Respuesta :
to get the midsegment, namely HN, well, we need H and N
hmm so.... notice the picture you have there, is just an "isosceles trapezoid", namely, it has two equal sides, the left and right one, namely JL and KM
the midpoint of JL is H and the midpoint of KM is N
thus
[tex]\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ \square}}\quad ,&{{ \square}})\quad % (c,d) &({{ \square}}\quad ,&{{ \square}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\\\\ -----------------------------\\\\[/tex]
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) J&({{ 4m}}\quad ,&{{ 4n}})\quad % (c,d) L&({{ 0}}\quad ,&{{ 0}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{0+4m}{2}\quad ,\quad \cfrac{0+4n}{2} \right) \\\\\\ \left( \cfrac{4m}{2},\cfrac{4n}{2} \right)\implies \boxed{(2m,2n)\impliedby H}\\\\ -----------------------------\\\\[/tex]
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ 4q}}\quad ,&{{ 4n}})\quad % (c,d) M&({{ 4p}}\quad ,&{{ 0}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{4p+4q}{2}\quad ,\quad \cfrac{0+4n}{2} \right) \\\\\\ \left( \cfrac{2(2p+2q)}{2},\cfrac{4n}{2} \right)\implies \boxed{[(2p+2q), 2n]\impliedby N}\\\\ -----------------------------\\\\[/tex]
[tex]\bf \textit{so, the midpoint of HN is } \\\\\\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) H&({{ 2m}}\quad ,&{{ 2n}})\quad % (c,d) N&({{ 2p+2q}}\quad ,&{{ 2n}}) \end{array}\\\\\\ % coordinates of midpoint \left(\cfrac{(2p+2q)+2m}{2}\quad ,\quad \cfrac{2n+2n}{2} \right) \\\\\\ \left( \cfrac{2(p+q+m)}{2},\cfrac{4n}{2} \right)\implies (p+q+m)\quad ,\quad 2n[/tex]
hmm so.... notice the picture you have there, is just an "isosceles trapezoid", namely, it has two equal sides, the left and right one, namely JL and KM
the midpoint of JL is H and the midpoint of KM is N
thus
[tex]\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ \square}}\quad ,&{{ \square}})\quad % (c,d) &({{ \square}}\quad ,&{{ \square}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\\\\ -----------------------------\\\\[/tex]
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) J&({{ 4m}}\quad ,&{{ 4n}})\quad % (c,d) L&({{ 0}}\quad ,&{{ 0}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{0+4m}{2}\quad ,\quad \cfrac{0+4n}{2} \right) \\\\\\ \left( \cfrac{4m}{2},\cfrac{4n}{2} \right)\implies \boxed{(2m,2n)\impliedby H}\\\\ -----------------------------\\\\[/tex]
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ 4q}}\quad ,&{{ 4n}})\quad % (c,d) M&({{ 4p}}\quad ,&{{ 0}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{4p+4q}{2}\quad ,\quad \cfrac{0+4n}{2} \right) \\\\\\ \left( \cfrac{2(2p+2q)}{2},\cfrac{4n}{2} \right)\implies \boxed{[(2p+2q), 2n]\impliedby N}\\\\ -----------------------------\\\\[/tex]
[tex]\bf \textit{so, the midpoint of HN is } \\\\\\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) H&({{ 2m}}\quad ,&{{ 2n}})\quad % (c,d) N&({{ 2p+2q}}\quad ,&{{ 2n}}) \end{array}\\\\\\ % coordinates of midpoint \left(\cfrac{(2p+2q)+2m}{2}\quad ,\quad \cfrac{2n+2n}{2} \right) \\\\\\ \left( \cfrac{2(p+q+m)}{2},\cfrac{4n}{2} \right)\implies (p+q+m)\quad ,\quad 2n[/tex]
The midpoint of the midsegment is H (2m, 2n) and N (2(q+p), 2n).
What is mid-point?
A point at or near the center or middle
Given:
J(4m, 4n), K(4q, 4n), M(4p, 0), and L(0, 0).
Using section formula
H (x- coordinate)= 1*4m + 0 /2
H (x- coordinate)=2m
H (Y- coordinate)= 4n/2
H(Y- coordinate)=2n
So, the coordinates of H (2m, 2n)
similarly,
N (x- coordinate)= (1*4q + 1*4p)/2
N (x- coordinate)=2(p+q)
N (Y- coordinate)= 4n/2
N(Y- coordinate)=2n
So, the coordinates of N (2(q+p), 2n)
Learn more about this concept here:
https://brainly.com/question/14169618
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