Question

A triangle has corners at `(6 ,2 )`, `(5 ,-8 )`, and `(-5 ,3 )`. If the triangle is dilated by a factor of `5` about point #(7 ,-2 ), how far will its centroid move?

Answer 1

The distance moved is `4sqrt(26)`

Explanation:

Let's begin by computing the current centroid, point `O_1`,

`O_(1x) = (6 + 5 - 5)/3 = 2`

`O_(1y) = (2 - 8 + 3)/3 = -1`

Point `O_1 = (2, -1)`

1. Scale point `(6, 2)`:

Compute the vector from point `(7, -2)` to point `(6, 2)`:

`(6 - 7)hati + (2 - -2)hatj = -hati + 4hatj`

Scale by 5:

`-5hati + 20hatj`

Find the new end point:

`(-5+ 7, 20 + -2) = (2, 18)`

1. Scale point `(5, -8)`:

Compute the vector from point `(7, -2)` to point `(5, -8)`:

`(5 - 7)hati + (-8 - -2)hatj = -2hati -6hatj`

Scale by 5:

`-10hati - 30hatj`

Find the new end point:

`(-10+ 7, -30 + -2) = (-3, -32)`

1. Scale point `(-5, 3)`:

Compute the vector from point `(7, -2)` to point `(-5, 3)`:

`(-5 - 7)hati + (3 - -2)hatj = -12hati + 5hatj`

Scale by 5:

`-60hati + 25hatj`

Find the new end point:

`(-60+ 7, 25 + -2) = (-53, 23)`

Our scaled triangle has the vertices, `(2, 18)`, `(-3, -32)`, and `(-53, 23)`. The new centroid, `O_2`, has coordinates:

`O_(2x) = (2 - 3 - 53)/3 = -18`

`O_(2y) = (18 - 32 + 23)/3 = 3`

`O_2 = (-18, 3)`

Compute the distance, d, between `O_1` and `O_2`:

`d = sqrt((-18 - 2)^2 + (3 - -1)^2)`

`d = sqrt((-20)^2 + 4^2)`

`d = sqrt((-20)^2 + 4^2)`

`d = sqrt(416)`

`d = 4sqrt(26)`