# How do you write an equation of a line given point (0,1) and (5,3)?

May 10, 2018

$y = \frac{2}{5} x + 1$

#### Explanation:

$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.

•color(white)(x)y=mx+b

$\text{where m is the slope and b the y-intercept}$

$\text{to calculate the slope use the "color(blue)"gradient formula}$

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)

$\text{let "(x_1,y_1)=(0,1)" and } \left({x}_{2} , {y}_{2}\right) = \left(5 , 3\right)$

$\Rightarrow m = \frac{3 - 1}{5 - 0} = \frac{2}{5}$

$\text{using "m=2/5" and } b = 1 \to \left(0 , \textcolor{red}{1}\right)$

$\Rightarrow y = \frac{2}{5} x + 1 \leftarrow \textcolor{red}{\text{equation of line}}$

May 10, 2018

$y = \frac{2}{5} x + 1$ or

$5 x - 2 y = 5$

#### Explanation:

The general line through $\left(a , b\right)$ and $\left(c , d\right)$ is

$\left(c - a\right) \left(y - b\right) = \left(d - b\right) \left(x - a\right)$

Can you see why? When $\left(x , y\right) = \left(a , b\right)$ both sides are zero, and when $\left(x , y\right) = \left(c , d\right)$ both sides are $\left(c - a\right) \left(d - b\right) .$

Substituting,

$\left(5 - 0\right) \left(y - 1\right) = \left(3 - 1\right) \left(x - 0\right)$

$5 y - 5 = 2 x$

$5 y = 2 x + 5$

$y = \frac{2}{5} x + 1$

Check:

2/5(0) + 1 = 1 quad sqrt

2/5 (5) + 1 = 3 quad sqrt