# What is the equation of the line passing through (5,12) and (14,2)?

Nov 18, 2015

$y = - \frac{1}{9} \left(10 x - 158\right)$

#### Explanation:

Assumption: Strait line passing through given points!
The left most point $\to \left(5 , 12\right)$

Standard form equation: $y = m x + c \text{ ............(1)}$
Where m is the gradient.

Let
$\left({x}_{1} , {y}_{1}\right) \to \left(5 , 12\right)$
$\left({x}_{2} , {y}_{2}\right) \to \left(14 , 2\right)$

Then $\textcolor{g r e e n}{m = \left(\text{Change in y-axis")/("Change in x-axis}\right) = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{2 - 12}{14 - 5} = \frac{- 10}{9}}$

As the gradient (m) is negative then the line 'slopes' downward from left to right.

Substitute value of $\left({x}_{1} , {y}_{1}\right)$ for the variables in equation (1) giving:

$12 = \left(- \frac{10}{9} \times 5\right) + c$

$c = 12 + \left(\frac{10}{9} \times 5\right)$

$\textcolor{g r e e n}{c = 12 + \frac{50}{9} \equiv \frac{158}{9}}$

So $y = m x + c \to \textcolor{b l u e}{y = \left(- \frac{10}{9}\right) x + \frac{158}{9}}$

$\textcolor{b l u e}{y = - \frac{1}{9} \left(10 x - 158\right)}$