What is the slope-intercept form of the equation of the line that passes through the points (2, -1) and (-3, 4)?

Apr 5, 2016

$\textcolor{b l u e}{y = - x + 1}$

Explanation:

$\text{standard form } \to y = m x + c$

Where $m$ is the gradient and $c$ is the ${y}_{\text{intercept}}$

$m = \left(\text{change in y-axis")/("change in x-axis}\right)$

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(2 , - 1\right)$
Let point 2 be${P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(- 3 , 4\right)$

Then $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{4 - \left(- 1\right)}{- 3 - 2}$

$\textcolor{b l u e}{\implies m = \frac{5}{- 5} = - 1}$

This means that as you move from left to right; for one along you go down 1 (negative incline).

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So the equation becomes

$\textcolor{b r o w n}{y = - x + c}$

At P_1"; "color(brown)(y=-x+c)color(green)( " "->" "-1=-2+c)

$\implies c = 2 - 1 = 1$

So the equation becomes

$\textcolor{b l u e}{y = - x + 1}$
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