Let 5 be the region that lies between the curves y=− xm ; y= − xn ; 0 < x < 1 where m and n are integers with 0 < n , m. (a) Sketch the region 5. (b) Find the coordinates of the centroid of 5. (c) Try to find values of m and n such that the centroid lie.

(c) due to the definition of the centroid of a triangle, this will always lie inside the triangle, therefore, for any value of [tex]m[/tex] and [tex]n[/tex], the centroid will lie

Step-by-step explanation:

Hi, let us first solve part (a). Since for any given values of [tex]n[/tex] and [tex]m[/tex] we will obtain two linear functions:

[tex]y=-mx[/tex] and

[tex]y=-nx[/tex]

with [tex]0\leq x\leq 1[/tex] we can assure that our region is going to be a triangle. To see this, please take a look at the plot I generated using Wolfram. In this case, I have used two specific values for m and n but keeping the condition [tex]0\leq n\leq m[/tex].

Now, for part (b) let me start remembering what the centroid is: the centroid of a triangle is the point where the three medians of the triangle meet. And a median of a triangle is a line segment from one vertex to the midpoint on the opposite side of the triangle (see the second figure where the medians are depicted in red and the centroid of the triangle, G is depicted in blue). For a given triangle [tex]\bigtriangleup \rm{ABC}[/tex], the coordinates of its centroid [tex]G[/tex] are given by:

[tex]G_x=\frac{A_x+B_x+C_x}{3}[/tex] and [tex]G_y=\frac{A_y+B_y+C_y}{3}[/tex]

Now let's apply this to our problem. Take a look at the first figure. The vertex A has clearly coordinates [tex](0,0)[/tex] for any value of [tex]m[/tex] and [tex]n[/tex] since the two lines have their intersection with y-axis in this point.

To obtain the coordinates of [tex]B[/tex] and [tex]C[/tex], let's use the given functions and the fact that the coordinate x is limited to 1. Then, we have:

For A:

[tex]y=-mx[/tex] then, when [tex]x=1[/tex], substituting in the formula [tex]y=-m[/tex]

For B and doing the same as for A:

[tex]y=-nx[/tex] then, when [tex]x=1[/tex], substituting in the formula [tex]y=-n[/tex]

Thus, the coordinates of the vertices of the triangle are: [tex]A(1, m)[/tex], [tex]B(1,3)[/tex] and [tex]C(0,0)[/tex] and the coordinates of the centroid are:

[tex]G_x=\frac{A_x+B_x+C_x}{3} = G_x=\frac{1+1+0}{3}\\G_x=\frac{2}{3}[/tex]

and

[tex]G_y=\frac{A_y+B_y+C_y}{3}=\frac{m+n+0}{3}\\G_y=\frac{m+n}{3}[/tex].

Summarizing: the coordinates of the centroid of the region are [tex]G(\frac{2}{3}, \frac{m+n}{3} )[/tex]

Now, for part (c), due to the definition of the centroid of a triangle, this will always lie inside the triangle, therefore, for any value of [tex]m[/tex] and [tex]n[/tex], the centroid will lie. Other important points of the triangle, like the orthocentre and circumcentre, can lie outside in obtuse triangles. In right triangles, the orthocentre always lies at the right-angled vertex.