A square ABCD has the vertices A(n, n), B(n, −n), C(−n, −n), and D(−n, n). Which vertex is in Quadrant II? Find the midpoint of the line segment defined by the points: (5, 4) and (−2, 1)
The one which lies in Quadrant II is D(-n,n). Quadrant II has positive y coordinate y and negative x coordinate.
Here's a way to find the midpoint x midpoint = [tex] \dfrac{x_{1}+x_{2}}{2}[/tex] y midpoint = [tex] \dfrac{y_{1}+y_{2}}{2}[/tex]
Find x midpoint, plug in the numbers x midpoint = [tex] \dfrac{x_{1}+x_{2}}{2}[/tex] x midpoint = [tex] \dfrac{5+(-2)}{2}[/tex] x midpoint = [tex] \dfrac{3}{2}[/tex] x midpoint = 1.5
Find y midpoint, plug in the numbers y midpoint = [tex] \dfrac{y_{1}+y_{2}}{2}[/tex] y midpoint = [tex] \dfrac{4+1}{2}[/tex] y midpoint = [tex] \dfrac{5}{2}[/tex] y midpoint = 2.5
