How do you write the equation of a line in slope intercept, point slope and standard form given P: (3, -4) and Q: (2,1)?

Aug 5, 2017

Slope intercept form: $y = - 5 x + 11$
Point slope form: $\left(y + 4\right) = - 5 \left(x - 3\right)$
Standard form: $- 5 x + y = 11$

Explanation:

First step is to find the slope (gradient), $m$:

$m = \frac{r i s e}{r u n} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{1 - \left(- 4\right)}{2 - 3} = \frac{5}{- 1} = - 5$

To write the line in 'point slope' form we can choose either point and substitute it into this form:

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

We'll choose the first point, P:

$\left(y - \left(- 4\right)\right) = - 5 \left(x - 3\right)$

Which simplifies to:

$\left(y + 4\right) = - 5 \left(x - 3\right)$

To write it in 'slope intercept' or 'standard' form, we need to find the y-intercept.

To do so we use one of the points in the form for slope intercept:

$y = m x + c$

Let's substitute in point Q to keep it interesting:

$1 = - 5 \left(2\right) + c$

Rearranging, $c = 11$.

We can immediately write the equation of the line in standard form:

$y = - 5 x + 11$

Standard form is:

$a x + b y = c$

We get to it by moving the $x$ term in slope intercept form to the left of the equals sign:

$- 5 x + y = 11$

And we're done!