Question 1082934
The general point slope form of a linear equation is {{{y-y[1]=m(x-x[1])}}} 


where *[Tex \large m] is the slope and *[Tex \large (x_1,y_1)] is the point the line goes through. 


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


*[Tex \large (x_1,y_1)=(3,4)]


So,


*[Tex \large x_1=3] and *[Tex \large y_1=4]


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


{{{y-y[1]=m(x-x[1])}}} 


{{{y-4=m(x-3)}}} 


Let's solve for y


{{{y-4=m(x-3)}}} 


{{{y-4+4=m(x-3)+4}}} 


{{{y=m(x-3)+4}}} 


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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(x-3)+4}}} 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(x-3)+4}}}


{{{y=m(0-3)+4}}}


{{{y=m(-3)+4}}}


{{{y=-3m+4}}}


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 *[Tex \large(0, -3m+4)] where m is is some negative number.


So the height is {{{h = -3m+4}}}


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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(x-3)+4}}}


{{{0=m(x-3)+4}}}


{{{0-4=m(x-3)+4-4}}}


{{{-4=m(x-3)}}}


{{{-4=mx-3m}}}


{{{-4+3m=mx}}}


{{{mx = -4+3m}}}


{{{(mx)/m = (-4+3m)/m}}}


{{{x = (-4+3m)/m}}}


The x intercept is *[Tex \large \left(\frac{-4+3m}{m}, 0\right)] where m is is some negative number.


The base is {{{b=(-4+3m)/m}}}


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From the last two sections above, we found the following


base = {{{b=(-4+3m)/m}}}


height = {{{h = -3m+4}}}


The area of the triangle is therefore


{{{A = (b*h)/2}}}


{{{A = (((-4+3m)/m)*(-3m+4))/2}}}


{{{A = ((-4+3m)*(-3m+4))/(2m)}}}


{{{A = ((-4)*(-3m)+(-4)*(4)+(3m)*(-3m)+3m*4)/(2m)}}}


{{{A = (12m-16-9m^2+12m)/(2m)}}}


{{{A = (-9m^2+24m-16)/(2m)}}}


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


{{{f(x) = (-9x^2+24x-16)/(2x)}}}


Graphing said function produces


<img src = "https://i.imgur.com/IXOIJux.png">
Image generated by <a href = "https://www.geogebra.org/home?ggbLang=en">GeoGebra</a> (free graphing software).


Now recall that {{{m < 0}}}. Because we replaced m with x, this means that {{{x < 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 = -4/3}}} is the slope which produces the smallest area 24.


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Plug in {{{m = -4/3}}} and simplify


{{{y=m(x-3)+4}}}


{{{y=(-4/3)(x-3)+4}}}


{{{y=(-4/3)(x)+(-4/3)*(-3)+4}}}


{{{y=(-4/3)x+4+4}}}


{{{y=(-4/3)x+8}}}


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The final answer is {{{y=(-4/3)x+8}}}