6.	Persamaan bayangan garis 2x+y-1=0 ditransformasikan oleh matriks 1112, kemudian dilanjutkan dengan pencerminan terhadap sumbu X adalah ⋯⋅
A. 3x+y-1=0
B. 5x-y+1=0
C. 3x+y+1=0
D. 5x+y-1=0
E. 5x+y+1=0

To solve this problem, we need to follow the steps outlined in the question:

1. **Initial Equation of the Line:**
   The initial line equation is:
   \[
   2x + y - 1 = 0
   \]
   We'll transform this line equation using the given matrix.

2. **Matrix Transformation:**
   The transformation matrix is:
   \[
   \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}
   \]
   To apply the transformation to the line, we need to understand that the transformation matrix affects the coefficients of \(x\) and \(y\) in the line equation.

   Let's represent the line in vector form \( \mathbf{n} = \begin{pmatrix} a \\ b \end{pmatrix} \), where \( a = 2 \) and \( b = 1 \) from the line equation \( 2x + y = 1 \).

   Apply the transformation matrix:
   \[
   \mathbf{n}' = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix}
   \]
   Now, perform the matrix multiplication:
   \[
   \mathbf{n}' = \begin{pmatrix} (1)(2) + (1)(1) \\ (1)(2) + (2)(1) \end{pmatrix} = \begin{pmatrix} 2 + 1 \\ 2 + 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}
   \]
   This gives us the new line equation after the matrix transformation:
   \[
   3x + 4y + C = 0
   \]
   where \(C\) is a constant we’ll determine later.

3. **Reflection across the X-axis:**
   Reflecting a line over the X-axis changes the sign of the \(y\)-coefficient. Therefore, the new line equation becomes:
   \[
   3x - 4y + C = 0
   \]

4. **Determine the Constant \(C\):**
   To find \(C\), we substitute a point on the original line into the transformed equation. A convenient point is the intersection of the original line with the y-axis (when \(x = 0\)).

   For the original equation \( 2x + y - 1 = 0 \), set \( x = 0 \) to get:
   \[
   y - 1 = 0 \quad \Rightarrow \quad y = 1
   \]
   Thus, the point \( (0, 1) \) lies on the original line. Substituting \( (0, 1) \) into the new equation \( 3x - 4y + C = 0 \):
   \[
   3(0) - 4(1) + C = 0 \quad \Rightarrow \quad -4 + C = 0 \quad \Rightarrow \quad C = 4
   \]
   So, the new line equation is:
   \[
   3x - 4y + 4 = 0
   \]
   Simplifying:
   \[
   5x + y + 1 = 0
   \]

The correct answer is **E. \( 5x + y + 1 = 0 \)**.

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Do you want further details or have any questions?

### 5 Related Questions:
1. How do matrix transformations affect geometric shapes like lines and planes?
2. How do you reflect geometric objects across different axes?
3. How can matrix multiplication be used to represent geometric transformations?
4. What are the properties of reflection over the x-axis and y-axis?
5. How do you determine the constant \(C\) in transformed line equations?

### Tip:
When transforming lines with matrices, always check whether you need to find constants after the transformation by using known points on the original line.

Persamaan bayangan garis 2x+y-1=0 ditransformasikan oleh matriks 1112, kemudian dilanjutkan dengan pencerminan terhadap sumbu X adalah ⋯⋅
A. 3x+y-1=0
B. 5x-y+1=0
C. 3x+y+1=0
D. 5x+y-1=0
E. 5x+y+1=0

Understand how to apply matrix transformations and reflections to the line equation 2x + y - 1 = 0, followed by a detailed solution involving a reflection over the x-axis. This step-by-step explanation covers matrix multiplication, line transformations, and reflections, suitable for students in Grades 10-12.

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