Question 182193: Given the line y=ax with a>0,and the line y= -ax+k with an x intercept of (n,0)
find the value of k in terms of a and n.find the point of intersection of the two lines by setting the equations for the lines equal to each other.
Find the area of the triangle with vertices at (0,0),(n,0)and the point of intersection.
this is for a homework project due on feb 23.My instructor has not taught us exactly how to do this and i am deffinatly one of those people who need step by step instructions and i'm totally lost as to how to do the problems and any help would be appreciated.Thank you
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website! 
In order to draw the picture above, I chose 2 for a and 4 for k, but that is nothing more than a convenience so that the diagram could be rendered and you must not read anything into the problem just by what you see in the picture except the general relationship of the lines and the points under discussion.
We are given that:
has an x-intercept of (n, 0).
That simply means that when x = n, y = 0, so:
Which is to say:
Now we can re-write the second equation thus:
and since this expression is equal to the expression given for y in the first equation, namely ax, we can set these two expressions equal to one another:
Multiply by 
Add x to both sides, then divide by 2:
Giving us the x-coordinate of the point of intersection of the two lines. Substituting this value into equation 1 (you could substitute in to equation 2, but this way the arithmetic is easier), you get:
 = \frac{an}{2})
Giving us the y-coordinate of the point of intersection of the two lines.
Hence, the lines intersect in the point )
Now, if we consider segment AC to be the base of the triangle, we can determine its length by inspection. Since the y-coordinates of points A and C do not vary from each other, the measure of the segment is the difference in the x-coordinates, or, in this case, n.
Now all we need to do is determine the x-coordinate of the x-axis end of the altitude, so that we can calculate its measure. It would be very handy if the altitude segment were to lie in a vertical line. This can only be true if the triangle is isoceles with the altitude bisecting the angle between the two equal-length sides. (That's what the diagram looks like, but never trust your eyes). If we can show that segment AB is equal in measure to segment BC, then we will have our vertical and easily measureable altitude. Using the distance formula:
^2 + (0 - \frac{an}{2})^2} = sqrt{(-\frac{n}{2})^2 + (- \frac{an}{2})^2} = sqrt{\frac{n^2}{4} + \frac{a^2n^2}{4}})
^2 + (0 - \frac{an}{2})^2} = sqrt{(\frac{2n}{2}-\frac{n}{2})^2 + (- \frac{an}{2})^2} = sqrt{\frac{n^2}{4} + \frac{a^2n^2}{4}})
So, now we know that AB = BC, so the triangle is isoceles and the altitude BD bisects the base. That means that the x-coordinate of point D is  .
Since the x-coordinate of point D is equal to the x-coordinate of point B, the altitude lies on a vertical line and the measure of the altitude is simply the difference of the y-coordinates of point D and B, or:
Area calculation
\left(\frac{an}{2}\right)=\frac{an^2}{8}
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