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Gnom [1K]
3 years ago
9

(02.02 MC) Triangle ABC is shown. A is at negative 2, 1. B is at negative 1, 4. C is at negative 4, 5. If triangle ABC is reflec

ted over the x‐axis, reflected over the y‐axis, and rotated 180 degrees, where will point B' lie? (−1, 4) (1, −4) (4, −1) (−4, 1)
Mathematics
1 answer:
Bezzdna [24]3 years ago
6 0

Given:

The vertices of a triangle ABC are A(-2,1), B(-1,4) and C(-4,5).

Triangle ABC is reflected over the x‐axis, reflected over the y‐axis, and rotated 180 degrees.

To find:

The point B'.

Solution:

If a figure reflected over x-axis, then

(x,y)\to (x,-y)

B(-1,4)\to B_1(-1,-4)

If a figure reflected over y-axis, then

(x,y)\to (-x,y)

B_1(-1,-4)\to B_2(-(-1),-4)

B_1(-1,-4)\to B_2(1,-4)

If a figure rotated 180 degrees about the origin, then

(x,y)\to (-x,-y)

B_2(1,-4)\to B'(-(1),-(-4))

B_2(1,-4)\to B'(-1,4)

So, the coordinate of point B' are (-1,4).

Therefore, the correct option is A.

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In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} to the standard basis of P2.
Serjik [45]

Complete Question:

In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.

Answer:

The change of coordinate matrix is :

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

Step-by-step explanation:

Let U =  {D, E, F} be any vector with respect to Basis B

U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)

U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)

In Matrix form;

\left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right] \left[\begin{array}{ccc}D\\E\\F\end{array}\right] = \left[\begin{array}{ccc}D+2E+F\\E+2F\\-3D-5E\end{array}\right]

The change of coordinate matrix is therefore,

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

To find D, E, F in (**) such that U = t²

D + 2E + F = 0.................(1)

E + 2F = 0.........................(2)

-3D -5E = 1........................(3)

Substituting eqn (2) into eqn (1 )

D=3F...................................(4)

Substituting equations (2) and (4) into eqn (3)

-9F+10F=1

F = 1

Put the value of F into equations (2) and (4)

E = -2(1) = -2

D = 3(1) = 3

Substituting the values of D, E, and F into (*)

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

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Step-by-step explanation:

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