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

1. ## 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???

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

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.

3. ## 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??

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

Originally Posted by sluggerbroth
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?