Question

A line goes through the points (2, b) and (c, 10). Find the y-coordinate of the point on the line that has an x-coordinate of 3.

Answer 1

The answer is in terms of c and b.

Answer 2

You would find the slope between the two points (we don't know b or c, so we'd just leave them as is), and then insert one of the points along with the slope into point-slope form of an equation.

\[ \begin{split}
y - y_1 = m(x - x_1)\\
m = \frac{y_2 - y_1}{x_2 - x_1}\\
(x_1,y_1), (x_2, y_2)\\
\end{split} \]

Answer 3

* that would define the line itself..
then we'd insert the x=3 to that equation and solve for y

Answer 4

Will you show me?

Answer 5

sure -- you need me to go through the whole problem step by step or a specific part?

Answer 6

This problem would play out much like any other "find the equation of a line between these points" and "find y at x=..." problem, except we're using two letters to represent numbers.
\[ \large \begin{split}
m &= \frac{y_2 - y_1}{x_2 - x_1}\\
&= \frac{10 - b}{c - 2}\\
\end{split} \]

Answer 7

\[ \large \begin{split}
y - y_1 &= m(x - x_1)\\
y - b &= (\frac{10 - b}{c - 2})(x - 2)\\
y &= (\frac{10 - b}{c - 2})(x - 2) + b\\
\end{split} \]

Does that all make sense?

Answer 8

yes, I got there but wasn't sure what to do next

Answer 9

Well, we want to find y when x=3. So, we can just substitute that into our equation and solve (I'd also get one fraction by using common denominators)

\[ \large \begin{split}
y &= (\frac{10 - b}{c - 2})(3 - 2) + b\\
&= (\frac{10 - b}{c - 2}) + b\\
&= (\frac{10 - b}{c - 2}) + b(\frac{c - 2}{c - 2})\\
&= (\frac{10 - b}{c - 2}) + \frac{bc - 2b}{c - 2}\\
&= (\frac{(10 - b) + (bc - 2b)}{c - 2}\\
&= \frac{10 - 3b + bc}{c - 2}\\
y &= \frac{bc - 3b + 10)}{c - 2}\\
\end{split} \]

Answer 10

Oh! That's what I didn't do, I didn't multiply b by c-2/c-2. Thanks acc and sorry for being away for awhile.

Answer 11

it's alright. as long as you got it now. :)