Question

The vertices of the trapezoid are J(4m, 4n), K(4q, 4n), M(4p, 0), and L(0, 0). Find the midpoint of the midsegment of the trapezoid.

Answer 1

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

$\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ \square}}\quad ,&{{ \square}})\quad % (c,d) &({{ \square}}\quad ,&{{ \square}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\\\\ -----------------------------$

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) J&({{ 4m}}\quad ,&{{ 4n}})\quad % (c,d) L&({{ 0}}\quad ,&{{ 0}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{0+4m}{2}\quad ,\quad \cfrac{0+4n}{2} \right) \\\\\\ \left( \cfrac{4m}{2},\cfrac{4n}{2} \right)\implies \boxed{(2m,2n)\impliedby H}\\\\ -----------------------------$

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ 4q}}\quad ,&{{ 4n}})\quad % (c,d) M&({{ 4p}}\quad ,&{{ 0}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{4p+4q}{2}\quad ,\quad \cfrac{0+4n}{2} \right) \\\\\\ \left( \cfrac{2(2p+2q)}{2},\cfrac{4n}{2} \right)\implies \boxed{[(2p+2q), 2n]\impliedby N}\\\\ -----------------------------$

$\bf \textit{so, the midpoint of HN is } \\\\\\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) H&({{ 2m}}\quad ,&{{ 2n}})\quad % (c,d) N&({{ 2p+2q}}\quad ,&{{ 2n}}) \end{array}\\\\\\ % coordinates of midpoint \left(\cfrac{(2p+2q)+2m}{2}\quad ,\quad \cfrac{2n+2n}{2} \right) \\\\\\ \left( \cfrac{2(p+q+m)}{2},\cfrac{4n}{2} \right)\implies (p+q+m)\quad ,\quad 2n$

Answer 2

m+q+p,2n

this is the answer bc i had it too.