# find possible slopes of line passing thru (4,3) so portion of line in first quadrant

#### sluggerbroth

...forms triangle of area 27 wit positive coordinate axes.

book has answer -3/2 or -3/8

i found first answeer but not second???

#### skeeter

MHF Helper
Re: find possible slopes of line passing thru (4,3) so portion of line in first quadr

...forms triangle of area 27 wit positive coordinate axes.

book has answer -3/2 or -3/8

i found first answeer but not second???
let $$\displaystyle x_i$$ and $$\displaystyle y_i$$ represent the positive x and y intercepts of the line

$$\displaystyle \frac{x_i \cdot y_i}{2} = 27 \implies x_i \cdot y_i = 54$$

slope of the line is $$\displaystyle m = -\frac{y_i}{x_i}$$

$$\displaystyle y = mx + b$$

$$\displaystyle 3 = -\frac{y_i}{x_i} \cdot 4 + y_i$$

$$\displaystyle 3 = -y_i\left(\frac{4}{x_i} -1\right)$$

$$\displaystyle 3 = -\frac{54}{x_i} \left(\frac{4}{x_i} -1\right)$$

$$\displaystyle 3 = -\frac{216}{x_i^2} + \frac{54}{x_i}$$

$$\displaystyle 3x_i^2 = -216 + 54x_i$$

$$\displaystyle 3x_i^2 - 54x_i + 216 = 0$$

$$\displaystyle x_i^2 - 18x_i + 72 = 0$$

$$\displaystyle (x_i - 6)(x_i - 12) = 0$$

$$\displaystyle x_i = 6 \implies y_i = 9$$ ... $$\displaystyle m = -\frac{9}{6} = -\frac{3}{2}$$

$$\displaystyle x_i = 12 \implies y_i = 4.5$$ ... $$\displaystyle m = -\frac{4.5}{12} = -\frac{9}{24} = -\frac{3}{8}$$

also, in future please post entire problem within the post ... not part in the title, rest in the post. thanks.

#### sluggerbroth

Re: find possible slopes of line passing thru (4,3) so portion of line in first quadr

how do know negative slope? ie m=-y/x??

#### skeeter

MHF Helper
Re: find possible slopes of line passing thru (4,3) so portion of line in first quadr

how do know negative slope? ie m=-y/x??
...forms triangle of area 27 wit positive coordinate axes
I always make a sketch before working on a problem ... I'm sure that you do, also, right?

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