Question 648398: I have tried solving four homework problems like this for hours but I just don't know what to do!
The problem reads: find the equation of the line perpendicular to the line y=3 and passing through the midpoint of the segment joining (2,4), (-6,10).
This is what I have tried: y=3 (2+-6)/2, (4+10)/2. I get (-2,7) and then I tries using the point-slope form: y-7=0(since y is horizontal and the slope is zero, right?) (x--2)
y=x+2-7 but then I don't know what to do. On the back of the textbook it says the answer is x=-2. I don't know how to get that!
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! You're almost right, except for your reasoning on the slope. The midpoint of the line between (-6,10) and (2,4) is as you correctly calculated (-2,7). Your line must go through that point. The other condition is that the line is perpendicular to y=3. And as you stated this has a zero slope (a horizontal line). However, you want a line that is PERPENDICULAR to a horizontal line, and that is a VERTICAL line that has an undefined slope (because it is the negative inverse of zero). The equation of a vertical line is
(1) x = constant
And since your line must go through (-2,7), the value of that constant is the x-coordinate of that point, giving you
(2) x = -2
Answer: The equation of the line perpendicular to y=3 and passes through the point (-2,7) is x = -2
See it?
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