None of the above.

Edit: wrong answer, I apologize. I’ll sit on this for a while.

How it is 1?

area of triangle (0.5)(2)(1)

take series of points and plot a small sample and solve, i followed this

The interception points in your graph is wrong, this is incorrect answer. Should be (1,0) and (0,1) for x-intercept and y-intercept

Edit: this is wrong message

you can verify the equation here (Graphing Calculator)

Oops, sorry I made an assumption here. Yes, you are right, I take it back.

But how to do this without graph?

by algebra

There’s no video solution yet.

So what I would do is reduce this to a pair of linear equations: you have 1 - x = y - 1, which would be x + y = 2. The other is 1 - x = 1 - y, which is x = y. Then find their intersection points and hence their area. Visualisation becomes easier once you do that.

Okay, I thought the same way, but where is \pm (1-x) ? I was thinking like we reduce it in a form which \pm(1-x)=\pm(y-1) , so we have 4 equations in total. Dose my method correct?

Never mind, just found that the result from -(1-x)=\pm(y-1) would still lead to

x=y and y=2-x

@Maverick mate, check this out.

By the way, when solving absolute-value-related equations, do we always just look for \pm for one side? @Leaderboard

What do you mean?

Seems like you just added negative sign to one side of this equation, so I wonder do we always just take care of RHS or LHS when comes to dealing with equations which contain absolute value in both sides?

It does not matter - you can show that for any two generic functions f(x) and g(x), the equation |f(x)| = |g(x)| can be reduced to two non-absolute equations. To do that, just consider each case:

When f(x) \geq 0 and g(x) \geq 0, we have f(x) = g(x)

When f(x) \geq 0 and g(x) \leq0, we have f(x) = -g(x)

When f(x) \leq0 and g(x) \geq 0, we have -f(x) = g(x) - which is the same as the second case

When f(x) \leq0 and g(x) \leq 0, we have -f(x) = -g(x) - which is the same as the first case

Hence we can show that it’s essentially a pair of equations despite it looking like four in the first.

Ah, this explanation is crystally clear now. Thank you so much for solving my question!

Hello everyone,

I stuck in this problem: https://www.gregmat.com/problems/problem/the-area-embedded-by-the-graph-of/

Please help me by explaining the strategy of this problem. Thanks in advance.

Hi,

Could anyone please advise on this question?

I wonder if there is any other (simpler) way to solve this other than my approach.

I basically divided into four situations:

(1) 1-x>=0, y-1>=0

(2) 1-x>=0, y-1<0

(3) 1-x<0, y-1>=0

(4) 1-x<0, y-1<0

Then, I solved each equation with an algebra.

(1) y=x ; when 1>=x, y>=1

(2) y=-x+2 ; when 1>= x, y<1

(3) y=x ; when 1<x, y>=1

(4) y=-x+2 ; when 1<x, y<1

And then, I drew a graph for each situation.

I got the question right, but it took me like 7 minutes, so my approach definitely was a time sinkhole.

Does anyone know a better approach to solve this than mine? Would really appreciate your tips!