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

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

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

i found first answeer but not second???

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

Quote:

Originally Posted by

**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???

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.

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??

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

Quote:

Originally Posted by

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

Quote:

...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?