# What is the equation of the line between (10,3) and (-4,12)?

Nov 27, 2015

$y = - \frac{9}{14} x + 9 \frac{3}{7}$

#### Explanation:

Working on the assumption that you are talking about a straight line graph

Standard equation form is: $y = m x + c$

$x$ is the independent variable

$y$ is the dependant variable (its value 'depends' on what you
assign to $x$

$m$ is the gradient of the line (slope)
$\textcolor{w h i t e}{X X X}$going from left to right, a positive
$\textcolor{w h i t e}{X X X}$slop is upwards and a negative slope is downwards.

$c$ is a constant value and is where the line intersects the y-axis.

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$\textcolor{b l u e}{\text{To find the gradient:}}$
The amount of up (or down) for the amount of along. That is:

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

Let $\left({x}_{1} , {y}_{1}\right) \to \left(10 , 3\right)$
Let x_2,y_2)->(-4,12)

so $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \to \frac{12 - 3}{\left(- 4\right) - 10} = \frac{9}{- 14}$

$\textcolor{b l u e}{m = - \frac{9}{14}}$ which is negative so the line descends from left to right
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$\textcolor{b l u e}{\text{To find the constant}}$

We can substitute known value for $x$ and $y$ to find $c$
I am choosing $\textcolor{b r o w n}{\left({x}_{1} , {y}_{1}\right) \to \left(10 , 3\right)}$

So ${y}_{1} = m {x}_{1} + c$ becomes:

$\textcolor{b r o w n}{3 = \textcolor{b l u e}{- \frac{9}{14}} \left(10\right) + \textcolor{b l a c k}{c}}$

$c = 3 + \frac{9 \times 10}{14}$

$\textcolor{b l u e}{c = 3 + 6 \frac{3}{7} = 9 \frac{3}{7}}$
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Putting it all together:

$y = - \frac{9}{14} x + 9 \frac{3}{7}$