Find the value of k so that the line through points (-3, 2k) and (k,6) has a slope of 4

Accepted Solution

Answer:-1Step-by-step explanation:To find the slope, I'm going to line up the points and subtract. I will then put 2nd difference over first difference. ( -3  ,   2k)-( k  ,    6)------------------3-k  ,   2k-6So the slope in terms of k is:[tex]\frac{2k-6}{-3-k}[/tex].We are also given the slope is 4 or 4/1.This means we have the following equation to solve for k such that the slope is 4:[tex]\frac{2k-6}{-3-k}=\frac{4}{1}[/tex]Cross multiply:[tex]1(2k-6)=4(-3-k)[/tex]Distribute:[tex]2k-6=-12-4k[/tex]Add 4k on both sides:[tex]6k-6=-12[/tex]Add 6 on both sides:[tex]6k=-6[/tex]Divide both sides by 6:[tex]k=-1[/tex]So k has to have a value of -1 for the slope to be 4.Let's check:(-3,-2)(-1,6)--------Subtracting-2,-8So -8/-2 is 4.The check is good and the value for k as -1 as been verified.