# What is the equation of the line passing through (11,17) and (23,11)?

Nov 18, 2015

$x + 2 y = 45$

#### Explanation:

1st point$= \left({x}_{1} , {y}_{1}\right) = \left(11 , 17\right)$
2nd point$= \left({x}_{2} , {y}_{2}\right) = \left(23 , 11\right)$

First, we will have to find the slope $m$ of this line:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{11 - 17}{23 - 11} = - \frac{6}{12} = - \frac{1}{2}$

Now, use point-slope formula with one of the given points:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 17 = - \frac{1}{2} \left(x - 11\right)$
$y - 17 = - \frac{1}{2} x + \frac{11}{2}$
$y = - \frac{1}{2} x + \frac{11}{2} + 17$
$y = \frac{- x + 11 + 34}{2}$
$2 y = - x + 45$
$x + 2 y = 45$

Nov 18, 2015

$y = - \frac{x}{2} + \frac{45}{2}$

#### Explanation:

Usung the formula $y - {y}_{1} = m \left(x - {x}_{1}\right)$
Considering
$\left(11 , 17\right) \mathmr{and} \left(23 , 11\right)$
$\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$

m (gradient) = $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

m = $\frac{11 - 17}{23 - 11}$

m = $- \frac{6}{12}$

m = $- \frac{1}{2}$

$y - 17 = - \frac{1}{2} \left(x - 11\right)$
$y - 17 = - \frac{x}{2} + \frac{11}{2}$
$y = - \frac{x}{2} + \frac{11}{2} + 17$
$y = - \frac{x}{2} + \frac{45}{2}$

This is the equation of the line