y – 4 = (x + 2)
Further explanation
We will determine the line equation. Later the equation will be arranged in slope-intercept, point-slope, and standard form.
Given:
- A line that passes through (–2, 4).
- A slope of 1.
The Process:
- Slope or gradient m = 2.
- Point (x₁, y₁) is (-2, 4).
Part-1
[tex]\boxed{ \ Slope-intercept \ form: y = mx + c \ }[/tex]
Substitution.
[tex]\boxed{ \ \ m = 1 \rightarrow y = x + c \ }[/tex]
The line passing through the point (-2, 4).
[tex]\boxed{ \ x_1 = -2, y_1 = 4 \ } \rightarrow \boxed{ \ 4 = (-2) + c \ } \rightarrow \boxed{ \ c = 6 \ }[/tex]
Thus the slope-intercept form is [tex]\boxed{\boxed{ \ y = x + 6 \ }}[/tex]
Part-2
[tex]\boxed{ \ Point-slope \ form: y - y_1 = m(x - x_1) \ }[/tex]
[tex] \boxed{ \ m = 1 \rightarrow y - y_1 = (x - x_1) \ } [/tex]
The line passing through the point (-2, 4).
[tex] \boxed{ \ x_1 = -2, y_1 = 4 \ } \rightarrow \boxed{ \ y - 4 = (x - (-2)) \ } [/tex]
Thus the point-slope form is [tex]\boxed{\boxed{ \ y - 4 = (x + 2) \ }}[/tex]
Part-3
[tex]\boxed{ \ Standard \ form: ax + by = k \ }[/tex]
From the results of Part-1, we set the position of the variables.
[tex]\boxed{ \ y = x + 6 \ }[/tex]
[tex]\boxed{ \ x + 6 = y \ }[/tex]
[tex]\boxed{ \ x - y = - 6 \ }[/tex]
Thus the standard form is [tex]\boxed{\boxed{ \ x - y = - 6 \ }} \ or \ \boxed{\boxed{ \ y - x = 6 \ }} \ or \ \boxed{\boxed{ \ x - y + 6 = 0 \ }}[/tex]
Learn more
- Find out ordered pairs could be points on a line that is perpendicular to another line https://brainly.com/question/2601054
- Determine the line equation, in slope-intercept form, that is parallel to the given line and passes through a point https://brainly.com/question/1473992
- Determine an equation represents Nolan’s line https://brainly.com/question/2657284
Keywords: the line equation, slope-intercept form, standard form, point-slope, gradien, a line that passes through (–2, 4)
