JK passes through the points J(–2,4) and K(1,10). J'K' is the image of JK after a dilation with a scale factor of 34 centered at (–4,0).What is the equation of J'K'?Write your answer in slope-intercept form.

To find the equation of the line J'K', we first need to find the coordinates of points J' and K' after the dilation.

The formula for dilation is:

X' = a * (X - h) + h
Y' = a * (Y - k) + k

where (X,Y) are the original coordinates, (X',Y') are the new coordinates after dilation, (h,k) is the center of dilation, and a is the scale factor.

Given that the center of dilation is (-4,0) and the scale factor is 3/4, we can substitute these values into the formula to find the new coordinates.

For point J(-2,4):

X' = 3/4 * (-2 - (-4)) - 4 = -3
Y' = 3/4 * (4 - 0) = 3

So, J' is (-3,3).

For point K(1,10):

X' = 3/4 * (1 - (-4)) - 4 = -1
Y' = 3/4 * (10 - 0) = 7.5

So, K' is (-1,7.5).

Now that we have the coordinates of J' and K', we can find the slope of the line J'K'. The formula for slope is (Y2 - Y1) / (X2 - X1).

Slope = (7.5 - 3) / (-1 - (-3)) = 4.5 / 2 = 2.25

The y-intercept (b) is found by substituting one of the points and the slope into the equation y = mx + b. Let's use point J'(-3,3):

3 = 2.25 * -3 + b
b = 3 + 6.75 = 9.75

So, the equation of the line J'K' in slope-intercept form is y = 2.25x + 9.75.