I am glad you fixed up this question, since the way you had it a while back made no sense.
The new equation must differ only in the constant, since the slope is the same.
so the new equation must be
5x - 3y = c, but (1,-2) is supposed to be on it, so
5 + 6 = c = 11
new equation: 5x - 3y = 11
if (3,k) is on it also, ...
15 - 3k = 11
-3k = -4
k = 4/3
find k so that the line through (3,k) and (1,-2) is parallel to 5x-3y=2. Find k so that the line is perpendicular to 3x+2y=6
2 answers
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