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The line whose equation is [tex]\(6x + 3y = 3\)[/tex] is dilated by a scale factor of 2 centered at the point [tex]\((0,0)\)[/tex]. An equation of its image is:

A. [tex]\(y = -2x + 1\)[/tex]

B. [tex]\(y = -2x + 2\)[/tex]

C. [tex]\(y = -4x + 1\)[/tex]

D. [tex]\(y = -4x + 2\)[/tex]

Sagot :

GinnyAnswer
Sure, let's go through this step by step to find the equation of the image of the given line after dilation.

### Step 1: Rewrite the original line equation in slope-intercept form (y = mx + b)

We start with the original line equation:
[tex]\[ 6x + 3y = 3 \][/tex]

First, we solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ 3y = -6x + 3 \][/tex]
[tex]\[ y = -2x + 1 \][/tex]

So the equation of the line in slope-intercept form is:
[tex]\[ y = -2x + 1 \][/tex]

### Step 2: Apply the dilation centered at (0,0)

Since the dilation is centered at [tex]\((0, 0)\)[/tex] with a scale factor of 2, it changes the slope [tex]\( m \)[/tex] to [tex]\( 2m \)[/tex]. The y-intercept remains unchanged because the dilation is centered at the origin.

Given the original slope [tex]\( m = -2 \)[/tex], the new slope [tex]\( m' \)[/tex] after dilation is:
[tex]\[ m' = 2 \times -2 = -4 \][/tex]

### Step 3: Substitute the new slope back into the line equation

We substitute the new slope [tex]\( m' = -4 \)[/tex] and the original y-intercept [tex]\( b = 1 \)[/tex] into the slope-intercept form of the line equation:
[tex]\[ y = -4x + 1 \][/tex]

### Final Equation

The equation of the image of the line after dilation is:
[tex]\[ y = -4x + 1 \][/tex]

Thus, the correct answer is:

c. [tex]\( y = -4x + 1 \)[/tex]
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