Answer:
Part 4) [tex]x=\frac{62}{3}\ ft[/tex]
Part 5a) [tex]J(-0.5,2)[/tex]
Part 5b) [tex]L(2.5,0)[/tex]
Part 5c) see the explanation
Step-by-step explanation:
Part 4) we know that
The Midpoint Theorem says that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the third side
so in this problem
Applying the midpoint theorem
[tex]x=\frac{124}{3}:2=\frac{124}{6}=\frac{62}{3}\ ft[/tex]
Part 5) we have
[tex]M(-4,1),A(3,3),N (2,-3)[/tex]
Part a) Determine the coordinates of J, the midpoint of MA
we know that
The formula to calculate the midpoint between two points is equal to
[tex](\frac{x1+x2}{2},\frac{y1+y2}{2})[/tex]
we have
[tex]M(-4,1),A(3,3)[/tex]
substitute
[tex]J(\frac{-4+3}{2},\frac{1+3}{2})[/tex]
[tex]J(-0.5,2)[/tex]
Part b) Determine the coordinates of L, the midpoint of AN
we know that
The formula to calculate the midpoint between two points is equal to
[tex](\frac{x1+x2}{2},\frac{y1+y2}{2})[/tex]
we have
[tex]A(3,3),N (2,-3)[/tex]
substitute
[tex]L(\frac{3+2}{2},\frac{3-3}{2})[/tex]
[tex]L(2.5,0)[/tex]
Part c) Prove that JL=1/2MN
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Find the distance JL
we have
[tex]J(-0.5,2),L(2.5,0)[/tex]
substitute the values in the formula
[tex]d=\sqrt{(0-2)^{2}+(2.5+0.5)^{2}}[/tex]
[tex]d=\sqrt{(-2)^{2}+(3)^{2}}[/tex]
[tex]JL=\sqrt{13}\ units[/tex]
Find the distance MN
we have
[tex]M(-4,1),N (2,-3)[/tex]
substitute the values in the formula
[tex]d=\sqrt{(-3-1)^{2}+(2+4)^{2}}[/tex]
[tex]d=\sqrt{(-4)^{2}+(6)^{2}}[/tex]
[tex]MN=\sqrt{52}\ units[/tex]
simplify
[tex]MN=2\sqrt{13}\ units[/tex]
Prove that
[tex]JL=\frac{1}{2} MN[/tex]
substitute the values
[tex]\sqrt{13}=\frac{1}{2} 2\sqrt{13}[/tex]
[tex]\sqrt{13}=\sqrt{13}[/tex] ---> is true
therefore
Is verified
