line n is represented by 3x+4y=20. Determine and state the equation of p, the image of line n, after a dilation of scale factor 1/3 centered at point 4,2

Explain your answer

since (4,2) is on the line, nothing changes.

All points in the plane will be 1/3 as far away from (4,2) as they started, but those on the line will still be on that line.

so line p, the image would just be on top of line n??

Well, if line n is represented by the equation 3x+4y=20, then we can rearrange it to the slope-intercept form y = mx + b, where m is the slope and b is the y-intercept.

To find the equation of p, the image of line n after the dilation, we need to apply the dilation equation:

(x', y') = (k * (x - h) + h, k * (y - v) + v),

where (x', y') is the image point, (x, y) is the original point, (h, v) is the center of dilation, and k is the scale factor.

In this case, the center of dilation is (4, 2) and the scale factor is 1/3. Plugging the values into the dilation equation, we get:

(x', y') = (1/3 * (x - 4) + 4, 1/3 * (y - 2) + 2).

Now, let's substitute x' and y' into the given equation of line n:

3x' + 4y' = 20.

Substituting the expressions for x' and y' from above, we have:

3((1/3 * (x - 4) + 4) + 4((1/3 * (y - 2) + 2) = 20.

Now, let's simplify this equation:

(x - 4) + 12(y - 2) + 24 = 60.

x + 12y - 36 = 60.

Finally, rearranging the equation, we get:

x + 12y = 96.

So, the equation of p, the image of line n after the dilation, is x + 12y = 96.

Now, isn't it funny how math can make us do all these calculations just to find an equation? I guess it's a bit like trying to solve a riddle!

To determine the equation of line p, we need to apply the dilation of scale factor 1/3 centered at point (4,2) to line n.

First, let's determine the new coordinates of the center point (4,2) after the dilation. Since the scale factor is 1/3, each coordinate should be multiplied by 1/3.

New x-coordinate: 4 * 1/3 = 4/3
New y-coordinate: 2 * 1/3 = 2/3

So, the new center point for the dilation is (4/3, 2/3).

Next, let's determine the new coordinates of a point on the original line n after the dilation. We can choose any point on line n, but for simplicity, let's use the x-intercept of line n where y = 0.

When y = 0, the equation of line n becomes:
3x + 4(0) = 20
3x = 20
x = 20/3

So, a point on line n is (20/3, 0).

Applying the dilation to this point:

New x-coordinate: (20/3 - 4/3) * 1/3 = 16/9
New y-coordinate: (0 - 2/3) * 1/3 = -2/9

So, the new point after the dilation is (16/9, -2/9).

Finally, we can use the new center point and the new point on line p to determine its equation. Using the point-slope form of a linear equation, the equation of line p is:

y - (2/3) = [(2/9 - 2/3) / (16/9 - 4/3)] * (x - (4/3))

Simplifying the equation:

y - (2/3) = [-4/9 / (4/9)] * (x - (4/3))
y - (2/3) = -1/3 * (x - (4/3))
y - (2/3) = -1/3 * (x - 4/3)
y - (2/3) = -1/3 * (x - 4/3)

Multiplying both sides by 3 to eliminate the fraction:

3y - 2 = -1(x - 4/3)
3y - 2 = -x + 4/3
3y + x = 4/3 + 2
3y + x = 10/3

Therefore, the equation of line p, after the dilation of scale factor 1/3 centered at point (4,2), is 3y + x = 10/3.

To find the equation of line p, the image of line n after a dilation with a scale factor of 1/3 centered at the point (4,2), we need to follow these steps:

Step 1: Find the image of the point (4,2).
To perform a dilation centered at (4,2) with a scale factor of 1/3, we need to scale the x-coordinate and y-coordinate of the point (4,2) by 1/3.

Scaled x-coordinate = (1/3) * 4 = 4/3
Scaled y-coordinate = (1/3) * 2 = 2/3

So, the image of the point (4,2) after dilation is (4/3, 2/3).

Step 2: Find the slope of line n.
The slope-intercept form of a line is given by y = mx + b, where m is the slope of the line. To find the slope, we can rewrite the equation of line n in slope-intercept form.

3x + 4y = 20
4y = -3x + 20
y = (-3/4)x + 5

From the above equation, we can see that the slope of line n is -3/4.

Step 3: Find the slope-intercept form of line p.
Since the line p is the image of line n after dilation, it will have the same slope as line n.

The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is any point on the line and m is the slope of the line.

Using the point (4/3, 2/3) and the slope -3/4, we can write the equation of line p as:

y - (2/3) = (-3/4)(x - (4/3))
y - (2/3) = (-3/4)(x - 4/3)
y - (2/3) = (-3/4)x + 3
y = (-3/4)x + 3 + (2/3)
y = (-3/4)x + 11/3

Therefore, the equation of line p, the image of line n after a dilation centered at (4,2) with a scale factor of 1/3, is y = (-3/4)x + 11/3.