Question

The triangle, ABC, shown on the coordinate plane below, is dilated from the origin by a scale factor of r = 1/3. What is the location of A'B'C'?​

Answer 1

After dilating triangle ABC from the origin by a scale factor
`\(r = (1)/(3)\)`, the new coordinates for A', B', and C' are
`(1, \((4)/(3)\)), (\((7)/(3)\), \((2)/(3)\))`, and
`(\((2)/(3)\), \((1)/(3)\))`, respectively.

When dilating a point (x, y) from the origin by a scale factor r, the new coordinates (x', y') are given by:

`\[ x' = r \cdot x \]\[ y' = r \cdot y \]`

Given that the scale factor
`\(r = (1)/(3)\)` and the original points are:

A (
`3`,
`4`)

B (
`7`,
`2`)

C (
`2`,
`1`)

The new coordinates A', B', and C' after dilation are:

`\[ A' = \left((1)/(3) \cdot 3, (1)/(3) \cdot 4\right) = (1, (4)/(3)) \]\[ B' = \left((1)/(3) \cdot 7, (1)/(3) \cdot 2\right) = \left((7)/(3), (2)/(3)\right) \]\[ C' = \left((1)/(3) \cdot 2, (1)/(3) \cdot 1\right) = \left((2)/(3), (1)/(3)\right) \]`

Therefore, the location of A'B'C' after the dilation is:

A'
`(1, \((4)/(3)\))`

B'
`(\((7)/(3)\), \((2)/(3)\))`

C'
`(\((2)/(3)\), \((1)/(3)\))`

Answer 2

Explanation:

Coordinates of the vertices of the triangle ABC are A(3, 4), B(-7, 2) and C(2, 1).

Rule for the dilation of the coordinates of the ends of a segment is,

(x, y) → k(x, y)

→ (kx, ky)

Where k = scale factor by which the segment is dilated

If the given triangle ABC is dilated by a scale factor =
`(1)/(3)`

Coordinates of the vertices of the image triangle will be,

A(3, 4) → A'(1,
`(4)/(3)`) Or A'(1, 1.3)

B(-7, 2) →
`B'((7)/(3),(2)/(3))` Or B'(2.3, 0.7)

C(2, 1) →
`C'((2)/(3),(1)/(3))` Or C'(0.7, 0.3)

Now we can plot these points on the given graph