For what value of k will the graph of 2x + ky = 6 be perpendicular to the graph of 6x - 4y = 12?

The problem wants to calculate the possible value of K that the first equation should be perpendicular to the second equation. First you must transform the both equation in to y slope intercept form or y = mx+b. By means of that you can identify its slope. The slope of the second equation is 6/4 so the first equation slope must be equal to the reciprocal of its slope and should be 3/2. So the value of K = 3.

Perpendicular is when the slopes multiply to negative 1

remember

the slope of the line in form

ax+by=c is - a/b

find slopes

2x+ky=6

slope is - 2/k

6x-4y=12

slope is - 6/-4=3/2

so

-2/k times 3/2=-1

solve for k

-6 / (2k) = - 1

times both sides by 2k

-6=-2k

divide by - 1

3=k

k has to be 3