Question

line ℓ has the equation ( y = 3x - 2 ). graph the image of ( ell ) after a dilation with a scale factor of 3, centered at the origin.

Explanation:

Step1: Recall dilation of a line

A dilation centered at the origin with scale factor \( k \) transforms a point \( (x,y) \) to \( (kx,ky) \). For a line \( y = mx + b \), if we take two points on the line, dilate them, and find the new equation.
First, find two points on \( y = 3x - 2 \). Let \( x = 0 \), then \( y=-2 \), so point \( A=(0, - 2) \). Let \( x = 1 \), then \( y=3(1)-2 = 1 \), so point \( B=(1,1) \).

Step2: Dilate the points

Dilate point \( A=(0,-2) \) with scale factor 3: \( (0\times3,-2\times3)=(0, - 6) \).
Dilate point \( B=(1,1) \) with scale factor 3: \( (1\times3,1\times3)=(3,3) \).

Step3: Find the equation of the new line

Now we have two points \( (0,-6) \) and \( (3,3) \) on the dilated line. The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{3-(-6)}{3 - 0}=\frac{9}{3}=3 \).
Using the slope - intercept form \( y=mx + b \), with \( x = 0,y=-6 \), we get \( b=-6 \). So the equation of the dilated line is \( y = 3x-6 \).

To graph it, plot the \( y \)-intercept at \( (0,-6) \) and use the slope (rise 3, run 1) to find another point (e.g., from \( (0,-6) \), moving 1 unit right and 3 units up gives \( (1,-3) \), but we already have \( (3,3) \) from dilation). Draw a line through these points.

Answer:

The equation of the image of line \( \ell \) after dilation is \( y = 3x-6 \), and its graph passes through \( (0,-6) \) and \( (3,3) \) (or other points on \( y = 3x - 6 \)).