SOLUTION: Find the equation of the line passing through the point (3,4) which cuts from the first quadrant of a triangle of minimum area.

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Question 1082934: Find the equation of the line passing through the point (3,4) which cuts from the first quadrant of a triangle of minimum area.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The general point slope form of a linear equation is y-y%5B1%5D=m%28x-x%5B1%5D%29

where is the slope and is the point the line goes through.

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We don't know the slope m, but we do know that



So,

and

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Plugging these x and y values into the point slope formula yields

y-y%5B1%5D=m%28x-x%5B1%5D%29

y-4=m%28x-3%29

Let's solve for y

y-4=m%28x-3%29

y-4%2B4=m%28x-3%29%2B4

y=m%28x-3%29%2B4

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In order for this line to form a triangle in quadrant 1, the slope m must be negative. So m < 0. If m > 0 then we end up with an unbounded figure of infinite area.

If m < 0, then y=m%28x-3%29%2B4 forms a triangle with base b and height h. The area of the triangle is A = (b*h)/2. We need to find the base and height.

To find the height h, plug in x = 0 to find the y intercept

y=m%28x-3%29%2B4

y=m%280-3%29%2B4

y=m%28-3%29%2B4

y=-3m%2B4

We stop here because we don't know m yet. If we did, we can replace m with the value and simplify.

So the y intercept is the point where m is is some negative number.

So the height is h+=+-3m%2B4

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To find the base b, we plug in y = 0 and solve for x. This yields the x intercept. The horizontal distance from the origin to the x intercept is equal to the base b.

y=m%28x-3%29%2B4

0=m%28x-3%29%2B4

0-4=m%28x-3%29%2B4-4

-4=m%28x-3%29

-4=mx-3m

-4%2B3m=mx

mx+=+-4%2B3m

%28mx%29%2Fm+=+%28-4%2B3m%29%2Fm

x+=+%28-4%2B3m%29%2Fm

The x intercept is where m is is some negative number.

The base is b=%28-4%2B3m%29%2Fm

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From the last two sections above, we found the following

base = b=%28-4%2B3m%29%2Fm

height = h+=+-3m%2B4

The area of the triangle is therefore

A+=+%28b%2Ah%29%2F2

A+=+%28%28%28-4%2B3m%29%2Fm%29%2A%28-3m%2B4%29%29%2F2

A+=+%28%28-4%2B3m%29%2A%28-3m%2B4%29%29%2F%282m%29



A+=+%2812m-16-9m%5E2%2B12m%29%2F%282m%29

A+=+%28-9m%5E2%2B24m-16%29%2F%282m%29

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Replace the A with f(x). Replace the m with x. We end up with this function.

f%28x%29+=+%28-9x%5E2%2B24x-16%29%2F%282x%29

Graphing said function produces


Image generated by GeoGebra (free graphing software).

Now recall that m+%3C+0. Because we replaced m with x, this means that x+%3C+0. Only focus on the portion that is to the left of the vertical y axis.

Use your calculator's "minimum" feature to find that the min point on the interval (-infinity, 0) is the point (-4/3, 24).
Note: -4/3 = -1.33 approximately.
The point P = (-1.33, 24) is marked on the graph above as that minimum point.

So m+=+-4%2F3 is the slope which produces the smallest area 24.

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Plug in m+=+-4%2F3 and simplify

y=m%28x-3%29%2B4

y=%28-4%2F3%29%28x-3%29%2B4

y=%28-4%2F3%29%28x%29%2B%28-4%2F3%29%2A%28-3%29%2B4

y=%28-4%2F3%29x%2B4%2B4

y=%28-4%2F3%29x%2B8

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The final answer is y=%28-4%2F3%29x%2B8