The base of triangle is on the X-axis, one side lies along the line y=3x, and the third side passes through the point (1,1).what is the slope of the third side if the area of the triangle is to be a minimum.

$\dfrac{y}{x} = \dfrac{y - 1}{2/3}$

$\frac{2}{3}y = xy - x$

$x = xy - \frac{2}{3}y$

$x = (x - \frac{2}{3})y$

$x = \left( \dfrac{3x - 2}{3} \right) y$

$y = \dfrac{3x}{3x - 2}$

$A = \frac{1}{2}xy$

$A = \dfrac{x}{2} \left( \dfrac{3x}{3x - 2} \right)$

$A = \dfrac{3x^2}{6x - 4}$

$\dfrac{dA}{dx} = \dfrac{(6x - 4)(6x) - 3x^2(6)}{(6x - 4)^2} = 0$

$6x(6x - 4) - 6(3x^2) = 0$

$(6x^2 - 4x) - 3x^2 = 0$

$3x^2 - 4x = 0$

$x(3x - 4) = 0$

$x = 0 ~ \text{and} ~ 4/3$

Use x = 4/3 $y = \dfrac{3(\frac{4}{3})}{3(\frac{4}{3}) - 2}$

$y = 2$

Required Slope $m = \dfrac{2 - 0}{\frac{2}{3} - \frac{4}{3}}$

$m = -3$ answer

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$\dfrac{y}{x} = \dfrac{y - 1}{2/3}$

$\frac{2}{3}y = xy - x$

$x = xy - \frac{2}{3}y$

$x = (x - \frac{2}{3})y$

$x = \left( \dfrac{3x - 2}{3} \right) y$

$y = \dfrac{3x}{3x - 2}$

$A = \frac{1}{2}xy$

$A = \dfrac{x}{2} \left( \dfrac{3x}{3x - 2} \right)$

$A = \dfrac{3x^2}{6x - 4}$

$\dfrac{dA}{dx} = \dfrac{(6x - 4)(6x) - 3x^2(6)}{(6x - 4)^2} = 0$

$6x(6x - 4) - 6(3x^2) = 0$

$(6x^2 - 4x) - 3x^2 = 0$

$3x^2 - 4x = 0$

$x(3x - 4) = 0$

$x = 0 ~ \text{and} ~ 4/3$

Use x = 4/3

$y = \dfrac{3(\frac{4}{3})}{3(\frac{4}{3}) - 2}$

$y = 2$

Required Slope

$m = \dfrac{2 - 0}{\frac{2}{3} - \frac{4}{3}}$

$m = -3$

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