Question

1. What are the endpoint coordinates for the midsegment of △JKL that is parallel to JL⎯⎯⎯⎯?

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( ) ( ), and ( ) ( )
2. In parallelogram ABCD , diagonals AC⎯⎯⎯⎯⎯ and BD⎯⎯⎯⎯⎯ intersect at point E, BE=2x2−3x , and DE=x2+10 .

What is BD ?
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units
3.JKLM is a parallelogram.

What is the measure of ∠KLJ?

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Answer 1

Part 1)
`(2.5,2.5)` and
`(2,0)`

Part 2)
`BD=70\ units`

Part 3) m∠KLJ=
`25\°`

Explanation:

Part 1)

we have

`J(1,4),K(4,1),L(0,-1)`

Find the coordinates of the midpoint JK

the x-coordinate of the midpoint JK is equal to

`x=(1+4)/(2)\\\\x=2.5`

the y-coordinate of the midpoint JK is equal to

`y=(4+1)/(2)\\\\y=2.5`

The midpoint JK is the point
`(2.5,2.5)`

Find the coordinates of the midpoint LK

the x-coordinate of the midpoint LK is equal to

`x=(0+4)/(2)\\\\x=2`

the y-coordinate of the midpoint LK is equal to

`y=(-1+1)/(2)\\\\y=0`

The midpoint LK is the point
`(2,0)`

The answer part 1) is

the endpoint coordinates for the midsegment of △JKL that is parallel to JL are the points
`(2.5,2.5)` and
`(2,0)`

Part 2)

we know that

The diagonals bisect the parallelogram into two congruent triangles

In the parallelogram ABCD

`BE=DE`

substitute the values

`2x^(2)-3x=x^(2)+10\\x^(2)-3x-10=0`

Solve the quadratic equation

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^(2)-3x-10=0`

so

`a=1\\b=-3\\c=-10`

substitute in the formula

`x=\frac{3(+/-)\sqrt{(-3)^(2)-4(1)(-10)}} {2(1)}`

`x=\frac{3(+/-)√(49)} {2}`

`x=\frac{3(+/-)7} {2}`

`x=\frac{3+7} {2}=5`

`x=\frac{3-7} {2}=-2`

Find the value of BD

`BD=2x^(2)-3x+x^(2)+10`

Substitute the value of
`x=5`

`BD=2(5)^(2)-3(5)+(5)^(2)+10=70\ units`

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=
`25\°`

therefore

m∠KLJ=
`25\°`