The answer is: " y = −[tex] \frac{5}{4}[/tex] x − 4 " .
_________________________________________________________
Explanation:
_________________________________________________________
Given a linear equation in "slope-intercept form" ; that is:
" y = mx + b " ;
________________________________________________
A line that is PARALLEL to the aforementioned equation has the same slope (i.e the same value for "m" ) ; and the given the [x and y coordinates of any particular point] on the parallel line; " (x₁ , y₁)" ; we can write the equation of the parallel line—in "slope-intercept format" — by using the following equation/formula:
y − y₁ = m(x − x₁) ;
in which: "m = the slope"
and plug in the values for: "m" ; and "x₁" and "y₁" ;
We are given the coordinates of a particular point on the line that is parallel:
" (-4, 1) " ;
as such: x₁ = -4 ; y₁ = 1 ;
& we are given: "m = − [tex] \frac{5}{4}[/tex]" .
_____________________________________________
So:
→ y − y₁ = m(x − x₁) ;
→ y − 1 = − [tex] \frac{5}{4}[/tex] [x − (-4) ] ;
→ y − 1 = − [tex] \frac{5}{4}[/tex] (x + 4) ;
→ y − 1 = − [tex] \frac{5}{4}[/tex] (x + 4) ;
Now; let us examine the "right-hand side of the equation" ;
We have: − [tex] \frac{5}{4}[/tex] (x + 4) ;
__________________________________________________
Note the "distributive property" of multiplication:__________________________________________a(b + c) = ab + ac ;
a(b – c) = ab – ac .__________________________________________
As such:
__________________________________________
− [tex] \frac{5}{4}[/tex] * x + (− [tex] \frac{5}{4}[/tex] * 4) ;
= − [tex] \frac{5}{4}[/tex] * x + (− [tex] \frac{5}{4}[/tex] * [tex] \frac{4}{1} [/tex]) ;
Note: Examine the " (− [tex] \frac{5}{4}[/tex] * [tex] \frac{4}{1} [/tex]) " ;
→ EACH of the 2 (TWO) "4's" cancel out to "1"s" ;
{ since: "4 ÷ 4 = 1" } ;
and we can rewrite the: "(− [tex] \frac{5}{4}[/tex] * [tex] \frac{4}{1} [/tex]) " ;
as: " (− [tex] \frac{5}{1}[/tex] * [tex] \frac{1}{1} [/tex]) " ;
Note that: "{-5 ÷ 1 = -5} ; and: "{1 ÷ 1 = 1} ;
so, rewrite the: "" (− [tex] \frac{5}{1}[/tex] * [tex] \frac{1}{1} [/tex]) " ;
as: "{-5 * 1}" → which equals: = " -5" ;
So:
− [tex] \frac{5}{4}[/tex] * x + (− [tex] \frac{5}{4}[/tex] * [tex] \frac{4}{1} [/tex]) ;
= - [tex] \frac{5}{4}[/tex] x + (-5) ;
= - [tex] \frac{5}{4}[/tex] x − 5 ;
______________________________________________
→ Now, bring down the "y −1" ; which goes on the left hand side;
→ y − 1 = - [tex] \frac{5}{4}[/tex] x − 5 ;
Add "1" to EACH SIDE of the equation; to isolate "y" as a single variable on the "left-hand side" of the equation ; & to write the equation of the particular parallel line in "slope-intercept format" ;
→ y − 1 + 1 = - [tex] \frac{5}{4}[/tex] x − 5 + 1 ;
_______________________________________________________
to get:
_______________________________________________________
→ " y = −[tex] \frac{5}{4}[/tex] x − 4 " .
_______________________________________________________