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Find the equation of the line passing through the point (3,4) which cuts the first quarter a triangle with minimum area

I do not know where to start, I just know I need to find the equation of the line to start with something, but I do not have the slope of the line.
Any Suggestions?
Thanks.

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maybe this would help you cut-the-knot.org/triangle/PointInAngleMinArea.shtml#solution –  dato datuashvili Apr 22 '13 at 7:20
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2 Answers

Hint: Let the slope of the line be $m$, then the point-slope equation for a line gives $$y - 4 = m(x - 3)$$ Now find the $x$-intercept (plug in $y = 0$ and solve for $x$, your answer will be a formula involving $m$), find the $y$-intercept (plug in $x = 0$), and use these two points to calculate the area of the triangle (area $= \frac{1}{2}$base$\cdot$height). This is again a formula involving $m$. You should now be able to find what value of $m$ minimizes this formula.

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in this case base and height are $y$ and $x$ right? –  dato datuashvili Apr 22 '13 at 7:22
    
They are the $x$ and $y$-intercepts. –  Jim Apr 22 '13 at 7:25
    
cant solve it.. –  Ofir Attia Apr 30 '13 at 17:08
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The equation of line with $x$ intercept $a$ and $y$ intercept $b$ is given by $$\dfrac{x}a + \dfrac{y}b = 1$$ The area of the triangle enclosed in the first quadrant is now given by $\Delta = \dfrac{ab}2$.

We are also given that the line passes through the points $(3,4)$. We hence have $$\dfrac3a + \dfrac4b = 1$$ Hence, we want to minimize $ab/2$ subject to the constraint $\dfrac3a + \dfrac4b = 1$. There are many ways to proceed. One easy way is to use AM-GM inequality. Move your mouse over the gray area for the complete solution.

By AM-GM, we have $$\underbrace{\dfrac{\dfrac3a + \dfrac4b}2}_{1/2} \geq \sqrt{\dfrac3a \cdot \dfrac4b} = \dfrac{2 \sqrt3}{\sqrt{ab}} \implies \sqrt{ab} \geq 4 \sqrt3 \implies \Delta \geq 24$$Hence, the minimum area is $24$ and the equation of the line is $$\dfrac{x}6 + \dfrac{y}8 =1$$

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