Slope-intercept form help?

**wrongnmbr** · August 15th 2010, 11:53 AM

Okay, I came up with a wild answer for this problem:

Find the slope-intercept form of the line passing through the points (5,-2) and (-3,5)

I came up with this fraction-ridden answer that I know is wrong...

y = -7/8x - 1 and -43/8

I'm trying to figure out what I did wrong and how to properly solve this. Thanks!

**masoug** · August 15th 2010, 12:22 PM

How did you do it?

What I did:
If the relation is linear,
$m=\frac{(y_{1}-y_{2})}{(x_{1}-x_{2})}$
so
$y=-\frac{7}{8}x + b$
and to find $b$ ,
plug in one of the coordinates,
$-2 = -\frac{7}{8}(5)+b$

$-2 = -\frac{35}{8}+b$

$-2 = -4\frac{3}{8}+b$

$2.375 = b$
therefore,
$y=-\frac{7}{8}x+2\frac{3}{8}$

I might be wrong, but this is a method for finding the equation of the line.

-Masoug

**wrongnmbr** · August 15th 2010, 03:56 PM

I used this method: y-y1 = m(x-x1), which I believe is why I'm messing up...

Now, I was following you up until 2.375 = b. How exactly did you come up with this? Not to mention your final solution?

**pickslides** · August 15th 2010, 04:03 PM

One of the original values (5,-2) were substituted to find the y-intercept.

**HallsofIvy** · August 16th 2010, 02:46 AM

Which did you use as $(x_1, y_1)$ and $(x_2, y_2)$ ? Either way works but you have to be consistent.

With $(x_1, y_1)= (5, -2)$ and $(x_2, y_2)= (-3, 5)$ , $m= \frac{y_2- y_1}{x_2- x_1}= \frac{5-(-2)}{-3- 5}$ $= \frac{7}{-8}= -\frac{7}{8}$ .

Taking $(x_1, y_1)= (-3, 5)$ and [tex](x_2, y_2)= (5, -2) just changes the sign in both numerator and denominator and so gives the same slope: $m= \frac{y_2- y_1}{x_2- x_1}= \frac{-2- 5}{5-(-3)}= \frac{-7}{i}= -\frac{7}{8}$ .

Taking $(x_1, y_1)= (5, -2)$ , $y- y_1= m(x- x_1)$ becomes $y-(-2)= -\frac{7}{8}(x- 5)$ so $y+ 2= -\frac{7}{8}x+ \frac{35}{8}$ . $y= -\frac{7}{8}x+ \frac{35}{8}- 2= -\frac{7}{8}x+ \frac{19}{8}$ .

Taking $(x_1, y_1)= (-3, 5)$ , $y- y_1= m(x- x_1)$ becomes $y- 5= -\frac{7}{8}(x- (-3))$ so $y- 5= -\frac{7}{8}x- \frac{21}{8}$ . $y= -\frac{7}{8}x- \frac{21}{8}+ 5= -\frac{7}{8}x+ \frac{19}{8}$ , exactly the same answer.

Of course, whichever way you choose, you can simplify by multiplying the whole equation, either here or, better, in the orginal form, by 8.

Multiplying both sides of $y- 5= -\frac{7}{8}(x- (-3))$ by 8 gives $8(y- 5)= -7(x+ 3)$ or 8y- 40= -7x- 21 so 7x+ 8y= -21+ 40= 19.

Multiplying both sides of $y-(-2)= -\frac{7}{8}(x- 5)$ by 8 gives $8(y+ 2)= -7(x- 5)$ or 8y+ 16= -7x+ 35 so 7x+ 8y= 40- 16= 19 again.

**HallsofIvy** · August 16th 2010, 02:53 AM

Originally Posted by wrongnmbr

I used this method: y-y1 = m(x-x1), which I believe is why I'm messing up...

Now, I was following you up until 2.375 = b. How exactly did you come up with this?

Masoug said
$-2= -\frac{35}{8}+ b$
and then said "b= 2.375"

He got that, of course, by adding $\frac{35}{8}$ to both sides of the equation:
$b= -2+ \frac{35}{8}= -\frac{16}{8}+ \frac{35}{8}= \frac{-16+ 35}{8}= \frac{19}{8}$
He then divided: 8 goes into 19 twice with remainder 3 and 3/8= 0.375: $\frac{19}{8}= 2.375$

Personally, I would have left it as an improper fraction, or, as I have already said, multiply the entire equation by 8 to get rid of the fractions.

Not to mention your final solution?

$y= -\frac{7}{8}x+ 2\frac{3}{8}= -\frac{7}{8}x+ \frac{19}{8}$ and, multiplying through by 8,

8y= -7x+ 19 or 7x+ 8y= 19.

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