Question

(3−k)x−4(3−k)y=−27 In the given linear equation, k is a constant and k≠3. When graphed in the xy-plane, what is the slope of the line?

Answer 1

Answer: $\frac{1}{4}$

Step-by-step explanation:

Given equation $(3-k)x-4(3-k)y=-27$ can be written as

$(3-k)x-4(3-k)y+27=0$

∵ Slope of the line of the form $ax+by+c=0$ is $-a/b$

∴ Slope of the given line is $-(\frac{(3-k)}{-4(3-k)} )$ $=-(\frac{1}{-4} ) =1/4 =0.25$