# What is this equation in point-slope form of the line passing through (-2,1) and (4,13)

QUESTION POSTED AT 16/04/2020 - 06:31 PM

QUESTION POSTED AT 16/04/2020 - 06:31 PM

The answer is: " y = 2x + 5 ".

_________________________________________

{Refer to the graph of the equation; as an "attached image".}.

___________________________________________

Explanation:

___________________________________________

Note: The equation of a line; in "point slope form"; also known as: "slope-intercept form"; is: " y = mx + b "

which is:

_________________________________

" y = mx + b " ;

in which:

"y" is a single, "stand-alone" variable on the "left-hand side of the equation"; "m" is the coefficient of "x"; also:

"m" is the slope of the line; which is what we want to solve for;

"b" is the "y-intercept"; or more precisely, the value of "x"

(that is; the "x-coordinate") of the point at which "y = 0";

that is, the value of "x" ; or the "x-coordinate" of the point at which

the graph of the equation crosses the "x-axis".

________________________________________

Given the coordinate of any TWO (2) points on the line:

Calculate the slope, "m" = (y₂ − y₁) / (x₂ − x₁) ;

Given the two points: "(-2,1)" and "(4,13)" ;

________________________________________

(x₁, y₁) ↔ (-2, 1) ; x₁, = -2 ; y₁ = 1 ;

(x₂, y₂) ↔ (4, 13); x₂ = 4 ; y₂ = 13 ;

__________________________________

→ m = (y₂ − y₁) / (x₂ − x₁) = (13 − 1) / [(4 − (-2)] ;

= 12 / (4+2) ;

= 12 / 6 ;

= 2 ;

__________________________________

So, we know that: "m = 2" .

_________________________

In "point-slope form" :

__________________________

→ " y = mx + b " ;

So, we can write: y = 2x + b ;

We need to solve for "b".

____________________________

→ When "y = 0" , what does "x" equal? The answer to this question, is the value for "b" (the "y-intercept" of the line of graph of this equation).

____________________________

One method is to plot these points on a graph; and, using our known slope, "m = 2" ; visually inspect the graph, plot other points; noting that the slope, "2", is "rise over the run"; i.e. the "change in y" over "the change in x";

(i.e. "Δy /Δxgraph the equation) ; and determine the "x-coordinate" of the y-intercept).

The other method is to solve for "b" ; as follows:

____________________________________________

→ y = 2x + b ;

Subtract "2x" from EACH SIDE of the equation; to isolate "b" on one side of the equation; and to solve for "b" ;

______________________________________

→ y − 2x = 2x + b − 2x ;

______________________________

to get: y − 2x = b ;

________________________

↔ b = y − 2x ;

↔ b = -2x + y ;

To solve for "b" ; take EITHER of the 2 points given; and plug in the values for "x" and "y" into the equation, to solve for "b"

________________________________________

Let us try: " (-2, 1) " : x = -2, y = 1 ;

________________________________________

→ b = -2x + y = -2(-2) + 1 = 4 + 1 = 5 ; → b = 5 .

________________________________________

Alternately, and to confirm that "b = 5" ; (Check our work):

________________________________________________

Let us try the second point: " (4, 13) " :

________________________________________________

→ b = -2x + y = -2(4) + 13 = -8 + 13 = 5 ; → b = 5 . Yes!

_________________________________________________

So; we have: → y = 2x + b ; and we know: " b = 5" ; so we can write the equation:

__________________________________

" y = 2x + 5 " .

__________________________________

The answer is: " y = 2x + 5 ".

__________________________________

{Refer to the graph of the equation; as an "attached image"}:

__________________________________

_________________________________________

{Refer to the graph of the equation; as an "attached image".}.

___________________________________________

Explanation:

___________________________________________

Note: The equation of a line; in "point slope form"; also known as: "slope-intercept form"; is: " y = mx + b "

which is:

_________________________________

" y = mx + b " ;

in which:

"y" is a single, "stand-alone" variable on the "left-hand side of the equation"; "m" is the coefficient of "x"; also:

"m" is the slope of the line; which is what we want to solve for;

"b" is the "y-intercept"; or more precisely, the value of "x"

(that is; the "x-coordinate") of the point at which "y = 0";

that is, the value of "x" ; or the "x-coordinate" of the point at which

the graph of the equation crosses the "x-axis".

________________________________________

Given the coordinate of any TWO (2) points on the line:

Calculate the slope, "m" = (y₂ − y₁) / (x₂ − x₁) ;

Given the two points: "(-2,1)" and "(4,13)" ;

________________________________________

(x₁, y₁) ↔ (-2, 1) ; x₁, = -2 ; y₁ = 1 ;

(x₂, y₂) ↔ (4, 13); x₂ = 4 ; y₂ = 13 ;

__________________________________

→ m = (y₂ − y₁) / (x₂ − x₁) = (13 − 1) / [(4 − (-2)] ;

= 12 / (4+2) ;

= 12 / 6 ;

= 2 ;

__________________________________

So, we know that: "m = 2" .

_________________________

In "point-slope form" :

__________________________

→ " y = mx + b " ;

So, we can write: y = 2x + b ;

We need to solve for "b".

____________________________

→ When "y = 0" , what does "x" equal? The answer to this question, is the value for "b" (the "y-intercept" of the line of graph of this equation).

____________________________

One method is to plot these points on a graph; and, using our known slope, "m = 2" ; visually inspect the graph, plot other points; noting that the slope, "2", is "rise over the run"; i.e. the "change in y" over "the change in x";

(i.e. "Δy /Δxgraph the equation) ; and determine the "x-coordinate" of the y-intercept).

The other method is to solve for "b" ; as follows:

____________________________________________

→ y = 2x + b ;

Subtract "2x" from EACH SIDE of the equation; to isolate "b" on one side of the equation; and to solve for "b" ;

______________________________________

→ y − 2x = 2x + b − 2x ;

______________________________

to get: y − 2x = b ;

________________________

↔ b = y − 2x ;

↔ b = -2x + y ;

To solve for "b" ; take EITHER of the 2 points given; and plug in the values for "x" and "y" into the equation, to solve for "b"

________________________________________

Let us try: " (-2, 1) " : x = -2, y = 1 ;

________________________________________

→ b = -2x + y = -2(-2) + 1 = 4 + 1 = 5 ; → b = 5 .

________________________________________

Alternately, and to confirm that "b = 5" ; (Check our work):

________________________________________________

Let us try the second point: " (4, 13) " :

________________________________________________

→ b = -2x + y = -2(4) + 13 = -8 + 13 = 5 ; → b = 5 . Yes!

_________________________________________________

So; we have: → y = 2x + b ; and we know: " b = 5" ; so we can write the equation:

__________________________________

" y = 2x + 5 " .

__________________________________

The answer is: " y = 2x + 5 ".

__________________________________

{Refer to the graph of the equation; as an "attached image"}:

__________________________________

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