Question

4. Determine the length across the river X to the nearest hundredth.

5. The coordinates of the vertices of a triangle are M(-4,1) , A(3,3) , and N (2,-3)

Part A : determine the coordinates of J, the midpoint of MA.


Part B: determine the coordinates of L, the midpoint of AN.

Part C: Prove that JL=1/2MN

How do I solve this?

Answer 1

Part 4)
`x=(62)/(3)\ ft`

Part 5a)
`J(-0.5,2)`

Part 5b)
`L(2.5,0)`

Part 5c) see the explanation

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

`x=(124)/(3):2=(124)/(6)=(62)/(3)\ ft`

Part 5) we have

`M(-4,1),A(3,3),N (2,-3)`

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

`((x1+x2)/(2),(y1+y2)/(2))`

we have

`M(-4,1),A(3,3)`

substitute

`J((-4+3)/(2),(1+3)/(2))`

`J(-0.5,2)`

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

`((x1+x2)/(2),(y1+y2)/(2))`

we have

`A(3,3),N (2,-3)`

substitute

`L((3+2)/(2),(3-3)/(2))`

`L(2.5,0)`

Part c) Prove that JL=1/2MN

we know that

the formula to calculate the distance between two points is equal to

`d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}`

Find the distance JL

we have

`J(-0.5,2),L(2.5,0)`

substitute the values in the formula

`d=\sqrt{(0-2)^(2)+(2.5+0.5)^(2)}`

`d=\sqrt{(-2)^(2)+(3)^(2)}`

`JL=√(13)\ units`

Find the distance MN

we have

`M(-4,1),N (2,-3)`

substitute the values in the formula

`d=\sqrt{(-3-1)^(2)+(2+4)^(2)}`

`d=\sqrt{(-4)^(2)+(6)^(2)}`

`MN=√(52)\ units`

simplify

`MN=2√(13)\ units`

Prove that

`JL=(1)/(2) MN`

substitute the values

`√(13)=(1)/(2) 2√(13)`

`√(13)=√(13)` ---> is true

therefore

Is verified