# How do you write the equation of the line through (4,-8) and (8,5)?

Feb 20, 2017

$13 x - 4 y = 84$

#### Explanation:

Step 1: Determine the slope of the line through the given points

"slope"=("change in "y)/("change in "x)

Given the points, $\left({x}_{1} , {y}_{1}\right) = \left(4 , - 8\right)$ and $\left({x}_{2} , {y}_{2}\right) = \left(8 , 5\right)$,
the slope, $m$ is given by:
$\textcolor{w h i t e}{\text{XXX}} m = \frac{5 - \left(- 8\right)}{8 - 4} = \frac{13}{4}$

Step 2: Write the equation in slope-point form
Given a slope $m$ and a point $\left({x}_{1} , {y}_{1}\right)$,
the slope-point form of the equation is:
$\textcolor{w h i t e}{\text{XXX}} y - {y}_{1} = m \left(x - {x}_{1}\right)$
In this case we have:
color(white)("XXX")m=13/4" and "(x_1,y_1)=(4,-8)
giving
$\textcolor{w h i t e}{\text{XXX}} y - \left(- 8\right) = \frac{13}{4} \left(x - 4\right)$

Step 3: Convert into standard form
Standard form of a linear equation is
$\textcolor{w h i t e}{\text{XXX}} A x + B y = C$
Starting from the slope-point form (above) we have:
$\textcolor{w h i t e}{\text{XXX}} 4 \left(y + 8\right) = 13 \left(x - 4\right)$

$\textcolor{w h i t e}{\text{XXX}} 4 y + 32 = 13 x - 52$

$\textcolor{w h i t e}{\text{XXX}} 13 x - 4 y = 84$

For verification purposes, here is a graph with the given points and the equation $13 x - 4 y = 84$

Feb 20, 2017

$y = \frac{13}{4} x - 21$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

consider the standardised equation format of: $y = m x + c$
where $m$ is the gradient.

$m = \left(\text{changing in up or down")/("change in along}\right)$

Very important$\to$the change in along (usually the x-axis) is read left to right.

Watch out for this because sometimes questions give the points in the revers order.

So the first point is at say ${x}_{1}$ and the second point is at say ${x}_{2}$. Then ${x}_{1} < {x}_{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

$\textcolor{b r o w n}{\text{Determine the gradient}}$

Let point 1 be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(4 , - 8\right)$
Let point 2 be ${P}_{2} \to \left({x}_{2} , {y}_{2}\right) = \left(8 , 5\right)$

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{5 - \left(- 8\right)}{8 - 4} = \frac{5 + 8}{8 - 4} = \frac{13}{4}$

So we now have:$\text{ } y = \frac{13}{4} x + c$

$\textcolor{b r o w n}{\text{Determine the value of the constant } c}$

I chose ${P}_{1} \to$using the value for this point substitute for $x \mathmr{and} y$

$y = \frac{13}{4} x + c \text{ "->" } - 8 = \frac{13}{4} \left(4\right) + c$

$c = - 8 - 13 = - 21$

So we now have:$\text{ } y = \frac{13}{4} x - 21$