Answer:
[tex]\textsf{Point-slope form}: \quad \sf y-2=4(x+1)[/tex]
[tex]\textsf{Slope-intercept form}: \quad \sf y=4x+6[/tex]
[tex]\textsf{Standard form}: \quad \sf 4x-y=-6[/tex]
Step-by-step explanation:
Given information:
- Slope = 4
- Point on line = (-1, 2)
Point-slope form of linear equation:
[tex]\sf y-y_1=m(x-x_1)[/tex]
(where m is the slope and (x₁, y₁) is a point on the line)
Substitute the given slope and point into the formula:
[tex]\implies \sf y-2=4(x-(-1))[/tex]
[tex]\implies \sf y-2=4(x+1)[/tex]
Slope-intercept form of a linear equation:
[tex]\sf y=mx+b[/tex]
(where m is the slope and b is the y-intercept)
Substitute the given slope and point into the formula and solve for b:
[tex]\implies \sf 2=4(-1)+b[/tex]
[tex]\implies \sf b=6[/tex]
Therefore:
[tex]\sf y=4x+6[/tex]
Standard form of a linear equation:
[tex]\sf Ax+By=C[/tex]
(where A, B and C are constants and A must be positive)
Rearrange the found slope-intercept form of the equation into standard form:
[tex]\implies \sf y=4x+6[/tex]
[tex]\implies \sf 4x-y+6=0[/tex]
[tex]\implies \sf 4x-y=-6[/tex]
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