Algebra question, help please? Serious answer too!?

Question:Find the equation that has an x intercept of 4 and is perpedicular to the line x-2y=4.

I know the answer is 2x+y=8 but I don't know how to get it. Please give me a thorough explanation and some other problems similiar to this question.

Answers:
To get this line, you must first figure out the slope. When a line is parallel to another line, their slopes are the same; however, when they are perpendicular, their slopes are negative reciprocals (flipped fractions) of each other. For example, a line perpendicular to a line with a slope of -3/2 would have a slope of +2/3, and a line perpendicular to a line with a slope of 4 would have a slope of -1/4. The slope of the given equaiton can be found this way:

x-2y=4
x=2y+4
2y=x-4
y=1/2x-2

Since y=mx+b with m being slope and b being y-intercept, then 1/2 is the slope of the given line. Therefore, the slope of the perpendicular line must be the negative reciprocal, which is -2. So far, you have the slope--now you just need the y-intercept. This can be found by substituting the point on the line that you know--the x-intercept--into the equation. An x-intercept of 4 would be found at (4,0) and by plugging this into y=mx+b we can figure out the y-intercept:

y=mx+b
(0)=(-2)(4)+b
0=-8+b
b=8

According to this, the y-intercept is 8. By putting this into the y=mx+b equation, we get the equation:

y=-2x+8

Now, to convert it to the form ax+by=c, we just have to do a bit more math:

y=-2x+8
2x+y=8

And voila, there is your answer. I hope it's not too confusing.

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