Question

A triangle has vertices at (4, 4), (-6, 2) and (2, 0).

a. Find the coordinates of the mid-points of eachside

b. Find the lengths of the sides of the triangle
formed by joining the mid-points.
each side.

Answer 1

a. To find the coordinates of endpoints we must add two x values and divide by 2 and then add 2 y- values and divide by 2.

(4-6)/2=-1 (4+2)/2=3

Repeat for other sides.

(2-6)/2=-2 (2+0)/2=1

(2+4)/2=3 (4+0)/2=2

Coordinates of midpoints are (-1,3), (-2,1), (3,2)

b. Now we use the distance formula for each midpoint to find the length of the inner- triangle.

sqrt((-1+2)^2 + (3-1)^2)

Sqrt(5)

Repeat.

Sqrt(17)

Sqrt(26)

The lengths of the inner triangle are as follows:

(-1,3), (-2,1) = Sqrt(5)

(-1,3), (3,2) = Sqrt(17)

(-2,1), (3,2) = Sqrt(26)

Answer 2

A. (-1, 3) (3, 2) (-2, 1)

B.
`√(17)`
`√(5)`
`√(26)`

Explanation:

The formula for finding the midpoints is
`(\frac{x_(1)+x_(2) }2}, \frac{y_(1)+y_(2) }2} )`.

MIDPOINTS

`(\frac{x_(1)+x_(2) }2}, \frac{y_(1)+y_(2) }2} )` =
`(\frac{4-6}2,\frac{4+2 }2} )` = (-1, 3)

`(\frac{x_(1)+x_(2) }2}, \frac{y_(1)+y_(2) }2} )` =
`(\frac{4+2}2,\frac{4+0 }2} )` = (3, 2)

`(\frac{x_(1)+x_(2) }2}, \frac{y_(1)+y_(2) }2} )` =
`(\frac{-6+2}2,\frac{2+0 }2} )` = (-2, 1)

Next, we will use the distance formula,
`\sqrt{(x_(1)-x_(2))^(2) + (y_(1)-y_(2))^(2) }`.

DISTANCE

`\sqrt{(x_(1)-x_(2))^(2) + (y_(1)-y_(2))^(2) }` =
`\sqrt{(-1-3)^(2) + (3-2)^(2) }` =
`√(17)`

`\sqrt{(x_(1)-x_(2))^(2) + (y_(1)-y_(2))^(2) }` =
`\sqrt{(-1+2)^(2) + (3-1)^(2) }` =
`√(5)`

`\sqrt{(x_(1)-x_(2))^(2) + (y_(1)-y_(2))^(2) }` =
`\sqrt{(3+2)^(2) + (2-1)^(2) }` =
`√(26)`