# What is the equation of the line with slope  m= 1 that passes through  (7,13) ?

Jan 21, 2016

$\left(y - 13\right) = 1 \left(x - 7\right)$ in explicit slope-point form
or
$x - y = - 6$ in standard form

#### Explanation:

General slope-point form is
$\textcolor{w h i t e}{\text{XXX}} \left(y - \overline{y}\right) = m \left(x - \overline{x}\right)$
for a line with slope $m$ through the point $\left(\overline{x} , \overline{y}\right)$

Substituting the given values results in
$\textcolor{w h i t e}{\text{XXX}} y - 13 = 1 \left(x - 7\right)$

To convert to standard form: $A x + B y = C$
$\textcolor{w h i t e}{\text{XXX}} - x + y - 13 = - 7$

$\textcolor{w h i t e}{\text{XXX}} - x + y = 6$

$\textcolor{w h i t e}{\text{XXX}} x - y = - 6$

or, in explicit standard form:
$\textcolor{w h i t e}{\text{XXX}} 1 x - 1 y = - 6$

Jan 22, 2016

A different approach to solving this one!

$y = x + 6$

Explained with a lot of detail!

#### Explanation:

$\textcolor{b l u e}{\text{Assumption "-> " this is a strait line graph}}$

Standard equation form for a strait line graph:

$y = m x + c$.....................(1)

Given: slop (gradient) $\to m = 1$

so equation (1) becomes:

$y = \left(1\right) x + c$

But $1 \times x = x$

$y = x + c$........................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{To find the value of } c}$
Using given point of $x = 7 \mathmr{and} y = 13$

Equation (2) becomes:

$y = x + c \textcolor{w h i t e}{. . x .} \to \textcolor{w h i t e}{. x . .} 13 = 7 + c$

$\textcolor{b l u e}{c = 13 - 7 = 6}$ as predicted.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$y = x + 6$