it is just the constant in a linear equation), and that is 2. You can then do 2 × 4 ÷ 2, (4 being the intersection X-coordinate) and that is 4.

[TheOriginalFayari]

Wait! You didn't do the intro! How are we supposed to remember your name, Fresh Tall Walker?

[rafadono]

The first thing I thought was integration, I'm mad :c

[chipsnguac]

Subtract Xs for both equations. Divide by two for the lower one. 5-x=3-.5x solve for x. Use x in a=1/2bh formula as h. B is difference of y intercepts.

[parthibhayat]

We have sort of a similar problem in our textbooks where there's like a direct formula if you manage to find coordinates of three points of triangle in a graph

[Snp-]

You could find the coordinates of the points and then calculate the area determinant

[TeamINWHex]

Man subscribing your channel after seeing 2 short videos

[I_am_brainlessrahulz]

First solving simulationeous equation
X+ y=5
X+2y=6
X=4 y=1
4×1=4

[jetx_]

Guess who also thought of integrating...

[el.koukouladakis]

I'm a retired civil engineer from Greece and my name is Lefteris. Ever since I was a student I have really liked maths, hence why I have been watching you for quite some time. I'd like to suggest what I consider to be an interesting geometry problem which is as follows: Let there be two parallel lines (e1) and (e2) and a point A beyond them. Both lines and the point are on the same plane. How can we construct two lines starting from A which intersect line (e1) on points B and C and line (e2) on points D and E, so that BC = α cm and DE = β cm, where β > α and α, β are given. I had to solve this problem during my first year at NTUA, back in 1975!

[batchrocketproject]

Thanks Presh, I missed the simplification at the end to directly calcualte the area, instead subtracting the lower right triangle from the larger one made by addition of the blue area. i.e. (4*4)/2 - (4*2)/2 = 4. I should have looked at it side-on where the 'sideways' height is obvious!