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What is the equation of the line that has a slope of 3 and goes through the point [tex]$(-3,-5)$[/tex]?

A. [tex]$y=3x+4$[/tex]
B. [tex]$y=3x-14$[/tex]
C. [tex]$y=3x-4$[/tex]
D. [tex]$y=3x+12$[/tex]

Sagot :

GinnyAnswer
To find the equation of a line that has a given slope and goes through a specified point, we can use the point-slope form of the equation of a line. The point-slope form is:

[tex]\[ y - y_1 = m(x - x_1) \][/tex]

where [tex]\(m\)[/tex] is the slope, and [tex]\((x_1, y_1)\)[/tex] is the point that the line passes through.

Given the slope [tex]\(m = 3\)[/tex] and the point [tex]\((-3, -5)\)[/tex], we can substitute these values into the point-slope form:

[tex]\[ y - (-5) = 3(x - (-3)) \][/tex]

Simplify the equation:

[tex]\[ y + 5 = 3(x + 3) \][/tex]

Distribute the 3 on the right-hand side:

[tex]\[ y + 5 = 3x + 9 \][/tex]

To get the equation in slope-intercept form ( [tex]\(y = mx + b\)[/tex] ), we need to isolate [tex]\(y\)[/tex] on one side of the equation. Subtract 5 from both sides:

[tex]\[ y = 3x + 9 - 5 \][/tex]

Simplify:

[tex]\[ y = 3x + 4 \][/tex]

Therefore, the equation of the line is:

[tex]\[ y = 3x + 4 \][/tex]

So, the correct answer is:

A. [tex]\(y = 3x + 4\)[/tex]
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