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I have a question that asks:

Give coordinates for a point that is $8$ units away from the line $y = 5$. Then find both points on the line $3x + 2y = 4$ that are $8$ units from the line $y = 5$.

So I don't really know how to go about doing this but I am having some thoughts regarding the fact that if the equation of the line is $y = 5$, then the $y$-coordinates for all of the points that the line will go through will be $5$.

And let's say if we have the coordinates of two points be $(x, y)$, then we can perhaps use the distance formula although I am not sure what I would use for the $x$-coordinate for the point that goes through line $y =$ $5$.

Can someone please help me with solving this problem? And please note that I am only in high school and haven't learned about these geometrical concepts so in-depth.

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    I may have the answer, although I feel like my approach was very simple and since the Exeter-math questions are known for their out-of-the-box thinking, but I found one point to be _(0, -3)_. This is just in relative to the point (0, 5), which is one point in the line _y = 5_. – 2017-01-25

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All the points which are $8$ units away from $y=5$ lie on parallels to that line with distance $8$, i.e. either on $y=5-8=-3$ or on $y=5+8=13$. Now intersect these two lines with the other line you have. I'll leave that to you as an exercise.