Answer:
Part 1) [tex](2.5,2.5)[/tex] and [tex](2,0)[/tex]
Part 2) [tex]BD=70\ units[/tex]
Part 3) m∠KLJ=[tex]25\°[/tex]
Step-by-step explanation:
Part 1)
we have
[tex]J(1,4),K(4,1),L(0,-1)[/tex]
Find the coordinates of the midpoint JK
the x-coordinate of the midpoint JK is equal to
[tex]x=\frac{1+4}{2}\\\\x=2.5[/tex]
the y-coordinate of the midpoint JK is equal to
[tex]y=\frac{4+1}{2}\\\\y=2.5[/tex]
The midpoint JK is the point [tex](2.5,2.5)[/tex]
Find the coordinates of the midpoint LK
the x-coordinate of the midpoint LK is equal to
[tex]x=\frac{0+4}{2}\\\\x=2[/tex]
the y-coordinate of the midpoint LK is equal to
[tex]y=\frac{-1+1}{2}\\\\y=0[/tex]
The midpoint LK is the point [tex](2,0)[/tex]
The answer part 1) is
the endpoint coordinates for the midsegment of △JKL that is parallel to JL are the points [tex](2.5,2.5)[/tex] and [tex](2,0)[/tex]
Part 2)
we know that
The diagonals bisect the parallelogram into two congruent triangles
In the parallelogram ABCD
[tex]BE=DE[/tex]
substitute the values
[tex]2x^{2}-3x=x^{2}+10\\x^{2}-3x-10=0[/tex]
Solve the quadratic equation
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-3x-10=0[/tex]
so
[tex]a=1\\b=-3\\c=-10[/tex]
substitute in the formula
[tex]x=\frac{3(+/-)\sqrt{(-3)^{2}-4(1)(-10)}} {2(1)}[/tex]
[tex]x=\frac{3(+/-)\sqrt{49}} {2}[/tex]
[tex]x=\frac{3(+/-)7} {2}[/tex]
[tex]x=\frac{3+7} {2}=5[/tex]
[tex]x=\frac{3-7} {2}=-2[/tex]
Find the value of BD
[tex]BD=2x^{2}-3x+x^{2}+10[/tex]
Substitute the value of [tex]x=5[/tex]
[tex]BD=2(5)^{2}-3(5)+(5)^{2}+10=70\ units[/tex]
Part 3)
we know that
The diagonals bisect the parallelogram into two congruent triangles
In the parallelogram JKLM
m∠KLJ=m∠MLJ
we have that
m∠MLJ=[tex]25\°[/tex]
therefore
m∠KLJ=[tex]25\°[/tex]
