I have the answer, but I am stuck in the middle of the problem

Through (4,2) and (1,3)

m= 3 - 2 / 1 - 4 = - 1/3

y - 2 = (-1/3)x (x-4)

y - 2 = this is where I get stuck. I know the answer is y = (-1/3)x + 11/3

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- Sep 8th 2008, 10:33 AMSammyjEquation in slop intercept forum
I have the answer, but I am stuck in the middle of the problem

Through (4,2) and (1,3)

m= 3 - 2 / 1 - 4 = - 1/3

y - 2 = (-1/3)x (x-4)

y - 2 = this is where I get stuck. I know the answer is y = (-1/3)x + 11/3 - Sep 8th 2008, 10:47 AMSimplicity
I get a different answer.

The general form is:

$\displaystyle y - y_1= m(x-x_1)$

Substituting values:

$\displaystyle y-2 = -\frac13 ( x - 4)$

Opening the bracket:

$\displaystyle y-2 = -\frac13x + \frac43$

Adding $\displaystyle 2$ to both sides:

$\displaystyle y = -\frac13 + \frac{10}{3}$ - Sep 8th 2008, 10:51 AMSammyj
Thanks

- Sep 8th 2008, 11:12 AMSammyj
How about through (1/2, 5/3) and (3, 1/16)

(1/16) - (5/3) / (3) - (1/2)

I understand how to do them until they are fractions. I know the answer is y = (2/3)x-2 but don't know how to work it out. - Sep 8th 2008, 11:52 AMSimplicity
$\displaystyle \Delta y = \frac{1}{16} - \frac{5}{3} = -\frac{77}{48}$

$\displaystyle \Delta x = 3 - \frac12 = \frac52$

$\displaystyle m = \frac{\Delta y}{\Delta x} = \frac{-\frac{77}{48}}{\frac52}$

$\displaystyle m = -\frac{77}{48} \div {\frac52} = -\frac{77}{48} \times \frac{2}{5} = -\frac{77}{120}$

Your gradient is $\displaystyle m = -\frac{77}{120}$

Use your normal apprach to find the equation. Substitute appropriate number into the general form, $\displaystyle y - y_1 = m(x-x_1)$.

**EDIT:**Sammyj, is your question posted correctly? The gradient is different to the answer that you posted.