Proving Convexity of the Set X = {(x, y) E R^2; ax + by <= c} in R^2

[bonildo]

This exercise is located in the vector space chapter of my book that's why I am posting it here.
Recently started with this kind of exercise, proof like exercises and I am a little bit lost
Proof that given a, b, c real numbers, the set X = {(x, y) E R^2; ax + by <= c} ´is convex at R^2

the definition of convex set in the book is given like that: u, v E X => [u, v] C X
and [u,v]={ (1-t)u+tv ; 0<=t<=1}Didnt do much, just that :
u=(x1,y1) and ax1+by1<c
v=(x2,y2) and ax2+by2<c

and that [u,v]={(1-t)x1+tx2,(1-t)y1+ty2)}

 

[Infrared]

Okay, you have $ax_1+by_1\leq c$ and $ax_2+by_2\leq c$. Try multiplying the two inequalities by $1-t$ and $t$, respectively, and then adding.

 

[bonildo]

Okay, you have $ax_1+by_1\leq c$ and $ax_2+by_2\leq c$. Try multiplying the two inequalities by $1-t$ and $t$, respectively, and then adding.

Ok, got to a((1-t)x1+tx2)+b((1-t)y1+ty2)<=c . But why it proof that X is convex ?

 

[zinq]

You just have to try to apply the exact definition of convex to the two arbitrary points of the set defined by that inequality. This will test whether you actually understand what that definition is saying.

 

[Infrared]

You're trying to show that $(1-t)u+tv$ is in the half-plane $ax+by\leq c$. You also have $(1-t)u+tv=((1-t)x_1+tx_2,(1-t)y_1+ty_2).$ So, you're trying to show that $a\left((1-t)x_1+tx_2\right)+b\left((1-t)y_1+ty_2\right)\leq c.$ But this is the inequality you just wrote

Ok, got to a((1-t)x1+tx2)+b((1-t)y1+ty2)<=c . But why it proof that X is convex ?

 

[bonildo]

You're trying to show that $(1-t)u+tv$ is in the half-plane $ax+by\leq c$. You also have $(1-t)u+tv=((1-t)x_1+tx_2,(1-t)y_1+ty_2).$ So, you're trying to show that $a\left((1-t)x_1+tx_2\right)+b\left((1-t)y_1+ty_2\right)\leq c.$ But this is the inequality you just wrote

got it , thanks