# Half right, missed the other half...

• Jul 19th 2007, 03:57 PM
earachefl
Half right, missed the other half...
"Find all values of $\displaystyle r$ such that the slope of the line through the points $\displaystyle (r, 4)$ and $\displaystyle (1, 3 - 2r)$ is less than 5."

I solved it by first finding the slope $\displaystyle \frac {2r + 1}{r - 1} = 5$ which simplifies to $\displaystyle r = 2$. Then I used the point-slope form to derive the equation $\displaystyle y = 5x + 16$. Then, I got lazy, and plugged in a couple of test numbers into the equation, found that numbers above 2 seemed to qualify, and decided that the answer was the set $\displaystyle (2, \infty)$. Unfortunately, the answer should have included also the set $\displaystyle (-\infty, 1)$.

Algebraically, what should I have done to get to that answer? What part did I skip?
• Jul 19th 2007, 04:43 PM
Soroban
Hello, earachefl!

Quote:

Find all values of $\displaystyle r$ such that the slope of the line
through the points $\displaystyle A(r,\,4)$ and $\displaystyle B(1,\,3\!-\!2r)$ is less than 5.

You should have solved the inequality . . . tricky, but safe.

You found the slope of $\displaystyle AB$ . . . good!

. . Then we have: .$\displaystyle \frac{2r + 1}{r - 1} \:<\:5$ . [1]

I'd like to multiply both sides by $\displaystyle (r-1)$
. . but the result depends on whether $\displaystyle (r-1)$ is positive or negative.

Recall: when multiplying or dividing an inequality by a negative quantity,
. . . . . .reverse the inequality.

Suppose $\displaystyle (r-1)$ is positive . . . that is: .$\displaystyle r > 1$

Multiply [1] by by positive $\displaystyle (r-1)\!:\;\;2r + 1 \:<\:5(r - 1)$

. . $\displaystyle 2r + 1 \:<\:5r-5\quad\Rightarrow\quad -3r \:<\:-6$

Divide both sides by -3: .$\displaystyle r \:>\:2$

We have two inequalities to satisfy: .$\displaystyle r > 1$ and $\displaystyle r > 2$

We take the "stronger" of the two: .$\displaystyle \boxed{r > 2}$

Suppose $\displaystyle (r-1)$ is negative . . . that is: .$\displaystyle r < 1$

Multiply [1] by negative $\displaystyle (r-1)\!:\;\;2r + 1 \;\;{\color{red}>} \;\;5(r-1)$

. . $\displaystyle 2r + 1 \:>\:5r - r\quad\Rightarrow\quad -3r \:>\:-6$

Divide both sides by -3: .$\displaystyle r \:<\:2$

We have two inequalities to satisfy: .$\displaystyle r < 1$ and $\displaystyle r < 2$

We take the "stronger" of the two: .$\displaystyle \boxed{r < 1}$

Therefore: .$\displaystyle (r < 1) \vee (r > 2)$ .. . . .or: .$\displaystyle (-\infty,\,1) \,\cup \,(2,\,\infty)$

• Jul 19th 2007, 04:56 PM
earachefl
Quote:

Originally Posted by Soroban
Hello, earachefl!

We have two inequalities to satisfy: .$\displaystyle r < 1$ and $\displaystyle r < 2$

We take the "stronger" of the two: .$\displaystyle \boxed{r < 1}$

Interesting approach. I don't think my book ever put it in quite these terms.

Thanks!