1. What are the endpoint coordinates for the midsegment of △JKL that is parallel to JL⎯⎯⎯⎯?Enter your answer, as a decimal or whole number, in the boxes.( ) ( ), and ( ) ( )2. In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, BE=2x2−3x , and DE=x2+10 .What is BD ?Enter your answer in the box. units3.JKLM is a parallelogram. What is the measure of ∠KLJ?Enter your answer in the box.

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 JKthe 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 LKthe 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 thatThe diagonals bisect the parallelogram into two congruent trianglesIn 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 thatThe diagonals bisect the parallelogram into two congruent trianglesIn the parallelogram JKLMm∠KLJ=m∠MLJwe have thatm∠MLJ=[tex]25\°[/tex]thereforem∠KLJ=[tex]25\°[/tex]