You want to find the equation of the image of 𝓁 after a dilation with a scale factor of 1/4 , centred at the origin. Call this image 𝓁’

To find the equation of 𝓁’ , you need two points that lie on  𝓁’ .Begin by finding two points that lie on 𝓁 .

Start with the y-intercept. Since the equation of 𝓁 in slope-intercept form is y=-2x-4 the y-intercept is (0 , -4).

Next, since the slope of 𝓁 is -2 , which can be written as -2/1 , move down 2 and right 1 from (0 , -4) to find a second point on 𝓁 , (1 , -6).

So, the points  (0 , -4) and (1 , -6) lie on 𝓁  To find two points on 𝓁‘, apply the dilation

(X , Y) ↦ (1/4 X , 1/4 Y )

(0 , -4) ↦ (0 , -4/4 ) = (0 ,-1)

(1 , -6) ↦ (1/4 , -6/4 ) = (1/4 , -3/2)

The image of the y -intercept of 𝓁 is (0 ,-1) , which is the  y-intercept of 𝓁’ . In general, the y-intercept of  a line’s image after a dilation centred at the origin is the image of the y-intercept  of the original line. This is because the x-coordinate of the y-intercept is 0, so multiplying by the scale factor of the dilation does not change its value.

Next, use the slope formula to find the slope of 𝓁‘.

Slope of 𝓁‘ = Y₂ -Y₁ / X₂ -X₁ Slope formula

= -3/2- -1 / 1/4-0 Plug in Y₂ = -3/2 , Y₁ = -1 ,X₂ = 1/4 and  X₁ =0

= -1/2 / 1/4 Subtract

= -1/2 . 4/1 To divide, multiply by the reciprocal

= -4/2 Multiply

= -2 Simplify

So, the slope of 𝓁‘  is -2 which is the same as the slope of 𝓁 . Since 𝓁‘ and 𝓁 have the same slope but different y-intercepts , they are parallel. In general, if a line does not pass through the centre of the dilation, then it is parallel to its image.

Finally, since 𝓁‘ has a slope of -2 and a y-intercept of – 1, the equation of 𝓁‘ in slope-intercept form is Y = -2X-1.

You want to find the equation of the image of 𝓁 after a dilation with a scale factor of 2 , centred at the origin. Call this image 𝓁’

To find the equation of 𝓁’ , you need two points that lie on  𝓁’ .Begin by finding two points that lie on 𝓁 .

Start with the y-intercept. Since the equation of 𝓁 in slope-intercept form is y=1/3x+3 the y-intercept is (0 , 3).

Next, since the slope of 𝓁 is 1/3 , move up  1 and right 3 from (0 , 3) to find a second point on 𝓁 , (3 , 4).

So, the points  (0 , 3) and (3 , 4) lie on 𝓁  To find two points on 𝓁‘, apply the dilation

(X , Y) ↦ (2 X , 2 Y )

(0 , 3) ↦ (0 , 6 )

(3 , 4) ↦ (6 , 8 )

The image of the y -intercept of 𝓁 is (0 ,6) , which is the  y-intercept of 𝓁’ . In general, the y-intercept of  a line’s image after a dilation centred at the origin is the image of the y-intercept  of the original line. This is because the x-coordinate of the y-intercept is 0, so multiplying by the scale factor of the dilation does not change its value.

Next, use the slope formula to find the slope of 𝓁‘.

Slope of 𝓁‘ = Y₂ -Y₁ / X₂ -X₁ Slope formula

= 8-6 / 6-0 Plug in Y₂ =8 , Y₁ = 6 ,X₂ = 6 and  X₁ =0

=2/6 Subtract

= -1/2 . 4/1 To divide, multiply by the reciprocal

= 1/3 Simplify

So, the slope of 𝓁‘  is 1/3 which is the same as the slope of 𝓁 . Since 𝓁‘ and 𝓁 have the same slope but different y-intercepts , they are parallel. In general, if a line does not pass through the centre of the dilation, then it is parallel to its image.

Finally, since 𝓁‘ has a slope of 1/3 and a y-intercept of 6, the equation of 𝓁‘ in slope-intercept form is Y = 1/3 X +6.