# What is the equation in point-slope form and slope intercept form for the line given (9, 1) and (4, 16)?

May 20, 2015

The point-slope form is $y - 1 = - 3 \left(x - 9\right)$, and the slope-intercept form is $y = - 3 x + 28$.

Determine the slope, $m$, using the two points.

Point 1: $\left(9 , 1\right)$
Point 2: $\left(4 , 16\right)$

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{16 - 1}{4 - 9} = \frac{15}{- 5} = - 3$

Point-slope form.

General equation: $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where ${x}_{1}$ and ${y}_{1}$ are one point on the line. I will use Point 1: $\left(9 , 1\right)$.

$y - 1 = - 3 \left(x - 9\right)$

Slope-intercept form.

General equation: $y = m x + b$, where $m$ is the slope and $b$ is the y-intercept.

Solve the point-slope equation for $y$.

$y - 1 = - 3 \left(x - 9\right)$

Distribute the $- 3$.

$y - 1 = - 3 x + 27$

Add $1$ to each side.

$y = - 3 x + 28$