# How do you find the equation of line L that passes through the points (1,3) and (-3,4)?

Feb 6, 2015

The answer is: $y = - \frac{1}{4} x + \frac{13}{4}$.

It is possible to use this formula, that is the formula to find a line given two points $A \left({x}_{A} , {y}_{A}\right)$ and $B \left({x}_{B} , {y}_{B}\right)$:

$\frac{y - {y}_{A}}{{y}_{B} - {y}_{A}} = \frac{x - {x}_{A}}{{x}_{B} - {x}_{A}}$.

So:

$\frac{y - 3}{4 - 3} = \frac{x - 1}{- 3 - 1} \Rightarrow y - 3 = - \frac{1}{4} \left(x - 1\right) \Rightarrow$

$y = - \frac{1}{4} x + \frac{1}{4} + 3 \Rightarrow y = - \frac{1}{4} x + \frac{13}{4}$.

Feb 7, 2015

I created a video that answers the question and shows you how to check you work.
http://www.frontporchmath.com/topic/equation-of-a-line-question-1/

To solve this problem you first find the slope of the line, using the two points $\left(1 , 3\right)$ and $\left(- 3 , 4\right)$.

$\frac{\Delta y}{\Delta x} = \frac{4 - 3}{- 3 - 1} = - \frac{1}{4}$

Then you use the y-intercept equation of a line using the slope above and one of the above sets of points to find the y-intercept.

$y = m x + b$
$4 = - \frac{1}{4} \left(- 3\right) + b$
$4 = \frac{3}{4} + b$
$3 \frac{1}{4} = b$

Then just plug that information into $y = m x + b$

$y = - \frac{1}{4} x + 3 \frac{1}{4}$
or
$y = \frac{1}{4} x + \frac{13}{4}$