# What is the slope-intercept form of the line passing through  (5, 1) and  (3, -2) ?

Jun 30, 2016

$y = \frac{3}{2} x - \frac{13}{2}$

#### Explanation:

Slope intercept form is:$\text{ } y = m x + c$
where $m$ is the gradient and $c$ is the y-intercept.

Gradient$\to \left(\text{change in y")/("change in x}\right)$

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(5 , 1\right)$
Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(3 , - 2\right)$

Thus Gradient $\to \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 2 - 1}{3 - 5} = \frac{- 3}{- 2} = + \frac{3}{2}$
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So now we have $y = \frac{3}{2} x + c$

To find the value of $c$ we substitute in the value of a known point so that there is only 1 unknown.

$\textcolor{b r o w n}{\implies {P}_{1} \to {y}_{1} = \frac{3}{2} {x}_{1} + c} \textcolor{b l u e}{\to 1 = \frac{3}{2} \left(5\right) + c}$

$\text{ } 1 = \frac{15}{2} + c$

Subtract $\textcolor{m a \ge n t a}{\frac{15}{2}}$ from both sides

" "color(blue)(1color(magenta)(-15/2)=15/2 color(magenta)(-15/2)+c

$c = - \frac{13}{2}$
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$\text{ } \overline{\underline{| \textcolor{w h i t e}{.} y = \frac{3}{2} x - \frac{13}{2} \textcolor{w h i t e}{.} |}}$