# How do you write an equation in standard form given a line that passes through (9,8) and (4,7)?

Mar 24, 2018

A lot of explanation given.

Standardised form type 1: $\textcolor{w h i t e}{\text{d}} x - 5 y = - 31$

Standardised form type 2: $\textcolor{w h i t e}{\text{d}} y = \frac{1}{5} x + \frac{31}{5}$

#### Explanation:

The gradient (slope) is the amount of up or down for a given amount of along reading left to right on the x-axis.

Given the two points $\left(x , y\right) \to \left(9 , 8\right) \mathmr{and} \left(4 , 7\right)$

The $x$ value of 4 comes before 9

Set point 1 as ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(4 , 7\right)$
Set point 2 as ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(9 , 8\right)$

So we are moving from ${P}_{1} \text{ to } {P}_{2}$

Thus any change is ${P}_{2} - {P}_{1}$
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$\textcolor{b l u e}{\text{Determine the gradient } \to m}$

Change in up or down is change in $y \to {y}_{2} - {y}_{1} = 8 - 7 = + 1$

As this is positive then the slope is up.

Change in along is change in $x \to {x}_{2} - {x}_{1} = 9 - 4 = 5$

Set gradient as $m = \left(\text{change in up or down")/("change in along}\right) = + \frac{1}{5}$
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$\textcolor{b l u e}{\text{Determine the full equation}}$

Gradient $= m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \left(\text{any "y-y_1)/("any } x - {x}_{1}\right) = \frac{y - 7}{x - 4}$

$m = \frac{1}{5} = \frac{y - 7}{x - 4}$

Multiply both sides by $\left(x - 4\right)$ giving:

$\frac{1}{5} \times \left(x - 4\right) = \left(y - 7\right)$

Multiply both sides by 5 giving

$x - 4 = 5 y - 35$

By further manipulation we have:

Standardised form type 1: $\textcolor{w h i t e}{\text{d}} x - 5 y = - 31$

Standardised form type 2: $\textcolor{w h i t e}{\text{d}} y = \frac{1}{5} x + \frac{31}{5}$