Answered

2.

(a) Using a scale of 2 cm to 2 units on both axes, draw on a sheet of graph paper two perpendicular axes [tex]\(OX\)[/tex] and [tex]\(OY\)[/tex] for the intervals [tex]\(-10 \leq x \leq 10\)[/tex] and [tex]\(-12 \leq y \leq 12\)[/tex].

(b) Draw on the same graph sheet, showing clearly the coordinates of all vertices:

(i) The triangle [tex]\(PQR\)[/tex] with [tex]\(P(4,8)\)[/tex], [tex]\(\overrightarrow{QP}=\binom{-2}{2}\)[/tex] and [tex]\(\overrightarrow{RP}=\binom{2}{4}\)[/tex];

(ii) The image [tex]\(\triangle P_1 Q_1 R_1\)[/tex] of [tex]\(\triangle PQR\)[/tex] under a reflection in the line [tex]\(y=-2\)[/tex], where [tex]\(P \rightarrow P_1\)[/tex], [tex]\(Q \rightarrow Q_1\)[/tex], and [tex]\(R \rightarrow R_1\)[/tex];

(iii) The image [tex]\(\triangle P_2 Q_2 R_2\)[/tex] of [tex]\(\triangle PQR\)[/tex] under translation by the vector [tex]\(\binom{-8}{2}\)[/tex], where [tex]\(P \rightarrow P_2\)[/tex], [tex]\(Q \rightarrow Q_2\)[/tex], and [tex]\(R \rightarrow R_2\)[/tex];

(iv) The image [tex]\(\triangle P_3 Q_3 R_3\)[/tex] of [tex]\(\triangle PQR\)[/tex] under a rotation through [tex]\(180^{\circ}\)[/tex] about the origin, where [tex]\(P \rightarrow P_3\)[/tex], [tex]\(Q \rightarrow Q_3\)[/tex], and [tex]\(R \rightarrow R_3\)[/tex].

(c) Find [tex]\(\overline{Q_2 Q_3}\)[/tex].



Answer :

GinnyAnswer
Sure, let's break down and solve each part of the question step-by-step.

### Part (a)
First, establish a coordinate system as described:

1. Draw two perpendicular axes: Draw the horizontal [tex]\(OX\)[/tex] axis and the vertical [tex]\(OY\)[/tex] axis with the origin [tex]\(O\)[/tex] at their intersection.
2. Set the scale: The scale is [tex]\(2\)[/tex] cm representing [tex]\(2\)[/tex] units. Therefore, each centimeter represents [tex]\(1\)[/tex] unit.
3. Mark intervals on the axes:
- On the [tex]\(OX\)[/tex] axis, mark intervals from [tex]\(-10\)[/tex] to [tex]\(10\)[/tex].
- On the [tex]\(OY\)[/tex] axis, mark intervals from [tex]\(-12\)[/tex] to [tex]\(12\)[/tex].

### Part (b)
For the following parts, you'll create different triangles on the graph paper based on the given transformations.

#### (i) Draw triangle [tex]\( PQR \)[/tex]:

1. Coordinates of [tex]\(P\)[/tex]: [tex]\((4,8)\)[/tex].
2. Find [tex]\(Q\)[/tex]:
[tex]\[ \overrightarrow{QP} = \begin{bmatrix}-2 \\ 2\end{bmatrix} \Rightarrow Q = P + \begin{bmatrix}-2 \\ 2\end{bmatrix} = \begin{bmatrix}4 \\ 8\end{bmatrix} + \begin{bmatrix}-2 \\ 2\end{bmatrix} = \begin{bmatrix}2 \\ 10\end{bmatrix} \][/tex]
3. Find [tex]\(R\)[/tex]:
[tex]\[ \overrightarrow{RP} = \begin{bmatrix}2 \\ 4\end{bmatrix} \Rightarrow R = P + \begin{bmatrix}2 \\ 4\end{bmatrix} = \begin{bmatrix}4 \\ 8\end{bmatrix} + \begin{bmatrix}2 \\ 4\end{bmatrix} = \begin{bmatrix}6 \\ 12\end{bmatrix} \][/tex]

So, the coordinates for triangle [tex]\(PQR\)[/tex] are:
- [tex]\(P(4,8)\)[/tex]
- [tex]\(Q(2,10)\)[/tex]
- [tex]\(R(6,12)\)[/tex]

#### (ii) Image [tex]\(\triangle P_1Q_1R_1\)[/tex] under reflection in [tex]\(y = -2\)[/tex]:

The reflection formula over the line [tex]\(y = c\)[/tex] is [tex]\( (x', y') = (x, 2c - y) \)[/tex].
Here, [tex]\(c = -2\)[/tex].

1. Reflect P:
[tex]\[ P_1 = (4, 2(-2) - 8) = (4, -12) \][/tex]
2. Reflect Q:
[tex]\[ Q_1 = (2, 2(-2) - 10) = (2, -14) \][/tex]
3. Reflect R:
[tex]\[ R_1 = (6, 2(-2) - 12) = (6, -16) \][/tex]

So, the coordinates for triangle [tex]\(\triangle P_1Q_1R_1\)[/tex] are:
- [tex]\(P_1(4, -12)\)[/tex]
- [tex]\(Q_1(2, -14)\)[/tex]
- [tex]\(R_1(6, -16)\)[/tex]

#### (iii) Image [tex]\(\triangle P_2Q_2R_2\)[/tex] under translation by vector [tex]\(\begin{bmatrix}-8 \\ 2\end{bmatrix}\)[/tex]:

1. Translate [tex]\(P\)[/tex]:
[tex]\[ P_2 = P + \begin{bmatrix}-8 \\ 2\end{bmatrix} = \begin{bmatrix}4 \\ 8\end{bmatrix} + \begin{bmatrix}-8 \\ 2\end{bmatrix} = \begin{bmatrix}-4 \\ 10\end{bmatrix} \][/tex]
2. Translate [tex]\(Q\)[/tex]:
[tex]\[ Q_2 = Q + \begin{bmatrix}-8 \\ 2\end{bmatrix} = \begin{bmatrix}2 \\ 10\end{bmatrix} + \begin{bmatrix}-8 \\ 2\end{bmatrix} = \begin{bmatrix}-6 \\ 12\end{bmatrix} \][/tex]
3. Translate [tex]\(R\)[/tex]:
[tex]\[ R_2 = R + \begin{bmatrix}-8 \\ 2\end{bmatrix} = \begin{bmatrix}6 \\ 12\end{bmatrix} + \begin{bmatrix}-8 \\ 2\end{bmatrix} = \begin{bmatrix}-2 \\ 14\end{bmatrix} \][/tex]

So, the coordinates for triangle [tex]\(\triangle P_2Q_2R_2\)[/tex] are:
- [tex]\(P_2(-4, 10)\)[/tex]
- [tex]\(Q_2(-6, 12)\)[/tex]
- [tex]\(R_2(-2, 14)\)[/tex]

#### (iv) Image [tex]\(\triangle P_3Q_3R_3\)[/tex] under rotation through [tex]\(180^\circ\)[/tex] about the origin:

The rotation formula for [tex]\(180^\circ\)[/tex] is [tex]\( (-x, -y) \)[/tex].

1. Rotate [tex]\(P\)[/tex]:
[tex]\[ P_3 = (-4, -8) \][/tex]
2. Rotate [tex]\(Q\)[/tex]:
[tex]\[ Q_3 = (-2, -10) \][/tex]
3. Rotate [tex]\(R\)[/tex]:
[tex]\[ R_3 = (-6, -12) \][/tex]

So, the coordinates for triangle [tex]\(\triangle P_3Q_3R_3\)[/tex] are:
- [tex]\(P_3(-4, -8)\)[/tex]
- [tex]\(Q_3(-2, -10)\)[/tex]
- [tex]\(R_3(-6, -12)\)[/tex]

### Part (c) Find [tex]\(\overline{Q_2 Q_3} \)[/tex]:

The distance between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \overline{Q_2 Q_3} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Given [tex]\(Q_2(-6, 12)\)[/tex] and [tex]\(Q_3(-2, -10)\)[/tex]:

[tex]\[ \overline{Q_2 Q_3} = \sqrt{((-2) - (-6))^2 + ((-10) - 12)^2} = \sqrt{(4)^2 + (-22)^2} = \sqrt{16 + 484} = \sqrt{500} = 22.360679774997898 \][/tex]

Thus, the length of [tex]\(\overline{Q_2 Q_3}\)[/tex] is approximately [tex]\(22.36\)[/tex].