Triangle BCD With Vertices B(4,4), C(8, 14), And P(12, 0); K = 1/4

Answer:

Hello, I am not sure if you mean a dilation by 1/4. If it is, this will be the answer:

B (1, 1)

C (2, 3.5)

D (3, 0)

Step-by-step explanation:

Just divide each number by 4. B (4, 4) = B (4/4, 4/4) = B (1, 1)

Continue to do this for each coordinate set given.

Hope this helps!

The vertices of triangle B'C'D' are:

B'(-1, -1/2)

C'(0, 2)

D'(1, -3/2)

A triangle is a 2-D figure with three sides and three angles.

The sum of the angles is 180 degrees.

We can have an obtuse triangle, an acute triangle, or a right triangle.

We have,

To find the vertices of triangle B'C'D', which is a dilation of triangle BCD with center at the origin and a scale factor of 1/4,

We can follow these steps:

Find the coordinates of the centroid of triangle BCD.

The coordinates of the centroid G are:

[tex]x_G = (x_B + x_C + x_D) / 3 = (4 + 8 + 12) / 3 = 8[/tex]

[tex]y_G = (y_B + y_C + y_D) / 3 = (4 + 14 + 0) / 3 = 6[/tex]

Translate triangle BCD so that its centroid is at the origin.

We can do this by subtracting the coordinates of the centroid from the coordinates of each vertex:

B(4, 4) → [tex]B_1(4 - 8, 4 - 6) = B_1(-4, -2)[/tex]

C(8, 14) → [tex]C_1(8 - 8, 14 - 6) = C_1(0, 8)[/tex]

D(12, 0) → [tex]D_1(12 - 8, 0 - 6) = D_1(4, -6)[/tex]

Dilate the translated triangle [tex]B_1C_1D_1[/tex] by a scale factor of 1/4.

To do this, we can multiply the coordinates of each vertex by 1/4:

[tex]B_1[/tex](-4, -2) → B'(-4 x 1/4, -2 x 1/4) = B'(-1, -1/2)

[tex]C_1[/tex](0, 8) → C'(0 x 1/4, 8 x 1/4) = C'(0, 2)

[tex]D_1[/tex](4, -6) → D'(4 x 1/4, -6 x 1/4) = D'(1, -3/2)

Therefore,

The vertices of triangle B'C'D' are:

B'(-1, -1/2)

C'(0, 2)

D'(1, -3/2)

Learn more about triangles here:

https://brainly.com/question/25950519

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