The coordinates of the vertices of cde are c(1,4), d(3,6), and e(7,4). if the triangle is reflected over the line y=3, what are the coordinates of the image of d?

 The required co-ordinates of the image of D are (3, 0).

Step-by-step explanation:  Given that the co-ordinates of the vertices of triangle CDE are C(1,4), D(3,6), and E(7,4). The triangle CDE is reflected over the line y=3.

We are to find the co-ordinates of the image of vertex D.

We know that

if a point with co-ordinates (x, y) is reflected across the line y = k, then the co-ordinates changes according to the following rule :

(x, y)    ⇒      (x, 2k - y).

So, the co-ordinates of the image of vertex D after reflection from the line y = 3 is

D(3, 6)    ⇒     D'(3, 2 × 3 - 6) = D'(3, 0).

The reflection is shown in the attached figure below.

Thus, the required co-ordinates of the image of D are (3, 0).

we are given vertices of CDE as

C(1,4), D(3,6), and E(7,4)

It is reflected about y=3 line

Let's assume reflected point as D'

D'=(a,b)

Mid-point will be at y=3

so, M=(3,3)

so, M is the mid-point of D and D'

now, we can solve for 'a' and 'b'

so, we will get reflected point as

D'=(3,0)...........Answer

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

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step-by-step explanation:

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