# How do you write an equation of a line going through (-5,-5), and (-2,3)?

Jan 22, 2016

$\textcolor{p u r p \le}{y = \frac{8}{3} x + \frac{25}{3}}$
Explanation given in detail to demonstrate methods. Shortcuts would reduce the method considerably!

#### Explanation:

Let the point be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(- 5 , - 5\right)$

Let the point be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(- 2 , 3\right)$

Let the gradient (slope) be $m \to \left(\text{change in y-axis")/("change in x-axis}\right)$

Let the constant be $c$

The standardised equation for a strait line graph is:

$\textcolor{w h i t e}{\text{xxxxxxxxxxxxxxxxxx}} y = m x + c$ ..................(1)

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$\textcolor{b l u e}{\text{To find the slope (gradient)}}$

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{3 - \left(- 5\right)}{- 2 - \left(- 5\right)} = \frac{8}{3}$

Substitute for m in equation (1) giving:

$\textcolor{b l u e}{y = \frac{8}{3} x + c}$.......................(2)

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$\textcolor{b l u e}{\text{To find the value of } c}$

At this stage if we substitute known value for $x$ and $y$ we end up with only one unknown. Which is c. Thus solvable.

Substitute ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(- 2 , 3\right)$ into equation (2) giving:

$3 = \frac{8}{3} \left(- 2\right) + c$

$\textcolor{b r o w n}{3 = - \frac{16}{3} + c}$

Add $\textcolor{g r e e n}{\frac{16}{3}}$ to both sides

$\textcolor{b r o w n}{3 \textcolor{g r e e n}{+ \frac{16}{3}} = - \frac{16}{3} \textcolor{g r e e n}{+ \frac{16}{3}} + c}$

$\frac{25}{3} = 0 + c$

$\textcolor{b l u e}{c = \frac{25}{3}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for $c$ in equation (2) giving:

$\textcolor{p u r p \le}{y = \frac{8}{3} x + \frac{25}{3}}$